Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 56: 49


The real solutions to the equation $4x^{2}-4x-15=0$ are $x=\frac{5}{2}$ and x=$-\frac{3}{2}$.

Work Step by Step

1. Factor $4x^{2}+4x-15$. The factors are $(-2x-3)(-2x+5)$ because $-2x\times-2x =4x^{2}$(the a value in $a^{2}+bx+c$) and, $-3\times5 = -15$, the c value in the equation. 2. Set the factors equal to zero: $-2x-3=0$ and $-2x+5 = 0$ 3. Solve: $x=-\frac{3}{2}$ and $x=\frac{5}{2}$
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