## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 10 - Exponents and Radicals - 10.6 Solving Radical Equations - 10.6 Exercise Set - Page 668: 39

#### Answer

$t=1$

#### Work Step by Step

Squaring both sides of the given equation, $\sqrt{3t+4}=\sqrt{4t+3} ,$ results to \begin{array}{l}\require{cancel} \sqrt{3t+4}=\sqrt{4t+3} \\\\ \left( \sqrt{3t+4} \right)^2=\left( \sqrt{4t+3} \right)^2 \\\\ 3t+4=4t+3 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 3t+4=4t+3 \\\\ 3t-4t=3-4 \\\\ -t=-1 \\\\ t=1 .\end{array} Upon checking, $t=1$ satisfies the original equation.

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