9.6 Solve equations with square roots (Page 8/6)

Elementary algebra Page 1 / 6

By the end of this section, you will be able to:

Solve radical equations
Use square roots in applications

Before you get started, take this readiness quiz.

Simplify: ⓐ $\sqrt{9}$ ⓑ $9^{2}$ .
If you missed this problem, review [link] and [link] .
Solve: $5 (x + 1) - 4 = 3 (2 x - 7)$ .
If you missed this problem, review [link] .
Solve: $n^{2} - 6 n + 8 = 0$ .
If you missed this problem, review [link] .

Solve radical equations

In this section we will solve equations that have the variable in the radicand of a square root. Equations of this type are called radical equations.

Radical equation

An equation in which the variable is in the radicand of a square root is called a radical equation .

As usual, in solving these equations, what we do to one side of an equation we must do to the other side as well. Since squaring a quantity and taking a square root are ‘opposite’ operations, we will square both sides in order to remove the radical sign and solve for the variable inside.

But remember that when we write $\sqrt{a}$ we mean the principal square root. So $\sqrt{a} \geq 0$ always. When we solve radical equations by squaring both sides we may get an algebraic solution that would make $\sqrt{a}$ negative. This algebraic solution would not be a solution to the original radical equation ; it is an extraneous solution. We saw extraneous solutions when we solved rational equations, too.

For the equation $\sqrt{x + 2} = x$ :

ⓐ Is $x = 2$ a solution? ⓑ Is $x = −1$ a solution?

Solution

ⓐ Is $x = 2$ a solution?


Let x = 2.
Simplify.

	2 is a solution.

ⓑ Is

x = −1

a solution?


Let x = −1.
Simplify.

	−1 is not a solution.
	−1 is an extraneous solution to the equation.

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For the equation $\sqrt{x + 6} = x$ :

ⓐ Is $x = −2$ a solution? ⓑ Is $x = 3$ a solution?

ⓐ no ⓑ $yes$

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For the equation $\sqrt{− x + 2} = x$ :

ⓐ Is $x = −2$ a solution? ⓑ Is $x = 1$ a solution?

ⓐ no ⓑ $yes$

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Now we will see how to solve a radical equation. Our strategy is based on the relation between taking a square root and squaring.

For a \geq 0, {(\sqrt{a})}^{2} = a

How to solve radical equations

Solve: $\sqrt{2 x - 1} = 7$ .

Solution

This table has three columns and four rows. The first row says, “Step 1. Isolate the radical on one side of equation. The square root of (2x minus 1) is already isolated on the left side.” It then shows the equation: the square root of (2x minus 1) equals 7.

The second row says, “Step 2. Square both sides of the equation. Remember, the square root of a squared equals a.” It then shows the equation: the square root of (2x minus 1) squared equals 7 squared.

The third row then says, “Step 3. Solve the new equation.” It indicates that 2x minus 1 equals 49 or 2x equals 50 which means that x equals 25.

The fourth row says, “Step 4. Check the answer. Check:” It then indicates the square root of (2x minus 1) equals 7. This becomes the square root of (2 times 25 minus 1) equals 7. This becomes the square root of (50 minus 1) equals 7. This becomes the square root of 49 equals 7, and thus 7 equals 7. The figure then states, “The solutions is x equals 25.”

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Solve: $\sqrt{3 x - 5} = 5$ .

$10$

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Solve: $\sqrt{4 x + 8} = 6$ .

$7$

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Solve a radical equation.

Isolate the radical on one side of the equation.
Square both sides of the equation.
Solve the new equation.
Check the answer.

Solve: $\sqrt{5 n - 4} - 9 = 0$ .

Solution


To isolate the radical, add 9 to both sides.
Simplify.
Square both sides of the equation.
Solve the new equation.


Check the answer.

	The solution is n = 17.

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Solve: $\sqrt{3 m + 2} - 5 = 0$ .

$\frac{23}{3}$

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Solve: $\sqrt{10 z + 1} - 2 = 0$ .

$\frac{3}{10}$

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Solve: $\sqrt{3 y + 5} + 2 = 5$ .

Solution


To isolate the radical, subtract 2 from both sides.
Simplify.
Square both sides of the equation.
Solve the new equation.


Check the answer.

	The solution is $y = \frac{4}{3}$ .

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Solve: $\sqrt{3 p + 3} + 3 = 5$ .

$\frac{1}{2}$

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Solve: $\sqrt{5 q + 1} + 4 = 6$ .

$\frac{3}{5}$

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When we use a radical sign, we mean the principal or positive root. If an equation has a square root equal to a negative number, that equation will have no solution.

Solve: $\sqrt{9 k - 2} + 1 = 0$ .

Solution


To isolate the radical, subtract 1 from both sides.
Simplify.
Since the square root is equal to a negative number, the equation has no solution.

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Solve: $\sqrt{2 r - 3} + 5 = 0$ .

$no solution$

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Solve: $\sqrt{7 s - 3} + 2 = 0$ .

$no solution$

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If one side of the equation is a binomial, we use the binomial squares formula when we square it.

Binomial squares

\begin{matrix} {(a + b)}^{2} = a^{2} + 2 a b + b^{2} & {(a - b)}^{2} = a^{2} - 2 a b + b^{2} \end{matrix}

Don’t forget the middle term!

Solve: $\sqrt{p - 1} + 1 = p$ .

Solution


To isolate the radical, subtract 1 from both sides.
Simplify.
Square both sides of the equation.
Simplify, then solve the new equation.
It is a quadratic equation, so get zero on one side.
Factor the right side.
Use the zero product property.
Solve each equation.
Check the answers.

	The solutions are p = 1, p = 2.

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Source: OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2

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