Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Section 8.3 - Solving Equations by Using Quadratic Methods - Exercise Set: 36

Answer

The solution of the given equation is $\frac{1}{2}$.

Work Step by Step

Square both sides to obtain: $(4x)^2=(\sqrt{2x+3})^2 \\16x^2=2x+3 \\16x^2-2x-3=0$ Factor the trinomial: $(8x+3)(2x-1)=0$ Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain: \begin{array}{ccc} \\&8x+3=0 &\text{or} &2x-1=0 \\&8x=-3 &\text{or} &2x=1 \\&x=-\frac{3}{8} &\text{or} &x=\frac{1}{2} \end{array} Note that when $x=-\frac{3}{8}$: the left side of the given equation becomes $-\frac{3}{2}$. Since the right side of the given equation involves a principal square root, its value must be non-negative. Thus, $-\frac{3}{8}$ is not a solution of the equation. Therefore, the solution of the given equation is $\frac{1}{2}$.
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