Solution:

Solve for [latex]x[/latex]: [latex]3x - 6 = 7x + 4[/latex]

The variable we are asked to solve for is [latex]x[/latex] so our first move is to gather all of the terms involving [latex]x[/latex] on one side and put the remaining terms on the other.^[3]

\[ \begin{array}{rclr}
3x – 6 & = & 7x + 4 & \\
(3x-6) – 7x + 6 & = & (7x+4) -7x +6 & \text{Subtract } 7x, \text{ add } 6 \\
3x – 7x – 6 + 6 & = & 7x – 7x + 4 + 6 & \text{Rearrange terms} \\
-4x & = & 10 & 3x-7x = (3-7)x = -4x \\ & & & \\
\dfrac{-4x}{-4} & = & \dfrac{10}{-4} & \text{Divide by the coefficient of } x \\ & & & \\
x & = & -\dfrac{5}{2} & \text{Reduce to lowest terms} \\
\end{array}\]

To check our answer, we substitute [latex]x = -\dfrac{5}{2}[/latex] into each side of the original equation to see the equation is satisfied. Sure enough, [latex]3\left(-\dfrac{5}{2}\right) - 6 = -\dfrac{27}{2}[/latex] and [latex]7\left(-\dfrac{5}{2}\right) + 4 = -\dfrac{27}{2}[/latex].

Solution:

Solve for [latex]t[/latex]: [latex]3 - 1.7t = \dfrac{t}{4}[/latex]

In our next example, the unknown is [latex]t[/latex] and we not only have a fraction but also a decimal to wrangle. Fortunately, with equations we can multiply both sides to rid us of these computational obstacles:

\[ \begin{array}{rclr} 3 – 1.7t & = & \dfrac{t}{4} & \\ & & & \\
40(3 – 1.7t) & = & 40 \left(\dfrac{t}{4}\right) & \text{Multiply by } 40 \\ & & & \\
40(3) – 40(1.7t) & = & \dfrac{40 t}{4} & \text{Distribute} \\
120 – 68t & = & 10 t & \\
(120 -68t) + 68 t & = & 10t + 68 t & \text{Add } 68t \text{ to both sides} \\
120 & = & 78 t & 68t + 10t = (68 + 10)t = 78 t\\
\dfrac{120}{78} & = & \dfrac{78t}{78} & \text{Divide by the coefficient of } t\\ & & & \\
\dfrac{120}{78} & = & t & \\ & & & \\
\dfrac{20}{13} & = & t & \text{Reduce to lowest terms} \\
\end{array}\]

To check, we again substitute [latex]t = \dfrac{20}{13}[/latex] into each side of the original equation.

We find that [latex]3 - 1.7 \left(\dfrac{20}{13}\right) = 3 - \left(\dfrac{17}{10}\right)\left(\dfrac{20}{13}\right) = \dfrac{5}{13}[/latex] and [latex]\dfrac{(20/13)}{4} = \dfrac{20}{13} \cdot \dfrac{1}{4} = \dfrac{5}{13}[/latex] as well.

Solution:

Solve for [latex]a[/latex]: [latex]\dfrac{1}{18}(7 - 4a) + 2 = \dfrac{a}{3} - \dfrac{4-a}{12}[/latex]

To solve this next equation, we begin once again by clearing fractions. The least common denominator here is 36:

\[ \begin{array}{rclr}
\dfrac{1}{18}(7 – 4a) + 2 & = & \dfrac{a}{3} – \dfrac{4-a}{12} & \\[8pt]
36 \left(\dfrac{1}{18}(7 – 4a) + 2\right) & = & 36 \left(\dfrac{a}{3} – \dfrac{4-a}{12}\right) & \text{Multiply by } 36\\[13pt]
\dfrac{36}{18} (7-4a) + (36)(2) & = & \dfrac{36a}{3} – \dfrac{36(4-a)}{12} & \text{Distribute} \\[5pt]
2(7-4a) + 72 & = & 12 a – 3(4-a) & \text{Distribute} \\
14 – 8a + 72 & = & 12a – 12 + 3a & \\
86 – 8a & = & 15 a – 12 & 12 a + 3a = (12+3)a = 15 \\
(86-8a)+8a+12 & = & (15a-12) + 8a + 12 & \text{Add } 8a \text{ and }12 \\
86 + 12 – 8a + 8a & = & 15a + 8a – 12 + 12 & \text{Rearrange terms} \\
98 & = & 23 a & 15a + 8a = (15+8)a = 23a \\ & & & \\
\dfrac{98}{23} & = & \dfrac{23a}{23} & \text{Divide by the coefficient of } a \\ & & & \\
\dfrac{98}{23} & = & a &
\end{array} \]

The check, as usual, involves substituting [latex]a = \dfrac{98}{23}[/latex] into both sides of the original equation. The reader is encouraged to work through the (admittedly messy) arithmetic. Both sides work out to [latex]\dfrac{199}{138}[/latex].

Solution:

Solve for [latex]y[/latex]: [latex]8 y \sqrt{3} + 1 = 7 - \sqrt{12}(5 - y)[/latex]

The square roots may dishearten you but we treat them just like the real numbers they are. Our strategy is the same: get everything with the variable (in this case [latex]y[/latex]) on one side, put everything else on the other and divide by the coefficient of the variable. We’ve added a few steps to the narrative that we would ordinarily omit just to help you see that this equation is indeed linear.

\[ \begin{array}{rclr}
8 y \sqrt{3} + 1 & = & 7 – \sqrt{12}(5 – y) & \\ & & & \\
8 y \sqrt{3} + 1 & = & 7 – \sqrt{12}(5) + \sqrt{12} y & \text{Distribute} \\ & & & \\
8 y \sqrt{3} + 1 & = & 7 – (2 \sqrt{3})5 + (2 \sqrt{3})y & \sqrt{12} = \sqrt{4\cdot 3} = 2 \sqrt{3} \\ & & & \\
8 y \sqrt{3} + 1 & = & 7 – 10 \sqrt{3} + 2y \sqrt{3} & \\ & & & \\
(8 y \sqrt{3} + 1) – 1 – 2y\sqrt{3} & = & (7 – 10 \sqrt{3} + 2y \sqrt{3}) – 1 – 2y\sqrt{3} & \text{Subtract 1 and } 2y\sqrt{3} \\ & & & \\
8 y \sqrt{3} – 2y\sqrt{3} + 1 – 1 & = & 7 – 1 – 10 \sqrt{3} + 2y \sqrt{3} – 2y\sqrt{3} & \text{Rearrange terms} \\ & & & \\
(8\sqrt{3}-2\sqrt{3})y & = & 6 – 10 \sqrt{3} & \\ & & & \\
6 y \sqrt{3} & = & 6 – 10 \sqrt{3} & \text{See note below} \\ & & & \\
\dfrac{6 y \sqrt{3}}{6 \sqrt{3}} & = & \dfrac{6 – 10 \sqrt{3}}{6 \sqrt{3}} & \text{Divide } 6\sqrt{3} \\ & & & \\
y & = & \dfrac{2 \cdot \sqrt{3} \cdot \sqrt{3} – 2 \cdot 5 \cdot \sqrt{3}}{2 \cdot 3 \cdot \sqrt{3}} & \\ & & & \\
y & = & \dfrac{\cancel{2} \cancel{\sqrt{3}}(\sqrt{3} – 5)}{\cancel{2} \cdot 3 \cdot \cancel{\sqrt{3}}} & \text{Factor and cancel} \\ & & & \\
y & = & \dfrac{\sqrt{3} – 5}{3} & \\
\end{array}\]

In the list of computations above we marked the row [latex]6 y \sqrt{3} = 6 - 10 \sqrt{3}[/latex] with a note. That’s because we wanted to draw your attention to this line without breaking the flow of the manipulations. The equation [latex]6 y \sqrt{3} = 6 - 10 \sqrt{3}[/latex] is in fact linear according to Definition 0.10: the variable is [latex]y[/latex], the value of [latex]A[/latex] is [latex]6\sqrt{3}[/latex] and [latex]B = 6 - 10 \sqrt{3}[/latex]. Checking the solution, while not trivial, is good mental exercise. Each side works out to be [latex]\dfrac{27 - 40 \sqrt{3}}{3}[/latex].

Solution:

Solve for [latex]x[/latex]: [latex]\dfrac{3x-1}{2} = x\sqrt{50} + 4[/latex]

Proceeding as before, we simplify radicals and clear denominators. Once we gather all of the terms containing [latex]x[/latex] on one side and move the other terms to the other, we factor out [latex]x[/latex] to identify its coefficient then divide to get our answer.

\[ \begin{array}{rclr}
\dfrac{3x-1}{2} & = & x\sqrt{50} + 4 & \\ & & & \\
\dfrac{3x-1}{2} & = & 5x\sqrt{2} + 4 & \sqrt{50} = \sqrt{25 \cdot 2} \\ & & & \\
2\left(\dfrac{3x-1}{2}\right) & = & 2 \left(5x\sqrt{2} + 4 \right) & \text{Multiply by 2} \\ & & & \\
\dfrac{\cancel{2} \cdot (3x-1)}{\cancel{2}} & = & 2 (5x \sqrt{2}) + 2\cdot 4 & \text{Distribute} \\ & & & \\
3x – 1 & = & 10x \sqrt{2} + 8 & \\
(3x-1) – 10x\sqrt{2} + 1 & = & (10x \sqrt{2} + 8) – 10x\sqrt{2} + 1 & \text{Subtract } 10x\sqrt{2}, \text{ add 1} \\
3x – 10x\sqrt{2} – 1 + 1 & = & 10x\sqrt{2} – 10x\sqrt{2} + 8 + 1 & \text{Rearrange terms} \\
3x – 10x\sqrt{2} & = & 9 & \\
(3 – 10\sqrt{2}) x & = & 9 & \text{Factor} \\ & & & \\
\dfrac{(3 – 10\sqrt{2}) x}{3 – 10\sqrt{2}} & = & \dfrac{9}{3 – 10\sqrt{2}} & \text{Divide by the coefficient of } x \\ & & & \\
x & = & \dfrac{9}{3 – 10\sqrt{2}} & \\
\end{array} \]

The reader is encouraged to check this solution – it isn’t as bad as it looks if you’re careful! Each side works out to be [latex]\dfrac{12 + 5\sqrt{2}}{3-10\sqrt{2}}[/latex].

Solution:

Solve for [latex]y[/latex]: [latex]x(4-y) = 8y[/latex]

If we were instructed to solve our last equation for [latex]x[/latex], we’d be done in one step: divide both sides by [latex](4-y)[/latex] – assuming [latex]4-y \neq 0[/latex], that is. Alas, we are instructed to solve for [latex]y[/latex], which means we have some more work to do.

\[ \begin{array}{rclr}
x(4-y) & = & 8y & \\
4x – xy & = & 8y & \text{Distribute} \\
(4x – xy) + xy & = & 8y + xy & \text{Add } xy \\
4x & = & (8+x)y & \text{Factor} \\
\end{array}\]

In order to finish the problem, we need to divide both sides of the equation by the coefficient of [latex]y[/latex] which in this case is [latex]8+x[/latex]. This expression contains a variable so we need to stipulate that we may perform this division only if [latex]8 + x \neq 0[/latex], or, in other words, [latex]x \neq -8[/latex]. Hence, we write our solution as:\[ y = \dfrac{4x}{8+x}, \quad \text{provided } x \neq -8\] What happens if [latex]x = -8[/latex]? Substituting [latex]x = -8[/latex] into the original equation gives [latex](-8)(4-y) = 8y[/latex] or [latex]-32 + 8y = 8y[/latex]. This reduces to [latex]-32 = 0[/latex], which is a contradiction. This means there is no solution when [latex]x = -8[/latex], so we’ve covered all the bases. Checking our answer requires some Algebra we haven’t reviewed yet in this text, but the necessary skills should be lurking somewhere in the mathematical mists of your mind. The adventurous reader is invited to plug [latex]y = \dfrac{4x}{8 + x}[/latex] into the original equation and show that both sides work out to [latex]\dfrac{32x}{x + 8}[/latex].

Solution:

Solve the equation: [latex]|3x-1| = 6[/latex].

The equation [latex]|3x-1| = 6[/latex] is of already in the form [latex]|X| = c[/latex], so we know that either [latex]3x-1=6[/latex] or [latex]3x-1 = -6[/latex]

Solving the equation [latex]3x-1=6[/latex] gives us [latex]x = \dfrac{7}{3}[/latex].

While solving the equation [latex]3x-1 = -6[/latex] yields [latex]x = -\dfrac{5}{3}[/latex].

We may check both of these solutions by substituting them into the original equation and showing that the arithmetic works out.

Solution:

Solve the equation: [latex]\dfrac{3 - |y+5|}{2} = 1[/latex].

We begin solving [latex]\dfrac{3 - |y+5|}{2} = 1[/latex] by isolating the absolute value to put it in the form [latex]|X| = c[/latex].

\[ \begin{array}{rclr}
\dfrac{3 – |y+5|}{2} & = & 1 & \\
3 – |y+5| & = & 2 & \text{Multiply by } 2\\
-|y+5| & = & -1 & \text{Subtract } 3 \\
|y+5| & = & 1 & \text{Divide by } -1 \\
\end{array} \]

At this point, we have [latex]y+5 = 1[/latex] or [latex]y+5 = -1[/latex], so our solutions are [latex]y = -4[/latex] or [latex]y = -6[/latex].

We leave it to the reader to check both answers in the original equation.

Solution:

Solve the equation: [latex]3|2t+1| - \sqrt{5} = 0[/latex].

As in the previous example, we first isolate the absolute value. Don’t let the [latex]\sqrt{5}[/latex] throw you off – it’s just another real number, so we treat it as such:

\[ \begin{array}{rclr}
3|2t+1| – \sqrt{5} & = & 0 & \\
3|2t+1| & = & \sqrt{5} & \text{Add }\sqrt{5} \\
|2t + 1| & = & \dfrac{\sqrt{5}}{3} & \text{Divide by } 3\\
\end{array} \]

From here, we have that [latex]2t+1 = \dfrac{\sqrt{5}}{3}[/latex] or [latex]2t+1 = -\dfrac{\sqrt{5}}{3}[/latex].

The first equation gives [latex]t = \dfrac{\sqrt{5}-3}{6}[/latex] while the second gives [latex]t = \dfrac{-\sqrt{5}-3}{6}[/latex] thus we list our answers as [latex]t = \dfrac{-3 \pm \sqrt{5}}{6}[/latex].

The reader should enjoy the challenge of substituting both answers into the original equation and following through the arithmetic to see that both answers work.

Solution:

Solve the equation: [latex]4 - |5w+3| = 5[/latex].

Upon isolating the absolute value in the equation [latex]4 - |5w+3| = 5[/latex], we get [latex]|5w+3| = -1[/latex].

At this point, we know there cannot be any real solution. By definition, the absolute value is a distance, and as such is never negative. We write “no solution” and carry on.

Solution:

Solve the equation: [latex]\left|3 - x \sqrt[3]{12}\right| = |4x+1|[/latex].

Our next equation already has the absolute value expressions (plural) isolated, so we work from the principle that if [latex]|x| = |y|[/latex], then [latex]x = y[/latex] or [latex]x = -y[/latex].

Thus from [latex]\left|3 - x \sqrt[3]{12}\right| = |4x+1|[/latex] we get two equations to solve: \[ 3 – x \sqrt[3]{12} = 4x+1, \qquad \text{and} \qquad 3 – x \sqrt[3]{12} = -(4x+1) \]

Notice that the right side of the second equation is [latex]-(4x+1)[/latex] and not simply [latex]-4x+1[/latex]. Remember, the expression [latex]4x+1[/latex] represents a single real number so in order to negate it we need to negate the entire expression [latex]-(4x+1)[/latex].

Moving along, when solving [latex]3 - x \sqrt[3]{12} = 4x+1[/latex], we obtain [latex]x = \dfrac{2}{4 + \sqrt[3]{12}}[/latex].

The solution to [latex]3 - x \sqrt[3]{12} = -(4x+1)[/latex] is [latex]x = \dfrac{4}{\sqrt[3]{12}-4}[/latex].

As usual, the reader is invited to check these answers by substituting them into the original equation.

Solution:

Solve the equation: [latex]|t-1| - 3|t+1| = 0[/latex].

We start by isolating one of the absolute value expressions: [latex]|t-1| - 3|t+1| = 0[/latex] gives [latex]|t-1| = 3|t+1|[/latex].

While this resembles the form [latex]|x| = |y|[/latex], the coefficient 3 in [latex]3|t+1|[/latex] prevents it from being an exact match. Not to worry – because 3 is positive, [latex]3 = |3|[/latex] so \[3|t+1| = |3| |t+1| = |3(t+1)| = |3t+3|.\]

Hence, our equation becomes [latex]|t-1| = |3t+3|[/latex] which results in the two equations: [latex]t-1 = 3t+3[/latex] and [latex]t-1 = -(3t+3)[/latex].

The first equation gives [latex]t = -2[/latex] and the second gives [latex]t = -\dfrac{1}{2}[/latex].

The reader is encouraged to check both answers in the original equation.

Solution:

Solve the equation: [latex]x^2 = 35 - 16x[/latex].

We begin by gathering all of the nonzero terms to one side getting 0 on the other. Then we proceed to factor and apply the Zero Product Property.

\[ \begin{array}{rclr}
3x^2 & = & 35 – 16x & \\
3x^2 + 16x – 35 & = & 0 & \text{Add } 16x, \text{ subtract } 35 \\
(3x-5)(x+7) & = & 0 & \text{Factor} \\
3x-5 = 0 & \text{or} & x+7 = 0 & \text{Zero Product Property} \\
x = \dfrac{5}{3} & \text{or} & x = -7 & \\
\end{array} \]

We check our answers by substituting each of them into the original equation. Plugging in [latex]x = \dfrac{5}{3}[/latex] yields [latex]\dfrac{25}{3}[/latex] on both sides while [latex]x = -7[/latex] gives 147 on both sides.

Solution:

Solve the equation: [latex]t = \dfrac{1+4t^2}{4}[/latex].

To solve [latex]t = \dfrac{1+4t^2}{4}[/latex], we first clear fractions. Then move all of the nonzero terms to one side of the equation, factor and apply the Zero Product Property.

\[ \begin{array}{rclr}
t & = & \dfrac{1+4t^2}{4} & \\
4t & = & 1+4t^2 & \text{Clear fractions (multiply by 4)} \\
0 & = & 1+4t^2 – 4t & \text{Subtract 4} \\
0 & = & 4t^2 – 4t + 1 & \text{Rearrange terms} \\
0 & = & (2t-1)^2 & \text{Factor (Perfect Square Trinomial)} \\
\end{array} \]

At this point, we get [latex](2t-1)^2 = (2t-1)(2t-1) = 0[/latex], so, the Zero Product Property gives us [latex]2t-1 =0[/latex] in both cases.^[6]

Our final answer is [latex]t = \dfrac{1}{2}[/latex], which we invite the reader to check.

Solution:

Solve the equation: [latex](y-1)^2 = 2(y-1)[/latex].

Following the strategy outlined above, the first step to solving [latex](y-1)^2 = 2(y-1)[/latex] is to gather the nonzero terms on one side of the equation with 0 on the other side and factor.

\[ \begin{array}{rclr}
(y-1)^2 & = & 2(y-1) & \\
(y-1)^2 – 2(y-1) & = & 0 & \text{Subtract } 2(y-1) \\
(y-1)[(y-1) – 2] & = & 0 & \text{Factor out G.C.F.} \\
(y-1)(y-3) & = & 0 & \text{Simplify} \\
y-1 = 0 & \text{or} & y – 3 = 0 & \\
y = 1 & \text{or} & y = 3 & \\
\end{array} \]

Both of these answers are easily checked by substituting them into the original equation.

An alternative method to solving this equation is to begin by dividing both sides by [latex](y-1)[/latex] to simplify things outright. As we saw in Example 0.5.1, however, whenever we divide by a variable quantity, we make the explicit assumption that this quantity is nonzero. Thus we must stipulate that [latex]y - 1 \neq 0[/latex].

\[ \begin{array}{rclr}
\dfrac{(y-1)^2}{(y-1)} & = & \dfrac{2(y-1)}{(y-1)} & \text{Divide by } (y-1) \text{ – this assumes } (y-1) \neq 0\\
y – 1 & = & 2 & \\
y & = & 3 & \\
\end{array} \]

Note that in this approach, we obtain the [latex]y=3[/latex] solution, but we lose the [latex]y = 1[/latex] solution. How did that happen? Assuming [latex]y - 1 \neq 0[/latex] is equivalent to assuming [latex]y \neq 1[/latex]. This is an issue because [latex]y = 1[/latex] is a solution to the original equation and it was divided out too early. The moral of the story? If you decide to divide by a variable expression, double check that you aren’t excluding any solutions.^[7]

Solution:

Solve the equation: [latex]\dfrac{w^4}{3} = \dfrac{8w^3-12}{12} - \dfrac{w^2-4}{4}[/latex].

Proceeding as before, we clear fractions, gather the nonzero terms on one side of the equation, have 0 on the other and factor.

\[ \begin{array}{rclr}
\dfrac{w^4}{3} & = & \dfrac{8w^3-12}{12} – \dfrac{w^2-4}{4} & \\  & & & \\
12 \left(\dfrac{w^4}{3}\right) & = & 12 \left(\dfrac{8w^3-12}{12} – \dfrac{w^2-4}{4} \right) & \text{Multiply by 12}\\ & & & \\
4w^4 & = & (8w^3 – 12) – 3(w^2-4) & \text{Distribute} \\
4w^4 & = & 8w^3 – 12 – 3w^2 + 12 & \text{Distribute} \\
0 & = & 8w^3 – 12 – 3w^2 + 12 – 4w^4 & \text{Subtract } 4w^4 \\
0 & = & 8w^3 – 3w^2 – 4w^4 & \text{Gather like terms} \\
0 & = & w^2(8w – 3 – 4w^2) & \text{Factor out G.C.F.} \\
\end{array} \]

At this point, we apply the Zero Product Property to deduce that [latex]w^2 = 0[/latex] or [latex]8w - 3 - 4w^2 = 0[/latex].

From [latex]w^2 = 0[/latex], we get [latex]w = 0[/latex].

To solve [latex]8w - 3 - 4w^2 = 0[/latex], we rearrange terms and factor:

\[-4w^2 + 8w – 3= (2w – 1)(-2w+3) = 0.\]

Applying the Zero Product Property again, we get [latex]2w - 1= 0[/latex] (which gives [latex]w = \dfrac{1}{2}[/latex]),

or [latex]-2w+3 = 0[/latex] (which gives [latex]w = \dfrac{3}{2}[/latex]).

Our final answers are [latex]w = 0[/latex], [latex]w = \dfrac{1}{2}[/latex] and [latex]w = \dfrac{3}{2}[/latex].

The reader is encouraged to check each of these answers in the original equation. (You need the practice with fractions!)

Solution:

Solve the equation: [latex]z(z(18z+9)-50) = 25[/latex].

For our next example, we begin by subtracting the 25 from both sides then work out the indicated operations before factoring by grouping.

\[ \begin{array}{rclr}
z(z(18z+9)-50) & = & 25 & \\
z(z(18z+9)-50) – 25 & = & 0 & \text{Subtract 25} \\
z(18z^2 + 9z – 50) – 25 & = & 0 & \text{Distribute} \\
18z^3 + 9z^2 – 50z – 25 & = & 0 & \text{Distribute} \\
9z^2(2z + 1) – 25(2z + 1) & = & 0 & \text{Factor} \\
(9z^2 – 25)(2z+1) & = & 0 & \text{Factor} \\
\end{array} \]

At this point, we use the Zero Product Property and get [latex]9z^2 - 25 = 0[/latex] or [latex]2z + 1 = 0[/latex].

The latter gives [latex]z = -\dfrac{1}{2}[/latex] whereas the former factors as [latex](3z - 5)(3z+5) = 0[/latex].

Applying the Zero Product Property again gives [latex]3z-5 = 0[/latex] (so [latex]z = \dfrac{5}{3}[/latex])

or [latex]3z+5 = 0[/latex] (so [latex]z = -\dfrac{5}{3}[/latex].)

Our final answers are [latex]z = -\dfrac{1}{2}[/latex], [latex]z = \dfrac{5}{3}[/latex] and [latex]z = -\dfrac{5}{3}[/latex], each of which is good fun to check.

Solution:

Solve the equation: [latex]x^4-8x^2 - 9 = 0[/latex].

The nonzero terms of the equation [latex]x^4-8x^2 - 9= 0[/latex] are already on one side of the equation so we proceed to factor.

This trinomial doesn’t fit the pattern of a perfect square so we attempt to reverse the F.O.I.L.ing process. With an [latex]x^4[/latex] term, we have two possible forms to try: [latex](ax^2 + b)(cx^2 + d)[/latex] and [latex](ax^3 + b)(cx +d)[/latex]. We leave it to you to show that [latex](ax^3 + b)(cx +d)[/latex] does not work and we show that [latex](ax^2 + b)(cx^2 + d)[/latex]does.

Due to the fact that the coefficient of [latex]x^4[/latex] is 1, we take [latex]a = c = 1[/latex].

The constant term is [latex]-9[/latex] so we know [latex]b[/latex] and [latex]d[/latex] have opposite signs and our choices are limited to two options: either [latex]b[/latex] and [latex]d[/latex] come from [latex]\pm 1[/latex] and [latex]\mp 9[/latex] OR one is 3 while the other is [latex]-3[/latex].

After some trial and error, we get [latex]x^4 - 8x^2 - 9 = (x^2 - 9)(x^2+1)[/latex].

Hence the equation [latex]x^4-8x^2 - 9= 0[/latex] may be rewritten as [latex](x^2 - 9)(x^2 + 1) = 0[/latex].

The Zero Product Property tells us that either [latex]x^2 - 9 = 0[/latex] or [latex]x^2+1 = 0.[/latex]

To solve [latex]x^2 - 9 = 0[/latex]

we factor: [latex](x-3)(x+3) = 0[/latex],

so [latex]x-3 = 0[/latex] or [latex]x+3 = 0[/latex],

so [latex]x=3[/latex] or [latex]x = -3[/latex].

The equation [latex]x^2 + 1 = 0[/latex] has no (real) solution, because for any real number [latex]x[/latex], [latex]x^2[/latex] is always 0 or greater. Thus [latex]x^2 + 1[/latex] is always positive.

Our final answers are [latex]x = 3[/latex] and [latex]x = -3[/latex].

As always, the reader is invited to check both answers in the original equation.

Solution:

Solve the equation: [latex](5x +3)^{4} = 16[/latex].

In our first equation, the quantity containing [latex]x[/latex] is already isolated, so we extract fourth roots. The exponent is even, so when the roots are extracted we need both the positive and negative roots

\[ \begin{array}{rclr}
(5x +3)^{4} & = & 16 & \\ & & & \\
5x+3 & = & \pm \sqrt[4]{16} & \text{Extract fourth roots} \\ & & &\\
5x + 3 & = & \pm 2 & \\  & & & \\
5x+3 = 2 & \text{or} & 5x+3 = -2 & \\
x = -\dfrac{1}{5} & \text{or} & x = -1 \\
\end{array} \]

We leave it to the reader to verify that both of these solutions satisfy the original equation.

Solution:

Solve the equation: [latex]1 - \dfrac{(5-2w)^3}{7} = 9[/latex].

In this example, we first need to isolate the quantity containing the variable [latex]w[/latex]. Here, third (cube) roots are required and the exponent (index) is odd indicating we do not need the [latex]\pm[/latex]:

\[ \begin{array}{rclr}
1 – \dfrac{(5-2w)^3}{7} & = & 9 & \\ & & &\\
– \dfrac{(5-2w)^3}{7} & = & 8 & \text{Subtract } 1\\ & & &\\
(5-2w) ^ 3 & = & -56 & \text{Multiply by } -7 \\ & & &\\
5 – 2w & = & \sqrt[3]{-56} & \text{Extract cube root} \\
5 – 2w & = & \sqrt[3]{(-8)(7)} & \\
5 – 2w & = & \sqrt[3]{-8} \sqrt[3]{7} & \text{Product Rule}\\ & & &\\
5 – 2w & = & -2\sqrt[3]{7} & \\ & & &\\
-2w & = & -5-2 \sqrt[3]{7} & \text{Subtract } 5 \\ & & &\\
w & = & \dfrac{-5 – 2\sqrt[3]{7}}{-2} & \text{Divide by } -2 \\ & & &\\
w & = & \dfrac{5 + 2\sqrt[3]{7}}{2} & \text{Properties of Negatives}
\end{array}\]

The reader should check the answer because it provides a hearty review of arithmetic.

Solution:

Solve the equation: [latex]t + \sqrt{2t+3} = 6[/latex].

To solve [latex]t + \sqrt{2t+3} = 6[/latex], we first isolate the square root, then proceed to square both sides of the equation. In doing so, we run the risk of introducing extraneous solutions therefore checking our answers here is a necessity.

\[ \begin{array}{rclr}
t + \sqrt{2t+3} & = & 6 & \\ & & &\\
\sqrt{2t+3} & = & 6 – t & \text{Subtract } t \\ & & &\\
(\sqrt{2t+3})^2 & = & (6-t)^2 & \text{Square both sides} \\ & & &\\
2t + 3 & = & 36-12t + t^2 & \text{F.O.I.L. / Perfect Square Trinomial} \\ & & &\\
0 & = & t^2 – 14t + 33 & \text{Subtract } 2t \text{ and } 3 \\ & & &\\
0 & = & (t-3)(t-11) & \text{Factor} \\
\end{array} \]

From the Zero Product Property, we know either [latex]t - 3 = 0[/latex] (which gives [latex]t=3[/latex]) or [latex]t-11 = 0[/latex] (which gives [latex]t=11[/latex]).

When checking our answers, we find [latex]t = 3[/latex] satisfies the original equation, but [latex]t = 11[/latex] does not.

It is worth noting that when [latex]t=11[/latex] is substituted into the original equation, we get [latex]11 + \sqrt{25} = 6[/latex]. If the [latex]+\sqrt{25}[/latex] were [latex]-\sqrt{25}[/latex], the solution would check.

Once again, when squaring both sides of an equation, we lose track of [latex]\pm[/latex], which is what lets extraneous solutions in the door.

So our final answer is [latex]t = 3[/latex] only.

Solution:

Solve the equation: [latex]\sqrt{2} - 3\sqrt[3]{2y+1} = 0[/latex].

In our next example, we locate the variable (in this case [latex]y[/latex]) beneath a cube root, so we first isolate that root and cube both sides.

\[ \begin{array}{rclr}
\sqrt{2} – 3\sqrt[3]{2y+1} & = & 0 & \\ & & &\\
– 3\sqrt[3]{2y+1} & = & – \sqrt{2} & \text{Subtract } \sqrt{2} \\ & & &\\
\sqrt[3]{2y+1} & = & \dfrac{-\sqrt{2}}{-3} & \text{Divide by } -3 \\
\sqrt[3]{2y+1} & = & \dfrac{\sqrt{2}}{3} & \text{Properties of Negatives} \\& & &\\
(\sqrt[3]{2y+1})^3 & = & \left( \dfrac{\sqrt{2}}{3} \right)^{3} & \text{Cube both sides} \\ & & &\\
2y + 1 & = & \dfrac{(\sqrt{2})^3}{3^3} & \\& & &\\
2y + 1 & = & \dfrac{2\sqrt{2}}{27} & \\ [& & &\\
2y & = & \dfrac{2 \sqrt{2}}{27} – 1 & \text{Subtract } 1 \\[8pt]
2y & = & \dfrac{2 \sqrt{2}}{27} – \dfrac{27}{27} & \text{Common denominators} \\ & & &\\
2y & = & \dfrac{2 \sqrt{2} – 27}{27} & \text{Subtract fractions} \\ & & &\\
y & = & \dfrac{2 \sqrt{2} – 27}{54} & \text{Divide by } \left(\text{multiply by } \dfrac{1}{2} \right) \\
\end{array}\]

As we raised both sides to an odd power, we don’t need to worry about extraneous solutions but we encourage the reader to check the solution just for the fun of it.

Solution:

Solve the equation: [latex]\sqrt{4x-1} + 2\sqrt{1 - 2x} = 1[/latex].

In the equation [latex]\sqrt{4x-1} + 2\sqrt{1 - 2x} = 1[/latex], we have not one but two square roots. We begin by isolating one of the square roots and squaring both sides.

\[ \begin{array}{rclr}
\sqrt{4x-1} + 2\sqrt{1 – 2x} & = & 1 & \\ & & &\\
\sqrt{4x-1} & = & 1 – 2\sqrt{1-2x} &  \text{Subtract } 2\sqrt{1 – 2x} \text{ from both sides} \\ & & &\\
(\sqrt{4x-1})^2 & = & (1 – 2\sqrt{1-2x})^2 & \text{Square both sides} \\ & & &\\
4x – 1 & = & 1 – 4\sqrt{1-2x} + (2\sqrt{1-2x})^2 & \text{F.O.I.L. / Perfect Square Trinomial} \\ & & &\\
4x – 1 & = & 1 – 4\sqrt{1-2x} + 4(1-2x) & \\ & & &\\
4x – 1 & = & 1 – 4\sqrt{1-2x} + 4 – 8x & \text{Distribute} \\ & & &\\
4x – 1 & = & 5 – 8x – 4\sqrt{1-2x} & \text{Gather like terms} \\
\end{array} \]

At this point, we have just one square root so we proceed to isolate it and square both sides a second time.^[10]

\[ \begin{array}{rclr}
4x – 1 & = & 5 – 8x – 4\sqrt{1-2x} & \\  & & &\\
12x – 6 & = & -4\sqrt{1-2x} & \text{Subtract 5, add } 8x\\ & & &\\
(12x-6)^2 & = & (-4\sqrt{1-2x})^2 & \text{Square both sides} \\ & & &\\
144x^2 – 144x + 36 & = & 16(1-2x) & \\ & & &\\
144x^2 – 144x + 36 & = & 16 – 32x & \\ & & &\\
144x^2 – 112x + 20 & = & 0 & \text{Subtract 16, add } 32x \\ & & &\\
4(36x^2 – 28x + 5) & = & 0 & \text{Factor} \\ & & &\\
4(2x-1)(18x – 5) & = & 0 & \text{Factor some more} \\
\end{array} \]

From the Zero Product Property, we know either [latex]2x-1 = 0[/latex] or [latex]18x - 5 = 0[/latex].

The former gives [latex]x = \dfrac{1}{2}[/latex] while the latter gives us [latex]x = \dfrac{5}{18}[/latex].

As we squared both sides of the equation (twice!), we need to check for extraneous solutions. We determine [latex]x = \dfrac{5}{18}[/latex] to be extraneous, so our only solution is [latex]x = \dfrac{1}{2}[/latex].

Solution:

Solve the equation: [latex]\sqrt[4]{n^2 + 2} + n = 0[/latex].

As usual, our first step in solving [latex]\sqrt[4]{n^2 + 2} + n = 0[/latex] is to isolate the radical. We then proceed to raise both sides to the fourth power to eliminate the fourth root:

\[ \begin{array}{rclr}
\sqrt[4]{n^2 + 2} + n & = & 0 & \\
\sqrt[4]{n^2 + 2} & = & -n & \text{Subtract } n\\
(\sqrt[4]{n^2 + 2})^4 & = & (-n)^4 & \text{Raise both sides to the } 4^{\text{th}} \text{ power} \\
n^2 + 2 & = & n^4 & \text{Properties of Negatives}\\
0 & = & n^{4} – n^2 – 2 & \text{Subtract } n^2 \text{ and } 2 \\
0 & = & (n^2 – 2)(n^2 + 1) & \text{Factor – this is a Quadratic in Disguise} \\
\end{array} \]

At this point, the Zero Product Property gives either [latex]n^2 - 2 = 0[/latex] or [latex]n^2 + 1 = 0[/latex].

From [latex]n^2 - 2 = 0[/latex], we get [latex]n^2 = 2[/latex], so [latex]n = \pm \sqrt{2}[/latex].

From [latex]n^2 + 1 = 0[/latex], we get [latex]n^2 = -1[/latex], which gives no real solutions.

As we raised both sides to an even (the fourth) power, we need to check for extraneous solutions. We determine that [latex]n = -\sqrt{2}[/latex] works but [latex]n = \sqrt{2}[/latex] is extraneous.

Solution:

Solve for [latex]r[/latex]: [latex]V = \dfrac{4\pi}{3}(R^3 - r^3)[/latex].

In this problem, we are asked to solve for [latex]r[/latex]. While there are a lot of letters in this equation^[12], [latex]r[/latex] appears in only one term: [latex]r^3[/latex]. Our strategy is to isolate [latex]r^3[/latex] then extract the cube root.

\[ \begin{array}{rclr}
V & = & \dfrac{4\pi}{3}(R^3 – r^3) & \\ & & &\\
3V & = & 4\pi (R^3 – r^3) & \text{Multiply by 3 to clear fractions}\\ & & &\\
3V & = & 4\pi R^3 – 4\pi r^3 & \text{Distribute} \\ & & &\\
3V – 4\pi R^3 & = & -4 \pi r^3 & \text{Subtract }4 \pi R^3 \\ & & &\\
\dfrac{3V – 4\pi R^3}{-4\pi} & = & r^3 & \text{Divide by } -4\pi \\ & & &\\
\dfrac{4\pi R^3 – 3V}{4\pi} & = & r^3 & \text{Properties of Negatives} \\ & & &\\
\sqrt[3]{\dfrac{4\pi R^3 – 3V}{4\pi}} & = & r & \text{Extract the cube root} \\
\end{array} \]

The check is, as always, left to the reader and highly encouraged.

Solution:

Solve for [latex]M_{1}[/latex]: [latex]\dfrac{r_{1}}{r_{2}} = \sqrt{\dfrac{M_{2}}{M_{1}}}[/latex].

The equation we are asked to solve in this example is from the world of Chemistry and is none other than Graham’s Law of Effusion.

Subscripts in Mathematics are used to distinguish between variables and have no arithmetic significance.  In this example, [latex]r_{1}[/latex], [latex]r_{2}[/latex], [latex]M_{1}[/latex] and [latex]M_{2}[/latex] are as different as [latex]x[/latex], [latex]y[/latex], [latex]z[/latex] and 117.

We are asked to solve for [latex]M_{1}[/latex], so we locate [latex]M_{1}[/latex] and see it is in the denominator of a fraction which is inside of a square root. We eliminate the square root by squaring both sides and proceed from there.

\[ \begin{array}{rclr}
\dfrac{r_{1}}{r_{2}} & = & \sqrt{\dfrac{M_{2}}{M_{1}}} & \\ & & &\\
\left(\dfrac{r_{1}}{r_{2}}\right)^2 & = & \left(\sqrt{\dfrac{M_{2}}{M_{1}}}\right)^2 & \text{Square both sides} \\ & & &\\
\dfrac{r_{1}^2}{r_{2}^2} & = & \dfrac{M_{2}}{M_{1}} & \\ & & &\\
r_{1}^2 M_{1} & = & M_{2}r_{2}^2 & \text{Multiply by } r_{2}^2 M_{1} \text{ to clear fractions, assume } r_{2}, \; M_{1} \neq 0 \\ & & &\\
M_{1} & = & \dfrac{M_{2}r_{2}^2}{r_{1}^2} & \text{Divide by } r_{1}^2, \text{ assume } r_{1} \neq 0 \\
\end{array} \]

As the reader may expect, checking the answer amounts to a good exercise in simplifying rational and radical expressions. The fact that we are assuming all of the variables represent positive real numbers comes in to play, as well.

Solution:

Solve for [latex]v[/latex]: [latex]m = \dfrac{m_{0}}{\sqrt{1 - \dfrac{v^2}{c^2}}}[/latex].

Our last equation to solve comes from Einstein’s Special Theory of Relativity and relates the mass of an object to its velocity as it moves.^[13]

\[ \begin{array}{rclr}
m & = & \dfrac{m_{0}}{\sqrt{1 – \dfrac{v^2}{c^2}}} & \\
m \sqrt{1 – \dfrac{v^2}{c^2}} & = & m_{0} & \text{Multiply by } \sqrt{1 – \dfrac{v^2}{c^2}} \text{ to clear fractions}\\ & & &\\
\left(m \sqrt{1 – \dfrac{v^2}{c^2}}\right)^{2} & = & m_{0}^{2} & \text{Square both sides}\\ & & &\\
m^2 \left(1 – \dfrac{v^2}{c^2}\right) & = & m_{0}^{2} & \text{Properties of Exponents}\\ & & &\\
m^2 – \dfrac{m^2 v^2}{c^2} & = & m_{0}^{2} & \text{Distribute} \\ & & &\\
– \dfrac{m^2 v^2}{c^2} & = & m_{0}^{2} – m^2 & \text{Subtract } m^2 \\ & & &\\
m^2 v^2 & = & -c^2 (m_{0}^{2} – m^2) & \text{Multiply by } -c^2\; (c^2 \neq 0) \\ & & &\\
m^2 v^2 & = & -c^2m_{0}^{2} + c^2 m^2 & \text{Distribute} \\& & &\\
v^2 & = & \dfrac{c^2 m^2 -c^2m_{0}^{2}}{m^2} & \text{Rearrange terms, divide by } m^2 \; (m^2 \neq 0) \\ & & &\\
v & = & \sqrt{\dfrac{c^2 m^2 -c^2m_{0}^{2}}{m^2}} & \text{Extract Square Roots, } v > 0 \text{ so no }\pm \\ & & &\\
v & = & \dfrac{\sqrt{c^2 (m^2 -m_{0}^{2})}}{\sqrt{m^2}} & \text{Properties of Radicals, factor} \\ & & &\\
v & = & \dfrac{|c|\sqrt{m^2 -m_{0}^{2}}}{|m|} & \\ & & &\\
v & = & \dfrac{c\sqrt{m^2 -m_{0}^{2}}}{m} & c > 0 \text{ and } m > 0 \text{ so } |c| = c \text{ and } |m| = m \\
\end{array}\]

Checking the answer algebraically would earn the reader great honor and respect on the Algebra battlefield, so it is highly recommended.

Solution:

Calculate all real number solutions to [latex]3 - (2w-1)^2 = 0[/latex].

As [latex]3 - (2w-1)^2 = 0[/latex] contains a perfect square, we isolate it first then extract square roots:

\[ \begin{array}{rclr}
3 – (2w-1)^2 & = & 0 & \\
3 & = & (2w-1)^2 & \text{Add } (2w-1)^2 \\
\pm \sqrt{3} & = & 2w – 1 & \text{Extract Square Roots} \\
1 \pm \sqrt{3} & = & 2w & \text{Add } 1 \\
\dfrac{1 \pm \sqrt{3}}{2} & = & w & \text{Divide by } 2 \\
\end{array} \]

We determine our two answers: [latex]w = \dfrac{1 \pm \sqrt{3}}{2}[/latex].

The reader is encouraged to check both answers by substituting each into the original equation.^[17]

Solution:

Calculate all real number solutions to [latex]5x - x(x-3) = 7[/latex].

To solve [latex]5x - x(x-3) = 7[/latex], we perform the indicated operations and set one side equal to 0.

\[ \begin{array}{rclr}
5x – x(x-3) & = & 7 & \\
5x – x^2 + 3x & = & 7 & \text{Distribute} \\
-x^2 + 8x & = & 7 & \text{Gather like terms} \\
-x^2 + 8x – 7 & = & 0& \text{Subtract } 7 \\
\end{array}\]

At this point, we attempt to factor and find [latex]-x^2 + 8x - 7 = (x-1)(-x+7)[/latex].

Using the Zero Product Property, we get [latex]x-1 = 0[/latex] or [latex]-x+7 = 0[/latex].

Our answers are [latex]x = 1[/latex] or [latex]x = 7[/latex], which are easily verified.

Solution:

Calculate all real number solutions to [latex](y-1)^2 = 2 - \dfrac{y+2}{3}[/latex].

Even though we have a perfect square in [latex](y-1)^2 = 2 - \dfrac{y+2}{3}[/latex], Extracting Square Roots won’t help matters as we have a [latex]y[/latex] on the other side of the equation. Our strategy here is to perform the indicated operations (and clear the fraction for good measure) and then get 0 on one side of the equation.

\[ \begin{array}{rclr}
(y-1)^2 & = & 2 – \dfrac{y+2}{3} & \\  & & &\\
y^2 – 2y + 1 & = & 2 – \dfrac{y+2}{3} & \text{Perfect Square Trinomial}\\ & & & \\
3(y^2 – 2y + 1) & = & 3\left(2 – \dfrac{y+2}{3} \right) & \text{Multiply by } 3 \\ & & & \\
3y^2 – 6y + 3 & = & 6 – 3\left(\dfrac{y+2}{3}\right) & \text{Distribute} \\ & & & \\
3y^2 – 6y + 3 & = & 6 – (y+2) & \\
3y^2 – 6y + 3 – 6 + (y+2) & = & 0 & \text{Subtract 6, Add } (y+2) \\
3y^2 – 5y – 1 & = & 0 & \\
\end{array}\]

A cursory attempt at factoring bears no fruit, so we complete the square on the left hand side of the equation.

\[ \begin{array}{rclr}
3y^2 – 5y – 1 & = & 0 & \\
3y^2 – 5y & = & 1 & \text{Subtract } c = -1 \text{ from both sides} \\
y^2 – \dfrac{5}{3}y & = & \dfrac{1}{3} & \text{Divide by } a = 3 \\ & & & \\
y^2 – \dfrac{5}{3}y + \dfrac{25}{36} & = & \dfrac{1}{3} + \dfrac{25}{36} & \text{Add } \left(\dfrac{b}{2a}\right)^2 = \left(-\dfrac{5}{6} \right)^2 = \dfrac{25}{36} \\ & & & \\
\left(y – \dfrac{5}{6} \right)^2 & = & \dfrac{37}{36} & \text{Factor: Perfect Square Trinomial} \\
y – \dfrac{5}{6} & = & \pm \sqrt{\dfrac{37}{36}} & \text{Extract Square Roots} \\ & & & \\
y & = & \dfrac{5}{6} \pm \sqrt{\dfrac{37}{36}} & \text{Add } \dfrac{5}{6} \\ & & & \\
y & = & \dfrac{5 \pm \sqrt{37}}{6} & \\
\end{array} \]

37 is prime, so we have no way to reduce [latex]\sqrt{37}[/latex].

Thus, our final answers are [latex]y = \dfrac{5 \pm \sqrt{37}}{6}[/latex].

The reader is encouraged to supply the details of the challenging verification of the answers.

Solution:

Calculate all real number solutions to [latex]5(25 - 21x) = \dfrac{59}{4} - 25x^2.[/latex]

We proceed as before; our goal is to gather the nonzero terms on one side of the equation.

\[ \begin{array}{rclr}
5(25 – 21x) & = & \dfrac{59}{4} – 25x^2 & \\ & & & \\
125 – 105x & = & \dfrac{59}{4} – 25x^2 & \text{Distribute} \\ & & & \\
4(125 – 105x) & = & 4\left(\dfrac{59}{4} – 25x^2 \right) & \text{Multiply by } 4 \\ & & & \\
500 – 420x & = & 59 – 100x^2 & \text{Distribute} \\ & & & \\
500 – 420x – 59 + 100x^2 & = & 0 & \text{Subtract 59, Add } 100x^2 \\ & & & \\
100x^2 – 420x + 441 & = & 0 & \text{Gather like terms} \\
\end{array} \]

With highly composite numbers like 100 and441, factoring seems inefficient at best,^[18] so we apply the Quadratic Formula with [latex]a = 100, \; b = -420[/latex]and [latex]c = 441[/latex]:

\[ \begin{array}{rclr}
x & = & \dfrac{-(-420) \pm \sqrt{(-420)^2 – 4(100)(441)}}{2(100)} & \\ & & & \\
& = & \dfrac{420 \pm \sqrt{176400 – 176400}}{200} & \\ & & & \\
& = & \dfrac{420 \pm \sqrt{0}}{200} & \\ & & & \\
& = & \dfrac{420 \pm 0}{200} & \\ & & & \\
& = & \dfrac{420}{200} & \\ & & & \\
& = & \dfrac{21}{10} & \\
\end{array} \]

To our surprise and delight we obtain just one answer, [latex]x = \dfrac{21}{10}[/latex].

Solution:

Calculate all real number solutions to [latex]-4.9t^2 + 10t\sqrt{3} + 2 = 0[/latex].

Our next equation [latex]-4.9t^2 + 10t\sqrt{3} + 2 = 0[/latex], already has 0 on one side of the equation, but with coefficients like [latex]-4.9[/latex] and [latex]10\sqrt{3}[/latex], factoring with integers is not an option.

We could make things a bit easier by clearing the decimal (by multiplying through by 10) to get [latex]-49t^2 + 100t\sqrt{3} + 20 = 0[/latex] but we simply cannot rid ourselves of the irrational number [latex]\sqrt{3}[/latex].

The Quadratic Formula is our only recourse. With [latex]a = -49, \; b = 100\sqrt{3}[/latex] and [latex]c = 20[/latex] we get:

\[ \begin{array}{rclr}
t & = & \dfrac{-100\sqrt{3} \pm \sqrt{(100\sqrt{3})^2 – 4(-49)(20)}}{2(-49)} & \\ & & & \\
& = & \dfrac{-100\sqrt{3} \pm \sqrt{30000 +3920}}{-98} & \\ & & & \\
& = & \dfrac{-100\sqrt{3} \pm \sqrt{33920}}{-98} & \\ & & & \\
& = & \dfrac{-100\sqrt{3} \pm 8\sqrt{530}}{-98} & \\ & & & \\
& = & \dfrac{2(-50\sqrt{3} \pm 4\sqrt{530})}{2(-49)} & \\ & & & \\
& = & \dfrac{-50\sqrt{3} \pm 4\sqrt{530}}{-49} & \text{Reduce} \\ & & & \\
& = & \dfrac{-(-50\sqrt{3} \pm 4\sqrt{530})}{49} & \text{Properties of Negatives} \\ & & & \\
& = & \dfrac{50\sqrt{3} \mp 4\sqrt{530}}{49} & \text{Distribute} \\
\end{array}\]

You’ll note that when we distributed the negative in the last step, we changed the [latex]\pm[/latex] to a [latex]\mp.[/latex] While this is technically correct, at the end of the day both symbols mean plus or minus,^[19] so we can write our answers as [latex]t = \dfrac{50\sqrt{3} \pm 4\sqrt{530}}{49}[/latex].

Checking these answers are a true test of arithmetic mettle.

Solution:

Calculate all real number solutions to [latex]2x^2 = 3x^4 - 6[/latex].

At first glance, the equation [latex]2x^2 = 3x^4 - 6[/latex] seems misplaced. The highest power of the variable [latex]x[/latex] here is 4, not 2, so this equation isn’t a quadratic equation – at least not in terms of the variable [latex]x[/latex]. It is, however, an example of an equation that is Quadratic in Disguise.^[20]

We introduce a new variable [latex]u[/latex] to help us see the pattern – specifically we let [latex]u = x^2[/latex]. Thus [latex]u^2 = (x^2)^2 = x^4[/latex]. So in terms of the variable [latex]u[/latex], the equation [latex]2x^2 = 3x^4 - 6[/latex] is [latex]2u = 3u^2 - 6[/latex]. The latter is a quadratic equation, which we can solve using the usual techniques:

\[ \begin{array}{rclr}
2u & = & 3u^2 – 6 & \\
0 & = & 3u^2 – 2u – 6 & \text{Subtract } 2u \\
\end{array}\]

After a few attempts at factoring, we resort to the Quadratic Formula with [latex]a = 3, \; b = -2[/latex] and [latex]c = -6[/latex] to get the following:

\[ \begin{array}{rclr}
u & = & \dfrac{-(-2) \pm \sqrt{(-2)^2 – 4(3)(-6)}}{2(3)} & \\ & & & \\
& = & \dfrac{2 \pm \sqrt{4 + 72}}{6} & \\ & & & \\
& = & \dfrac{2 \pm \sqrt{76}}{6} & \\ & & & \\
& = & \dfrac{2 \pm \sqrt{4 \cdot 19}}{6} & \\ & & & \\
& = & \dfrac{2 \pm 2\sqrt{19}}{6} & \text{Properties of Radicals} \\ & & & \\
& = & \dfrac{2(1 \pm \sqrt{19})}{2(3)} & \text{Factor} \\ & & & \\
& = & \dfrac{1 \pm \sqrt{19}}{3} & \text{Reduce} \\
\end{array} \]

We’ve solved the equation for [latex]u[/latex], but what we still need to solve the original equation – which means we need to find the corresponding values of [latex]x[/latex]. We stated [latex]u = x^2[/latex], thus we now have we have two equations:

\[ \begin{array}{rclr}
x^2 =\dfrac{1 + \sqrt{19}}{3} & \text{or} & x^2 =\dfrac{1 – \sqrt{19}}{3} & \\
\end{array}\]

We can solve the first equation by extracting square roots to get [latex]x = \pm \sqrt{\dfrac{1 + \sqrt{19}}{3}}[/latex].

The second equation, however, has no real number solutions because [latex]\dfrac{1 - \sqrt{19}}{3}[/latex] is a negative number.

For our final answers we can rationalize the denominator to get:

\[ \begin{array}{rcl} x &=& \pm \sqrt{\dfrac{1 + \sqrt{19}}{3}} \\ & & \\ &=& \pm \sqrt{\dfrac{1 + \sqrt{19}}{3} \cdot \dfrac{3}{3}} \\ & & \\ &=& \pm \dfrac{\sqrt{3 + 3\sqrt{19}}}{3} \end{array}\]

As with the previous exercise, the very challenging check is left to the reader.

Solution:

Simplify [latex](1-2i) - (3+4i)[/latex].

As mentioned earlier, we treat expressions involving [latex]i[/latex] as we would any other radical. We distribute and combine like terms:

\[ \begin{array}{rclr}
(1-2i) – (3+4i) & = & 1-2i-3-4i & \text{Distribute} \\
& = & -2 – 6i & \text{Gather like terms} \\
\end{array}\]

Technically, we’d have to rewrite our answer [latex]-2-6i[/latex]as [latex](-2) + (-6)i[/latex] to be (in the strictest sense) in the form [latex]a+bi[/latex].

That being said, even pedants have their limits, so [latex]-2-6i[/latex] is good enough.

Solution:

Simplify [latex](1-2i)(3+4i)[/latex].

Using the Distributive Property (a.k.a. F.O.I.L.), we get

\[ \begin{array}{rclr}
(1-2i)(3+4i) & = & (1)(3) + (1)(4i) – (2i)(3) – (2i)(4i) & \text{F.O.I.L.} \\
& = & 3+4i-6i-8i^2 & \\
& = & 3 – 2i – 8(-1) & i^2=-1 \\
& = & 3 – 2i + 8 & \\
& = & 11 – 2i & \\
\end{array}\]

Solution:

Simplify [latex]\dfrac{1-2i}{3-4i}[/latex].

How are we supposed to simplify [latex]\dfrac{1-2i}{3-4i}[/latex]? Well, we deal with the denominator [latex]3-4i[/latex] as we would any other denominator containing two terms, one of which is a square root. We multiply both numerator and denominator by [latex]3+4i[/latex], the (complex) conjugate of [latex]3 - 4i[/latex]. Doing so produces

\[\begin{array}{rclr}
\dfrac{1-2i}{3-4i} & = & \dfrac{(1-2i)(3+4i)}{(3-4i)(3+4i)} & \text{Equivalent Fractions} \\ & & & \\
& = & \dfrac{3 + 4i – 6i – 8i^2}{9 – 16i^2} & \text{F.O.I.L.}\\ & & & \\
& = & \dfrac{3 – 2i – 8(-1)}{9 – 16(-1)} & i^2 = -1\\ & & & \\
& = & \dfrac{11 – 2i}{25} & \\ & & & \\
& = & \dfrac{11}{25} – \dfrac{2}{25} \, i & \\
\end{array}\]

Solution:

Simplify [latex]\sqrt{-3} \sqrt{-12}.[/latex]

We use property 2 of Definition 0.14 first, then apply the rules of radicals applicable to real numbers to get

\[ \begin{array}{rcl} \sqrt{-3} \sqrt{-12} &=& \left(i \sqrt{3}\right) \left(i \sqrt{12}\right) \\ & & \\ &=& i^2 \sqrt{3\cdot 12} \\ &=& -\sqrt{36} \\ &=& -6 \end{array} \]

Solution:

Simplify [latex]\sqrt{(-3)(-12)}[/latex].

We adhere to the order of operations here and perform the multiplication before the radical to get

\[ \begin{array}{rcl} \sqrt{(-3)(-12)} &=& \sqrt{36} \\ &=& 6 \end{array} \]

Solution:

Simplify [latex](x-[1+2i])(x-[1-2i])[/latex].

We brute force multiply using the distributive property and find that

\[\begin{array}{rclr} (x-[1+2i])(x-[1-2i]) & = & x^2 -x[1-2i]-x[1+2i]+[1-2i][1+2i] & \text{F.O.I.L.}\\ & & & \\
&= & x^2-x+2ix-x-2ix+1-2i+2i-4i^2 & \text{Distribute} \\ & & & \\
&= & x^2-2x + 1-4(-1)& \text{Gather like terms}\\ & & &\\
& = & x^2 -2x +5 & i^2=-1 \\ \end{array}\]

Solution:

Determine the complex solutions to [latex]\dfrac{2x}{x+1} = x+3[/latex].

Clearing fractions yields a quadratic equation so we then proceed as in Section 0.5.5.

\[ \begin{array}{rclr}
\dfrac{2x}{x+1} & = & x+3 & \\ & & &\\
2x & = & (x+3)(x+1) & \text{Multiply by } (x+1) \text{ to clear denominators} \\
2x & = & x^2 + x + 3x + 3 & \text{F.O.I.L.} \\
2x & = & x^2 + 4x + 3 & \text{Gather like terms} \\
0 & = & x^2 + 2x + 3 & \text{Subtract } 2x \\
\end{array}\]

From here, we apply the Quadratic Formula

\[ \begin{array}{rclr}
x & = & \dfrac{-2 \pm \sqrt{2^2 – 4(1)(3)}}{2(1)} & \text{Quadratic Formula}\\ & & &\\
& = & \dfrac{-2 \pm \sqrt{-8}}{2} & \text{Simplify}\\ & & &\\
& = & \dfrac{-2 \pm i \sqrt{8}}{2} & \text{Definition of } i \\ & & &\\
& = & \dfrac{-2 \pm i 2\sqrt{2}}{2} & \text{Product Rule for Radicals}\\ & & &\\
& = & \dfrac{\cancel{2}(-1 \pm i\sqrt{2})}{\cancel{2}} & \text{Factor and reduce}\\ & & &\\
& = & -1 \pm i \sqrt{2} & \\
\end{array} \]

We get two answers: [latex]x = -1 + i\sqrt{2}[/latex] and its conjugate [latex]x = -1 - i\sqrt{2}[/latex].

Checking both of these answers reviews all of the salient points about complex number arithmetic and is therefore strongly encouraged.

Solution:

Determine the complex solutions to [latex]2t^4 = 9t^2 + 5[/latex].

We have a Quadratic in Disguise, because we have three terms and the exponent on one term (4 on [latex]t^4[/latex]) is exactly twice the exponent on the other (2 on [latex]t^2[/latex]). We proceed accordingly.

\[ \begin{array}{rclr}
2t^4 & = & 9t^2 + 5 & \\
2t^4 – 9t^2 – 5 & = & 0 & \text{Subtract } 9t^2 \text{ and } 5 \\
(2t^2 + 1)(t^2 – 5) & = & 0 & \text{Factor} \\
2t^2 + 1 = 0 & \text{or} & t^2 = 5 & \text{Zero Product Property} \\
\end{array}\]

From [latex]2t^2 + 1 = 0[/latex] we get [latex]2t^2 = -1[/latex], or [latex]t^2 = -\dfrac{1}{2}[/latex].

We extract square roots as follows:

\[ \begin{array}{rcl} t &=& \pm \sqrt{-\dfrac{1}{2}} \\ & & \\ &=& \pm i \sqrt{\dfrac{1}{2}} \\ & & \\ &=& \pm i \dfrac{\sqrt{1}}{\sqrt{2}} \\ & & \\ &=& \pm i \dfrac{1}{\sqrt{2}} \\ & & \\ &=& \pm \dfrac{i \sqrt{2}}{2} \end{array}\]

where we have rationalized the denominator per convention.

From [latex]t^2 = 5[/latex], we get [latex]t = \pm \sqrt{5}[/latex].

In total, we have four complex solutions

– two real: [latex]t = \pm \sqrt{5}[/latex] and two non-real: [latex]t = \pm \dfrac{i \sqrt{2}}{2}[/latex].

Solution:

Determine the complex solutions to [latex]z^3 + 1 = 0[/latex].

To find the real solutions to [latex]z^3 + 1 = 0[/latex], we can subtract the 1 from both sides and extract cube roots: [latex]z^3 = -1[/latex], so [latex]z = \sqrt[3]{-1} = -1[/latex].

It turns out there are two more non-real complex number solutions to this equation. To get at these, we factor:

\[ \begin{array}{rclr}
z ^ 3 + 1 & = & 0 & \\
(z + 1)(z^2 – z + 1) & = & 0 & \text{Factor (Sum of Two Cubes)} \\
z + 1 = 0 & \text{or} & z^2 – z + 1 = 0 & \\
\end{array} \]

From [latex]z+1 = 0[/latex], we get our real solution [latex]z = -1[/latex].

From [latex]z^2 -z + 1 = 0[/latex], we apply the Quadratic Formula to get:

\[ \begin{array}{rcl} z &=& \dfrac{-(-1) \pm \sqrt{(-1)^2 – 4(1)(1)}}{2(1)} \\ & & \\ &=& \dfrac{1 \pm \sqrt{-3}}{2} \\ & & \\ &=& \dfrac{1 \pm i\sqrt{3}}{2} \end{array} \]

Thus, we get three solutions to [latex]z^3 + 1 = 0[/latex]

– one real: [latex]z = -1[/latex] and two non-real: [latex]z = \fdrac{1 \pm i\sqrt{3}}{2}[/latex].

As always, the reader is encouraged to test their algebraic mettle and check these solutions.