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

 Page 3 / 10

## 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: $\text{\hspace{0.17em}}2.$

the third and the seventh terms of a G.P are 81 and 16, find the first and fifth terms.
if a=3, b =4 and c=5 find the six trigonometric value sin
pls how do I factorize x⁴+x³-7x²-x+6=0
in a function the input value is called
how do I test for values on the number line
if a=4 b=4 then a+b=
a+b+2ab
Kin
commulative principle
a+b= 4+4=8
Mimi
If a=4 and b=4 then we add the value of a and b i.e a+b=4+4=8.
Tariq
what are examples of natural number
an equation for the line that goes through the point (-1,12) and has a slope of 2,3
3y=-9x+25
Ishaq
show that the set of natural numberdoes not from agroup with addition or multiplication butit forms aseni group with respect toaaddition as well as multiplication
x^20+x^15+x^10+x^5/x^2+1
evaluate each algebraic expression. 2x+×_2 if ×=5
if the ratio of the root of ax+bx+c =0, show that (m+1)^2 ac =b^2m
By the definition, is such that 0!=1.why?
(1+cosA+IsinA)(1+cosB+isinB)/(cos@+isin@)(cos$+isin$)
hatdog
Mark
jaks
Ryan
how we can draw three triangles of distinctly different shapes. All the angles will be cutt off each triangle and placed side by side with vertices touching