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14.4 Equations with Radicals

Solving by Squaring Both Sides • Isolating a Radical • Solving a Nested Radical Equation

Equations containing radicals are normally solved by squaring both sides of the equation if the radical represents a square root or by a similar operation for the other roots. However, when we do this, we often introduce extraneous roots.

CAUTION

Thus, it is very important that all solutions be checked in the original equation.

EXAMPLE 1 Solve by squaring both sides

Solve the equation $\sqrt{x - 4} = 2.$ $\sqrt{x - 4} = 2.$

By squaring both sides of the equation, we have

\begin{array}{rcl} {(\sqrt{x - 4})}^{2} & = & 2^{2} \\ x - 4 & = & 4 \\ x & = & 8 \end{array}

$\begin{array}{rcl} {(\sqrt{x - 4})}^{2} & = & 2^{2} \\ x - 4 & = & 4 \\ x & = & 8 \end{array}$

This solution checks when put into the original equation.

EXAMPLE 2 Solve by squaring both sides

Solve the equation $2 \sqrt{3 x - 1} = 3 x .$ $2 \sqrt{3 x - 1} = 3 x .$

Squaring both sides of the equation gives us

The process of solving the equation, 2 square root of start expression 3 x minus 1 end expression = 3 x.

.4-2 Full Alternative Text

Checking this solution in the original equation, we have

\begin{array}{l} 2 \sqrt{3 (\frac{2}{3}) - 1} \overset{?}{=} 3 (\frac{2}{3}), & 2 \sqrt{2 - 1} \overset{?}{=} 2, & 2 = 2 \end{array}

$\begin{array}{l} 2 \sqrt{3 (\frac{2}{3}) - 1} \overset{?}{=} 3 (\frac{2}{3}), & 2 \sqrt{2 - 1} \overset{?}{=} 2, & 2 = 2 \end{array}$

Therefore, the solution $x = \frac{2}{3}$ $x = \frac{2}{3}$ checks.

We can check this solution graphically by letting $y_{1} = 2 \sqrt{3 x - 1}$ $y_{1} = 2 \sqrt{3 x - 1}$ and $y_{2} = 3 x .$ $y_{2} = 3 x .$ The calculator display is shown in Fig. 14.18. The intersection feature shows that the only x-value that the curves have in common is $x = 0.6667,$ $x = 0.6667,$ which agrees with the solution of $x = 2/3.$ $x = 2/3.$ This also means the line $y_{1}$ $y_{1}$ is tangent to the curve of $y_{2} .$ $y_{2} .$

A curve begins near (one-third, one-third) and rises with decreasing steepness. The line y sub 2 = 3 x rises through (0, 0), intersecting the line at (0.667, 2). All data are approximate. — Fig. 14.18
Graphing calculator keystrokes: bit.ly/2OiC2lw

EXAMPLE 3 Solve by cubing both sides

Solve the equation $\sqrt[3]{x - 8} = 2.$ $\sqrt[3]{x - 8} = 2.$

Cubing both sides of the equation, we have

\begin{array}{rcl} x - 8 & = & 8 \\ x & = & 16 \end{array}

$\begin{array}{rcl} x - 8 & = & 8 \\ x & = & 16 \end{array}$

Checking this solution in the original equation, we get

\begin{array}{l} \sqrt[3]{16 - 8} \overset{?}{=} 2, & 2 = 2 \end{array}

$\begin{array}{l} \sqrt[3]{16 - 8} \overset{?}{=} 2, & 2 = 2 \end{array}$

Therefore, the solution checks.

If a radical and other terms are on one side of the equation, we first isolate the radical. That is, we rewrite the equation with the radical on one side and all other terms on the other side.

EXAMPLE 4 Solve by isolating the radical

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

We first isolate the radical by subtracting 3 from each side. This gives us

\sqrt{x - 1} = x - 3

$\sqrt{x - 1} = x - 3$

We now square both sides and proceed with the solution:

The process of solving the equation, square root of start expression x minus 1 end expression = x minus 3 has 5 steps.

.4-2 Full Alternative Text

The solution $x = 5$ $x = 5$ checks, but the solution $x = 2$ $x = 2$ gives $4 = 2.$ $4 = 2.$ Thus, the only solution is $x = 5.$ $x = 5.$ The value $x = 2$ $x = 2$ is an extraneous root. The graph of $y = \sqrt{x - 1} + 3 - x,$ $y = \sqrt{x - 1} + 3 - x,$ shown in Fig. 14.19, verifies that $x = 5$ $x = 5$ is a solution but $x = 2$ $x = 2$ is not.

The curve y sub 1 = square root of, x minus 1, plus 3 minus x, falls through the zero (5, 0). — Fig. 14.19
Graphing calculator keystrokes: bit.ly/2CeJWW7

EXAMPLE 5 Solve by isolating the radical

Solve the equation $\sqrt{x + 1} + \sqrt{x - 4} = 5.$ $\sqrt{x + 1} + \sqrt{x - 4} = 5.$

This is most easily solved by first isolating one of the radicals by placing the other radical on the right side of the equation. We then square both sides of the resulting equation.

.4-2 Full Alternative Text

Now, isolating the radical on one side of the equation and squaring again, we have

\begin{array}{rcl} 10 \sqrt{x - 4} & = & 20 \\ \sqrt{x - 4} & = & 2 & divide by 10 \\ x - 4 & = & 4 & square both sides \\ x & = & 8 \end{array}

$\begin{array}{rcl} 10 \sqrt{x - 4} & = & 20 \\ \sqrt{x - 4} & = & 2 & divide by 10 \\ x - 4 & = & 4 & square both sides \\ x & = & 8 \end{array}$

This solution checks.

We note again that in squaring $5 - \sqrt{x - 4},$ $5 - \sqrt{x - 4},$ we do not simply square 5 and $\sqrt{x - 4} .$ $\sqrt{x - 4} .$ We must square the entire expression.

EXAMPLE 6 Nested radical equation

Solve the equation $\sqrt{7 + \sqrt{x}} - 1 = \sqrt{x} .$ $\sqrt{7 + \sqrt{x}} - 1 = \sqrt{x} .$

We first isolate the outer radical and then square both sides. We then isolate the remaining radical and square both sides again. The steps are shown below.

\begin{array}{rcl} \sqrt{7 + \sqrt{x}} & = & \sqrt{x} + 1 & isolate the outer radical \\ {(\sqrt{7 + \sqrt{x}})}^{2} & = & {(\sqrt{x} + 1)}^{2} & square the expression on each side \\ 7 + \sqrt{x} & = & x + 2 \sqrt{x} + 1 \\ 6 - x & = & \sqrt{x} & isolate the remaining radical \\ {(6 - x)}^{2} & = & {(\sqrt{x})}^{2} & square the expression on each side \\ 36 - 12 x + x^{2} & = & x \\ x^{2} - 13 x + 36 & = & 0 \\ (x - 9) (x - 4) & = & 0 \\ x & = & 9 or x = 4 \end{array}

$\begin{array}{rcl} \sqrt{7 + \sqrt{x}} & = & \sqrt{x} + 1 & isolate the outer radical \\ {(\sqrt{7 + \sqrt{x}})}^{2} & = & {(\sqrt{x} + 1)}^{2} & square the expression on each side \\ 7 + \sqrt{x} & = & x + 2 \sqrt{x} + 1 \\ 6 - x & = & \sqrt{x} & isolate the remaining radical \\ {(6 - x)}^{2} & = & {(\sqrt{x})}^{2} & square the expression on each side \\ 36 - 12 x + x^{2} & = & x \\ x^{2} - 13 x + 36 & = & 0 \\ (x - 9) (x - 4) & = & 0 \\ x & = & 9 or x = 4 \end{array}$

When checking, $x = 4$ $x = 4$ satisfies the original equation but $x = 9$ $x = 9$ does not. Therefore, the only solution is $x = 4$ $x = 4$ ( $x = 9$ $x = 9$ is an extraneous root). The calculator graph in Fig. 14.20 verifies that $x = 4$ $x = 4$ is the only solution.

The curve y sub 1 = square root of 7, plus square root of, x, minus 1 minus square root of x, begins at approximately (0, 1.3) and falls through the zero (4, 0). — Fig. 14.20

EXAMPLE 7 System with a radical—holograph dimensions

Each cross section of a holographic image is in the shape of a right triangle. The perimeter of the cross section is 60 cm, and its area is ${120 cm}^{2} .$ ${120 cm}^{2} .$ Find the length of each of the three sides.

If we let the two legs of the triangle be x and y, as shown in Fig. 14.21, from the formulas for the perimeter p and the area A of a triangle, we have

\begin{array}{l} p = x + y + \sqrt{x^{2} + y^{2}} & and & A = \frac{1}{2} x y \end{array}

$\begin{array}{l} p = x + y + \sqrt{x^{2} + y^{2}} & and & A = \frac{1}{2} x y \end{array}$

The cross section of a right triangle with leg x, leg y, hypotenuse square root of, x squared plus y squared. The triangle has perimeter p = 60 centimeters, and area Ay = 120 centimeters squared. — Fig. 14.21

where the hypotenuse was found by use of the Pythagorean theorem. Using the information given in the statement of the problem, we arrive at the equations

\begin{array}{l} x + y + \sqrt{x^{2} + y^{2}} = 60 & and & x y = 240 \end{array}

$\begin{array}{l} x + y + \sqrt{x^{2} + y^{2}} = 60 & and & x y = 240 \end{array}$

Isolating the radical in the first equation and then squaring both sides, we have

\begin{array}{rcl} \sqrt{x^{2} + y^{2}} & = & 60 - x - y \\ x^{2} + y^{2} & = & 3600 - 120 x - 120 y + x^{2} + 2 x y + y^{2} \\ 0 & = & 3600 - 120 x - 120 y + 2 x y \end{array}

$\begin{array}{rcl} \sqrt{x^{2} + y^{2}} & = & 60 - x - y \\ x^{2} + y^{2} & = & 3600 - 120 x - 120 y + x^{2} + 2 x y + y^{2} \\ 0 & = & 3600 - 120 x - 120 y + 2 x y \end{array}$

Solving the second of the original equations for y, we have $y = 240 / x .$ $y = 240 / x .$ Substituting, we have

\begin{array}{rcl} 0 & = & 3600 - 120 x - 120 (\frac{240}{x}) + 2 x (\frac{240}{x}) \\ 0 & = & 3600 x - 120 x^{2} - 120 (240) + 480 x & multiply by x \\ 0 & = & 30 x - x^{2} - 240 + 4 x & divide by 120 \\ x^{2} - 34 x + 240 & = & 0 & collect terms on left \\ (x - 10) (x - 24) & = & 0 \\ x = 10 cm or x & = & 24 cm \end{array}

$\begin{array}{rcl} 0 & = & 3600 - 120 x - 120 (\frac{240}{x}) + 2 x (\frac{240}{x}) \\ 0 & = & 3600 x - 120 x^{2} - 120 (240) + 480 x & multiply by x \\ 0 & = & 30 x - x^{2} - 240 + 4 x & divide by 120 \\ x^{2} - 34 x + 240 & = & 0 & collect terms on left \\ (x - 10) (x - 24) & = & 0 \\ x = 10 cm or x & = & 24 cm \end{array}$

If $x = 10 cm,$ $x = 10 cm,$ then $y = 24 cm,$ $y = 24 cm,$ or if $x = 24 cm,$ $x = 24 cm,$ then $y = 10 cm .$ $y = 10 cm .$ Therefore, the legs of the holographic cross section are 10 cm and 24 cm, and the hypotenuse is 26 cm. For these sides, $p = 60 cm$ $p = 60 cm$ and $A = {120 cm}^{2} .$ $A = {120 cm}^{2} .$ We see that these values check with the statement of the problem.

EXERCISES 14.4

In Exercises 1–4, make the given changes in the indicated examples of this section, and then solve the resulting equations.

In Example 2, change the 3x on the right to 3.
In Example 3, change the 8 under the radical to 19.
In Example 4, change the 3 on the left to 7.
In Example 5, change the 4 under the second radical to 14.

In Exercises 5–34, solve the given equations. In Exercises 19 and 22, explain how the extraneous root is introduced.

$\sqrt{x - 8} = 2$ $\sqrt{x - 8} = 2$
$\sqrt{x + 4} = 3$ $\sqrt{x + 4} = 3$
$\sqrt{15 - 2 x} = x$ $\sqrt{15 - 2 x} = x$
$2 \sqrt{2 P + 5} = P$ $2 \sqrt{2 P + 5} = P$
$2 \sqrt{3 x + 2} = 6 x$ $2 \sqrt{3 x + 2} = 6 x$
$\sqrt{5 x - 1} + 3 = x$ $\sqrt{5 x - 1} + 3 = x$
$2 \sqrt{3 - x} - x = 5$ $2 \sqrt{3 - x} - x = 5$
$x - 3 \sqrt{2 x + 1} = - 5$ $x - 3 \sqrt{2 x + 1} = - 5$
$\sqrt[3]{y - 7} = 2$ $\sqrt[3]{y - 7} = 2$
$\sqrt[4]{5 - x} = 2$ $\sqrt[4]{5 - x} = 2$
$5 \sqrt{s} - 6 = s$ $5 \sqrt{s} - 6 = s$
$6 \sqrt{x} + 4 = 4 x$ $6 \sqrt{x} + 4 = 4 x$
$\sqrt{x^{2} - 11} = 5$ $\sqrt{x^{2} - 11} = 5$
$t^{2} = 3 - \sqrt{2 t^{2} - 3}$ $t^{2} = 3 - \sqrt{2 t^{2} - 3}$
$\sqrt{x + 4} + 8 = x$ $\sqrt{x + 4} + 8 = x$
$\sqrt{x^{2} - x - 4} = x + 2$ $\sqrt{x^{2} - x - 4} = x + 2$
$\sqrt{5 + \sqrt{x}} = \sqrt{x} - 1$ $\sqrt{5 + \sqrt{x}} = \sqrt{x} - 1$
$\sqrt{13 + \sqrt{x}} = \sqrt{x} + 1$ $\sqrt{13 + \sqrt{x}} = \sqrt{x} + 1$
$x - 3 x \sqrt{x} = 0$ $x - 3 x \sqrt{x} = 0$
$\sqrt{3 - 2 \sqrt{x}} = \sqrt{x}$ $\sqrt{3 - 2 \sqrt{x}} = \sqrt{x}$
$2 \sqrt{x + 2} - \sqrt{3 x + 4} = 1$ $2 \sqrt{x + 2} - \sqrt{3 x + 4} = 1$
$\sqrt{x - 1} + \sqrt{x + 2} = 3$ $\sqrt{x - 1} + \sqrt{x + 2} = 3$
$\sqrt{5 x + 1} - 1 = 3 \sqrt{x}$ $\sqrt{5 x + 1} - 1 = 3 \sqrt{x}$
$\sqrt{x - 7} = \sqrt{x} - 7$ $\sqrt{x - 7} = \sqrt{x} - 7$
$\sqrt{6 x - 5} - \sqrt{x + 4} = 2$ $\sqrt{6 x - 5} - \sqrt{x + 4} = 2$
$\sqrt{5 x - 4} - \sqrt{x} = 2$ $\sqrt{5 x - 4} - \sqrt{x} = 2$
$\sqrt{x - 9} = \frac{36}{\sqrt{x - 9}} - \sqrt{x}$ $\sqrt{x - 9} = \frac{36}{\sqrt{x - 9}} - \sqrt{x}$
$\sqrt[4]{x + 10} = \sqrt{x - 2}$ $\sqrt[4]{x + 10} = \sqrt{x - 2}$
$\sqrt{x - \sqrt{2 x}} = 2$ $\sqrt{x - \sqrt{2 x}} = 2$
$\sqrt{3 x + \sqrt{3 x + 4}} = 4$ $\sqrt{3 x + \sqrt{3 x + 4}} = 4$

In Exercises 35–38, solve the given equations algebraically and check the solutions with a graphing calculator.

$\sqrt{3 x + 4} = x$ $\sqrt{3 x + 4} = x$
$\sqrt{x - 2} + 3 = x$ $\sqrt{x - 2} + 3 = x$
$\sqrt{2 x + 1} + 3 \sqrt{x} = 9$ $\sqrt{2 x + 1} + 3 \sqrt{x} = 9$
$\sqrt{2 x + 1} - \sqrt{x + 4} = 1$ $\sqrt{2 x + 1} - \sqrt{x + 4} = 1$

In Exercises 39–52, solve the given problems.

If $f (x) = \sqrt{x + 3},$ $f (x) = \sqrt{x + 3},$ find x if $f (x + 6) = 5.$ $f (x + 6) = 5.$
Solve $\sqrt{x - 2} = \sqrt[4]{x - 2} + 12$ $\sqrt{x - 2} = \sqrt[4]{x - 2} + 12$ by first writing it in quadratic form as shown in Section 14.3.
Solve $\sqrt{x - 1} + x = 3$ $\sqrt{x - 1} + x = 3$ algebraically. Then compare the solution with that of Example 4. Noting that the algebraic steps after isolating the radical are identical, why is the solution different?
The resonant frequency f in an electric circuit with an inductance L and a capacitance C is given by $f = \frac{1}{2 π \sqrt{L C}} .$ $f = \frac{1}{2 π \sqrt{L C}} .$ Solve for L.
A formula used in calculating the range R for radio communication is $R = \sqrt{2 r h + h^{2}} .$ $R = \sqrt{2 r h + h^{2}} .$ Solve for h.
The distance d (in km) to the horizon from a height h (in km) above the surface of Earth is $d = \sqrt{1.28 \times 10^{4} h + h^{2}} .$ $d = \sqrt{1.28 \times 10^{4} h + h^{2}} .$ Find h for $d = 980 km .$ $d = 980 km .$
In the study of spur gears in contact, the equation $k C = \sqrt{R_{1}^{2} - R_{2}^{2}} + \sqrt{r_{1}^{2} - r_{2}^{2}} - A$ $k C = \sqrt{R_{1}^{2} - R_{2}^{2}} + \sqrt{r_{1}^{2} - r_{2}^{2}} - A$ is used. Solve for $r_{1}^{2} .$ $r_{1}^{2} .$
The speed s (in m/s) at which a tsunami wave moves is related to the depth d (in m) of the ocean according to $s = \sqrt{g d},$ $s = \sqrt{g d},$ where g is the acceleration of gravity $(9.8 m / s^{2}) .$ $(9.8 m / s^{2}) .$ If a wave from the 2004 Indian Ocean tsunami was traveling at 195 m/s, estimate the depth of the ocean at that point.
If the value of a home increases from $v_{1}$ $v_{1}$ to $v_{2}$ $v_{2}$ over n years, the average annual rate of growth (as a decimal) is given by $r = \sqrt[n]{\frac{v_{2}}{v_{1}}} - 1.$ $r = \sqrt[n]{\frac{v_{2}}{v_{1}}} - 1.$ Suppose the value of a home increases on average by 3.6% per year over 10 years. If its value at the end of the 10-year period is $325,000, find its value at the beginning of the period.
A smaller of two cubical boxes is centered on the larger box, and they are taped together with a wide adhesive that just goes around both boxes (see Fig. 14.22). If the edge of the larger box is 1.00 in. greater than that of the smaller box, what are the lengths of the edges of the boxes if 100.0 in. of tape is used?

Fig. 14.22
A freighter is 5.2 km farther from a Coast Guard station on a straight coast than from the closest point A on the coast. If the station is 8.3 km from A, how far is it from the freighter?
The velocity v of an object that falls through a distance h is given by $v = \sqrt{2 g h},$ $v = \sqrt{2 g h},$ where g is the acceleration due to gravity. Two objects are dropped from heights that differ by 10.0 m such that the sum of their velocities when they strike the ground is 20.0 m/s. Find the heights from which they are dropped if $g = 9.80 m / s^{2} .$ $g = 9.80 m / s^{2} .$
A point D on Denmark’s largest island is 2.4 mi from the nearest point S on the coast of Sweden (assume the coast is straight, which is nearly the case). A person in a motorboat travels straight from D to a point on the beach x mi from S and then travels x mi farther along the beach away from S. Find x if the person traveled a total of 4.5 mi. See Fig. 14.23.

Fig. 14.23

Figure 14.23 Full Alternative Text
The length of the roller belt in Fig. 14.24 is 28.0 ft. Find x.

Fig. 14.24

Figure 14.24 Full Alternative Text

Answers to Practice Exercises

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Table of Contents for 14.4 Equations with Radicals

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Table of Contents for
14.4 Equations with Radicals