# 2.6 Other types of equations  (Page 3/10)

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## Solving an equation with one radical

Solve $\text{\hspace{0.17em}}\sqrt{15-2x}=x.$

The radical is already isolated on the left side of the equal side, so proceed to square both sides.

$\begin{array}{ccc}\hfill \sqrt{15-2x}& =& x\hfill \\ \hfill {\left(\sqrt{15-2x}\right)}^{2}& =& {\left(x\right)}^{2}\hfill \\ \hfill 15-2x& =& {x}^{2}\hfill \end{array}$

We see that the remaining equation is a quadratic. Set it equal to zero and solve.

$\begin{array}{ccc}\hfill 0& =& {x}^{2}+2x-15\hfill \\ & =& \left(x+5\right)\left(x-3\right)\hfill \\ \hfill x& =& -5\hfill \\ \hfill x& =& 3\hfill \end{array}$

The proposed solutions are $-5\text{\hspace{0.17em}}$ and $3.\text{\hspace{0.17em}}$ Let us check each solution back in the original equation. First, check $\text{\hspace{0.17em}}x=-5.$

$\begin{array}{ccc}\hfill \sqrt{15-2x}& =& x\hfill \\ \hfill \sqrt{15-2\left(-5\right)}& =& -5\hfill \\ \hfill \sqrt{25}& =& -5\hfill \\ \hfill 5& \ne & -5\hfill \end{array}$

This is an extraneous solution. While no mistake was made solving the equation, we found a solution that does not satisfy the original equation.

Check $\text{\hspace{0.17em}}x=3.$

$\begin{array}{ccc}\hfill \sqrt{15-2x}& =& x\hfill \\ \hfill \sqrt{15-2\left(3\right)}& =& 3\hfill \\ \hfill \sqrt{9}& =& 3\hfill \\ \hfill 3& =& 3\hfill \end{array}$

The solution is $\text{\hspace{0.17em}}3.$

Solve the radical equation: $\text{\hspace{0.17em}}\sqrt{x+3}=3x-1$

$x=1;$ extraneous solution $\text{\hspace{0.17em}}x=-\frac{2}{9}$

Solve $\text{\hspace{0.17em}}\sqrt{2x+3}+\sqrt{x-2}=4.$

As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.

Use the perfect square formula to expand the right side: $\text{\hspace{0.17em}}{\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}.$

Now that both radicals have been eliminated, set the quadratic equal to zero and solve.

The proposed solutions are $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}83.\text{\hspace{0.17em}}$ Check each solution in the original equation.

$\begin{array}{ccc}\hfill \sqrt{2x+3}+\sqrt{x-2}& =& 4\hfill \\ \hfill \sqrt{2x+3}& =& 4-\sqrt{x-2}\hfill \\ \hfill \sqrt{2\left(3\right)+3}& =& 4-\sqrt{\left(3\right)-2}\hfill \\ \hfill \sqrt{9}& =& 4-\sqrt{1}\hfill \\ \hfill 3& =& 3\hfill \end{array}$

One solution is $\text{\hspace{0.17em}}3.$

Check $\text{\hspace{0.17em}}x=83.$

$\begin{array}{ccc}\hfill \sqrt{2x+3}+\sqrt{x-2}& =& 4\hfill \\ \hfill \sqrt{2x+3}& =& 4-\sqrt{x-2}\hfill \\ \hfill \sqrt{2\left(83\right)+3}& =& 4-\sqrt{\left(83-2\right)}\hfill \\ \hfill \sqrt{169}& =& 4-\sqrt{81}\hfill \\ \hfill 13& \ne & -5\hfill \end{array}$

The only solution is $\text{\hspace{0.17em}}3.\text{\hspace{0.17em}}$ We see that $\text{\hspace{0.17em}}x=83\text{\hspace{0.17em}}$ is an extraneous solution.

Solve the equation with two radicals: $\text{\hspace{0.17em}}\sqrt{3x+7}+\sqrt{x+2}=1.$

$x=-2;$ extraneous solution $\text{\hspace{0.17em}}x=-1$

## Solving an absolute value equation

Next, we will learn how to solve an absolute value equation    . To solve an equation such as $\text{\hspace{0.17em}}|2x-6|=8,$ we notice that the absolute value will be equal to 8 if the quantity inside the absolute value bars is $\text{\hspace{0.17em}}8\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}-8.\text{\hspace{0.17em}}$ This leads to two different equations we can solve independently.

$\begin{array}{ccccccc}\hfill 2x-6& =& 8\hfill & \phantom{\rule{2em}{0ex}}\text{or}\phantom{\rule{2em}{0ex}}& \hfill 2x-6& =& -8\hfill \\ \hfill 2x& =& 14& & \hfill 2x& =& -2\hfill \\ \hfill x& =& 7\hfill & & \hfill x& =& -1\hfill \end{array}$

Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.

## Absolute value equations

The absolute value of x is written as $\text{\hspace{0.17em}}|x|.\text{\hspace{0.17em}}$ It has the following properties:

For real numbers $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}B,$ an equation of the form $\text{\hspace{0.17em}}|A|=B,$ with $\text{\hspace{0.17em}}B\ge 0,$ will have solutions when $\text{\hspace{0.17em}}A=B\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}A=-B.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}B<0,$ the equation $\text{\hspace{0.17em}}|A|=B\text{\hspace{0.17em}}$ has no solution.

An absolute value equation    in the form $\text{\hspace{0.17em}}|ax+b|=c\text{\hspace{0.17em}}$ has the following properties:

Given an absolute value equation, solve it.

1. Isolate the absolute value expression on one side of the equal sign.
2. If $\text{\hspace{0.17em}}c>0,$ write and solve two equations: $\text{\hspace{0.17em}}ax+b=c\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}ax+b=-c.$

## Solving absolute value equations

Solve the following absolute value equations:

• (a) $|6x+4|=8$
• (b) $|3x+4|=-9$
• (c) $|3x-5|-4=6$
• (d) $|-5x+10|=0$
• (a) $|6x+4|=8$

Write two equations and solve each:

$\begin{array}{ccccccc}\hfill 6x+4& =& 8\hfill & \phantom{\rule{5em}{0ex}}& \hfill 6x+4& =& -8\hfill \\ \hfill 6x& =& 4\hfill & \phantom{\rule{5em}{0ex}}& \hfill 6x& =& -12\hfill \\ \hfill x& =& \frac{2}{3}\hfill & \phantom{\rule{5em}{0ex}}& \hfill x& =& -2\hfill \end{array}$

The two solutions are $\text{\hspace{0.17em}}\frac{2}{3}$ and $-2.$

• (b) $|3x+4|=-9$

There is no solution as an absolute value cannot be negative.

• (c) $|3x-5|-4=6$

Isolate the absolute value expression and then write two equations.

$\begin{array}{ccccccccc}& & & \hfill |3x-5|-4& =& 6\hfill & & & \\ & & & \hfill |3x-5|& =& 10\hfill & & & \\ \hfill 3x-5& =& 10\hfill & & & & \hfill 3x-5& =& -10\hfill \\ \hfill 3x& =& 15\hfill & & & & \hfill 3x& =& -5\hfill \\ \hfill x& =& 5\hfill & & & & \hfill x& =& -\frac{5}{3}\hfill \end{array}$

There are two solutions: $\text{\hspace{0.17em}}5,$ and $-\frac{5}{3}.$

• (d) $|-5x+10|=0$

The equation is set equal to zero, so we have to write only one equation.

$\begin{array}{ccc}\hfill -5x+10& =& 0\hfill \\ \hfill -5x& =& -10\hfill \\ \hfill x& =& 2\hfill \end{array}$

There is one solution: $\text{\hspace{0.17em}}2.$

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