Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 6 - Radical Functions and Rational Exponents - 6-5 Solving Square Root and Other Radical Equations - Practice and Problem-Solving Exercises - Page 395: 37

Answer

$x = 1$

Work Step by Step

Rewrite the problem in radical form. Remember that $a^{\frac{m}{n}}$ can be rewritten in radical form as $\sqrt[n] {a^{m}}$: $\sqrt {7x + 6} - \sqrt {9 + 4x} = 0$ Add $\sqrt {9 + 4x}$ to both sides of the equation: $\sqrt {7x + 6} = \sqrt {9 + 4x}$ Since the indices are the same for both radicals, we can square both sides of the equation to eliminate the radicals: $7x + 6 = 9 + 4x$ Subtract $4x$ from each side of the equation to move the variables to the left side of the equation: $3x + 6 = 9$ Subtract $6$ from each side of the equation to move the constants to the right side of the equation: $3x = 3$ Divide each side of the equation by $3$: $x = 1$ To check if we have an extraneous solution, we substitute the solution into the original equation to see if the two sides equal one another: $\sqrt {7(1) + 6} - \sqrt {9 + 4(1)} = 0$ Simplify the radicands: $\sqrt {13} - \sqrt {13}=0\\ 0=0$ Both sides are equal; therefore, this solution is valid.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.