2.6 Other types of equations (Page 3/10)

College algebra

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Solving an equation with one radical

Solve $\sqrt{15 - 2 x} = x .$

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

\begin{matrix} \sqrt{15 - 2 x} & = & x \\ {(\sqrt{15 - 2 x})}^{2} & = & {(x)}^{2} \\ 15 - 2 x & = & x^{2} \end{matrix}

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

\begin{matrix} 0 & = & x^{2} + 2 x - 15 \\ = & (x + 5) (x - 3) \\ x & = & −5 \\ x & = & 3 \end{matrix}

The proposed solutions are $x = −5$ and $x = 3.$ Let us check each solution back in the original equation. First, check $x = −5.$

\begin{matrix} \sqrt{15 - 2 x} & = & x \\ \sqrt{15 - 2 (- 5)} & = & −5 \\ \sqrt{25} & = & −5 \\ 5 & \neq & −5 \end{matrix}

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 $x = 3.$

\begin{matrix} \sqrt{15 - 2 x} & = & x \\ \sqrt{15 - 2 (3)} & = & 3 \\ \sqrt{9} & = & 3 \\ 3 & = & 3 \end{matrix}

The solution is $x = 3.$

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

$x = 1;$ extraneous solution $x = - \frac{2}{9}$

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Solving a radical equation containing two radicals

Solve $\sqrt{2 x + 3} + \sqrt{x - 2} = 4.$

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

\begin{matrix} \sqrt{2 x + 3} + \sqrt{x - 2} & = & 4 \\ \sqrt{2 x + 3} & = & 4 - \sqrt{x - 2} & Subtract \sqrt{x - 2} from both sides . \\ {(\sqrt{2 x + 3})}^{2} & = & {(4 - \sqrt{x - 2})}^{2} & Square both sides . \end{matrix}

Use the perfect square formula to expand the right side: ${(a - b)}^{2} = a^{2} −2 a b + b^{2} .$

\begin{matrix} 2 x + 3 & = & {(4)}^{2} - 2 (4) \sqrt{x - 2} + {(\sqrt{x - 2})}^{2} \\ 2 x + 3 & = & 16 - 8 \sqrt{x - 2} + (x - 2) \\ 2 x + 3 & = & 14 + x - 8 \sqrt{x - 2} & Combine like terms . \\ x - 11 & = & −8 \sqrt{x - 2} & Isolate the second radical . \\ {(x - 11)}^{2} & = & {(−8 \sqrt{x - 2})}^{2} & Square both sides . \\ x^{2} - 22 x + 121 & = & 64 (x - 2) \end{matrix}

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

\begin{matrix} x^{2} - 22 x + 121 & = & 64 x - 128 \\ x^{2} - 86 x + 249 & = & 0 \\ (x - 3) (x - 83) & = & 0 & Factor and solve . \\ x & = & 3 \\ x & = & 83 \end{matrix}

The proposed solutions are $x = 3$ and $x = 83.$ Check each solution in the original equation.

\begin{matrix} \sqrt{2 x + 3} + \sqrt{x - 2} & = & 4 \\ \sqrt{2 x + 3} & = & 4 - \sqrt{x - 2} \\ \sqrt{2 (3) + 3} & = & 4 - \sqrt{(3) - 2} \\ \sqrt{9} & = & 4 - \sqrt{1} \\ 3 & = & 3 \end{matrix}

One solution is $x = 3.$

Check $x = 83.$

\begin{matrix} \sqrt{2 x + 3} + \sqrt{x - 2} & = & 4 \\ \sqrt{2 x + 3} & = & 4 - \sqrt{x - 2} \\ \sqrt{2 (83) + 3} & = & 4 - \sqrt{(83 - 2)} \\ \sqrt{169} & = & 4 - \sqrt{81} \\ 13 & \neq & - 5 \end{matrix}

The only solution is $x = 3.$ We see that $x = 83$ is an extraneous solution.

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

$x = −2;$ extraneous solution $x = −1$

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Solving an absolute value equation

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

\begin{matrix} 2 x - 6 & = & 8 & or & 2 x - 6 & = & −8 \\ 2 x & = & 14 & 2 x & = & −2 \\ x & = & 7 & x & = & −1 \end{matrix}

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.

A General Note

Absolute value equations

The absolute value of x is written as $| x | .$ It has the following properties:

\begin{array}{l} If x \geq 0, then | x | = x . \\ If x < 0, then | x | = - x . \end{array}

For real numbers $A$ and $B,$ an equation of the form $| A | = B,$ with $B \geq 0,$ will have solutions when $A = B$ or $A = - B .$ If $B < 0,$ the equation $| A | = B$ has no solution.

An absolute value equation in the form $| a x + b | = c$ has the following properties:

\begin{array}{l} If c < 0, | a x + b | = c has no solution . \\ If c = 0, | a x + b | = c has one solution . \\ If c > 0, | a x + b | = c has two solutions . \end{array}

How To

Given an absolute value equation, solve it.

Isolate the absolute value expression on one side of the equal sign.
If $c > 0,$ write and solve two equations: $a x + b = c$ and $a x + b = - c .$

Solving absolute value equations

Solve the following absolute value equations:

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

(a) $| 6 x + 4 | = 8$

Write two equations and solve each:

$\begin{matrix} 6 x + 4 & = & 8 & 6 x + 4 & = & −8 \\ 6 x & = & 4 & 6 x & = & −12 \\ x & = & \frac{2}{3} & x & = & −2 \end{matrix}$

The two solutions are $x = \frac{2}{3},$ $x = −2.$
(b) $| 3 x + 4 | = −9$

There is no solution as an absolute value cannot be negative.
(c) $| 3 x - 5 | - 4 = 6$

Isolate the absolute value expression and then write two equations.

$\begin{matrix} | 3 x - 5 | - 4 & = & 6 \\ | 3 x - 5 | & = & 10 \\ 3 x - 5 & = & 10 & 3 x - 5 & = & −10 \\ 3 x & = & 15 & 3 x & = & −5 \\ x & = & 5 & x & = & - \frac{5}{3} \end{matrix}$

There are two solutions: $x = 5,$ $x = - \frac{5}{3} .$
(d) $| −5 x + 10 | = 0$

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

$\begin{matrix} −5 x + 10 & = & 0 \\ −5 x & = & −10 \\ x & = & 2 \end{matrix}$

There is one solution: $x = 2.$

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Source: OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3

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