Int-Alg Radical Equations

Section 6.5 Radical Equations

Subsection 6.5.1 Solving a Radical Equation

A radical equation is one in which the variable appears under a square root or other radical. We solve simple radical equations by raising both sides to the appropriate power. For example, to solve the equation

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\sqrt{x + 3} = 4

we square both sides to find

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\begin{aligned} (\sqrt{x + 3})^{2} & = 4^{2} \\ x + 3 & = 16 \end{aligned}

You can check that

x = 13

is the solution for this equation.

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Example 6.5.1.

Solve

4 \sqrt[3]{x - 9} = 12

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Solution.

We first divide both sides of the equation by 4 to isolate the radical.

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\sqrt[3]{x - 9} = 3

Next, we cube both sides of the equation.

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\begin{aligned} (\sqrt[3]{x - 9})^{3} & = 3^{3} \\ x - 9 & = 27 \\ x & = 36 \end{aligned}

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The solution is 36. We can also solve the equation graphically by graphing

y = 4 \sqrt[3]{x - 9},

as shown in the figure. The point

(36, 12)

lies on the graph, so

x = 36

is the solution of the equation

4 \sqrt[3]{x - 9} = 12

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$transformed cube root function$

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Checkpoint 6.5.2. Practice 1.

Solve

6 + 2 \sqrt[4]{12 - v} = 10

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Hint:

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Isolate the radical.
Raise each side to the fourth power.
Complete the solution.

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Solution.

v = - 4

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Subsection 6.5.2 Extraneous Solutions

Whenever we raise both sides of an equation to an even power, it is possible to introduce false or extraneous solutions. For example, the equation

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\sqrt{x} = - 5

has no solution, because

\sqrt{x}

is never a negative number. However, if we try to solve the equation by squaring both sides, we find

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\begin{aligned} (\sqrt{x})^{2} & = (- 5)^{2} \\ x & = 25 \end{aligned}

You can check that

25

is not a solution to the original equation,

\sqrt{x} = - 5,

because

\sqrt{25}

does not equal

- 5 .

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Raising both sides of an equation to an odd power does not introduce extraneous solutions. However, if we raise both sides to an even power, we should check each solution in the original equation.

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Checkpoint 6.5.3. QuickCheck 1.

When we raise both sides of an equation to an power, it is possible to introduce solutions.

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Example 6.5.4.

Solve the equation

\sqrt{x + 2} + 4 = x

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Solution.

First, we isolate the radical expression on one side of the equation. (This will make it easier to square both sides.)

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\begin{aligned} \sqrt{x + 2} & = x - 4 & Square both sides of the equation. \\ {(\sqrt{x + 2})}^{2} & = (x - 4)^{2} \\ x + 2 & = x^{2} - 8 x + 16 & Subtract x + 2 from both sides. \\ x^{2} - 9 x + 14 & = 0 & Factor the left side. \\ x = 2 & or x = 7 \end{aligned}

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Check

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Does

\sqrt{2 + 2} + 4 = 2 ?

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No; 2 is not a solution.

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Does

\sqrt{7 + 2} + 4 = 7 ?

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Yes; 7 is  a solution.

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The apparent solution

2

is extraneous. The only solution to the original equation is

7 .

We can verify the solution by graphing the equations

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y_{1} = \sqrt{x + 2} and y_{2} = x - 4

as shown at right. The graphs intersect in only one point,

(7, 3),

so there is only one solution,

x = 7 .

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graphs of both sides of radical equation

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Caution 6.5.5.

When we square both sides of an equation, it is not correct to square each term of the equation separately. Thus, in Example 6.5.4, the original equation is not equivalent to

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(\sqrt{x + 2})^{2} + 4^{2} = x^{2} Incorrect!

This is because

(a + b)^{2} \neq a^{2} + b^{2} .

Instead, we must square the entire left side of the equation as a binomial, like this,

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(\sqrt{x + 2} + 4)^{2} = x^{2}

or we may proceed as shown in Example 6.5.4.

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Checkpoint 6.5.6. Practice 2.

Solve

2 x - 5 = \sqrt{40 - 3 x}

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Hint:

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Square both sides.
Solve the quadratic equation.
Check for extraneous roots.

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Solution.

x = 5

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Checkpoint 6.5.7. QuickCheck 2.

True or false.

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At most one of the solutions of a radical equation can be extraneous.
We don’t have to check for extraneous solutions if we cube both sides of an equation.
We can square each term of an equation separately without changing its solutions.
It is helpful to isolate the radical before squaring both sides.

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Subsection 6.5.3 Solving Formulas

We can also solve formulas involving radicals for one variable in terms of the others.

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Example 6.5.8.

Solve the formula

t = \sqrt{1 + s^{2}}

for

s .

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Solution.

Because the variable we want is under a square root, we square both sides of the equation, to get

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\begin{aligned} t^{2} & = 1 + s^{2} & Subtract 1 from both sides. \\ t^{2} - 1 & = s^{2} & Take square roots. \\ s & = \pm \sqrt{t^{2} - 1} \end{aligned}

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Checkpoint 6.5.9. Practice 3.

Solve the formula

r - 2 = \sqrt[3]{V - B h}

for

h .

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Solution.

h = \frac{V - (r - 2)^{3}}{B}

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Subsection 6.5.4 Equations with More than One Radical

Sometimes we need to square both sides of an equation more than once in order to eliminate all the radicals.

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Example 6.5.10.

Solve

\sqrt{x - 7} + \sqrt{x} = 7 .

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Solution.

First, we isolate the more complicated radical on one side of the equation. (This will make it easier to square both sides.) We subtract

\sqrt{x}

from both sides to get

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\sqrt{x - 7} = 7 - \sqrt{x}

Now we square each side to remove one radical. Be careful when squaring the binomial

7 - \sqrt{x} .

[TK]

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\begin{aligned} (\sqrt{x - 7})^{2} & = (7 - \sqrt{x})^{2} \\ x - 7 & = 49 - 14 \sqrt{x} + x \end{aligned}

We collect like terms, and isolate the radical on one side of the equation.

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\begin{aligned} - 56 & = - 14 \sqrt{x} & Divide both sides by - 14 . \\ 4 & = \sqrt{x} \end{aligned}

Finally, we square both sides again to obtain

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\begin{aligned} (4)^{2} & = (\sqrt{x})^{2} \\ 16 & = x \end{aligned}

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Check

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Does

\sqrt{16 - 7} + \sqrt{16} = 7 ?

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Yes. The solution is 16.

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Caution 6.5.11.

Recall that we cannot solve a radical equation by squaring each term separately. So it is incorrect to begin Example 10 by writing

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(\sqrt{x - 7})^{2} + (\sqrt{x})^{2} = 7^{2} Incorrect!

We must square the entire expression on each side of the equal sign as one piece.

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Checkpoint 6.5.12. Practice 4.

Solve

\sqrt{3 x + 1} = 6 - \sqrt{9 - x}

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Hint:

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Square both sides.
Isolate the radical term.
Divide both sides by 4.
Square both sides again.
Solve the quadratic equation.
Check for extraneous roots.

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Solution.

x = 5;

x = 8

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Subsection 6.5.5 Simplifying $\sqrt[n]{a^{n}}$

We have seen that raising to a power is the inverse operation for extracting roots, so that

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(\sqrt[n]{a})^{n} = a

For example,

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(\sqrt[4]{16})^{4} = 16 and (\sqrt[5]{x})^{5} = x

What if the power and root operations occur in the opposite order? Is it always true that

\sqrt[n]{a^{n}} = a ?

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First, consider the case where the index is an odd number. For example,

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\sqrt[3]{2^{3}} = \sqrt[3]{8} = 2 and \sqrt[3]{(- 2)^{3}} = \sqrt[3]{- 8} = - 2

Because every real number has exactly one

n

th root if

n

is odd, we see that,

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\sqrt[n]{a^{n}} = a, for n odd

However, if

n

is even, then

a^{n}

is positive, regardless of whether

a

itself is positive or negative, and hence

\sqrt[n]{a^{n}}

is positive also. For example, if

a = - 3,

then

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\sqrt{(- 3)^{2}} = \sqrt{9} = 3

In this case,

\sqrt{a^{2}}

does not equal

a,

because

a

is negative but

\sqrt{a^{2}}

is positive. We must be careful when taking even roots of powers. We have the following special relationship for even roots.

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\sqrt[n]{a^{n}} = | a |, for n even

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We summarize our results in the box below.

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Roots of Powers.

\begin{aligned} \sqrt[n]{a^{n}} & = a & If n is odd. \\ \sqrt[n]{a^{n}} & = | a | & If n is even. \end{aligned}

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In particular, note that it is not always true that

\sqrt{a^{2}} = a,

unless we know that

a \geq 0 .

Otherwise, we can only assume that

\sqrt{a^{2}} = | a | .

[TK]

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Example 6.5.13.

$\sqrt{16 x^{2}} = 4 | x |$
$\sqrt{(x - 1)^{2}} = | x - 1 |$

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Checkpoint 6.5.14. Practice 5.

Simplify.

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$\sqrt[3]{- 125 x^{6}}$
$\sqrt[4]{16 x^{12}}$

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Solution.

$- 5 x^{2}$
$2 | x^{3} |$

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Checkpoint 6.5.15. QuickCheck 3.

True or false.

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If an equation contains more than one radical, we square each term of the equation.
$(7 - \sqrt{x})^{2} = 49 - x$
If $\sqrt{x^{2}} \neq x,$ then $\sqrt{x^{2}} = - x .$
If $n$ is even and $a \neq 0,$ then $a^{n}$ is always positive.

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Exercises 6.5.6 Problem Set 6.5

Warm Up

1.

Solve the equation. Isolate the radical first, then square both sides of the equation.

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$- 3 \sqrt{z} + 14 = 8$
$8 - 3 \sqrt{9 + 2 w} = - 7$

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2.

Square the binomial.

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$(3 x + 4)^{2}$
$(2 \sqrt{x} - 3)^{2}$
$(4 - \sqrt{x + 3})^{2}$
$(5 - 2 \sqrt{2 x - 5})^{2}$

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3.

Complete the table of values and graph $y = \sqrt{x - 4} .$

$x$ $y$

$4$

$5$

$6$

$10$

$16$

$19$

$24$

$grid$
Solve $\sqrt{x - 4} = 3$ graphically and algebraically. Do your answers agree?

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4.

Complete the table of values and graph $y = 4 - \sqrt{x + 3} .$

$x$ $y$

$- 3$

$- 2$

$0$

$1$

$4$

$8$

$16$

$grid$
Solve $4 - \sqrt{x + 3} = 1$ graphically and algebraically. Do your answers agree?

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5.

Complete the table of values and graph $y = 4 - \sqrt[3]{x} .$

$x$ $y$

$- 25$

$- 20$

$- 15$

$- 10$

$- 5$

$0$

$5$

$10$

$15$

$20$

$grid$
Solve $4 - \sqrt[3]{x} = 6$ graphically and algebraically. Do your answers agree?

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6.

Complete the table of values and graph $y = 3 + \sqrt[3]{x - 3} .$

$x$ $y$

$- 8$

$- 6$

$- 4$

$- 2$

$0$

$2$

$4$

$6$

$8$

$grid$
Solve $3 + \sqrt[3]{x - 3} = 1$ graphically and algebraically. Do your answers agree?