Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 6 - Section 6.6 - Solving Quadratic Equations by Factoring - Exercise Set: 52

Answer

The solutions are - $\frac{4}{3}$ and $\frac{3}{4}$.

Work Step by Step

12$x^{2}$+7x-12=0 12$x^{2}$+16x-9x-12=0 4x(3x+4)-3(3x+4)=0 (4x-3)(3x+4)=0 4x-3 =0 or 3x+4=0 x=$\frac{3}{4}$ 0r x=-$\frac{4}{3}$ The solutions are - $\frac{4}{3}$ and $\frac{3}{4}$. Check Let x=- $\frac{4}{3}$ 12$x^{2}$+7x-12=0 12$(-4/3)^{2}$+7$\frac{-4}{3}$-12=0 12*16/9-$\frac{28}{3}$-12=0 4*16/3-$\frac{28}{3}$-2=0 64/3-$\frac{28}{3}$-12=0 $\frac{36}{3}$-12=0 12-12=0 0=0 Let x= $\frac{3}{4}$ 12$x^{2}$+7x-12=0 12$(3/4)^{2}$+7$\frac{3}{4}$-12=0 12*9/16+$\frac{21}{4}$-12=0 108/16+$\frac{21}{4}$-12=0 27/4+$\frac{21}{4}$-12=0 $\frac{48}{4}$-12=0 12-12=0 0=0
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