Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Appendix A - Equations and Inequalities - Page 420: 35



Work Step by Step

$-4x^{2}+x=-3$ Adding 3 on both sides, $-4x^{2}+x+3=-3+3$ Simplifying, $-4x^{2}+x+3=0$ $4x^{2}-x-3=0$ $4x^{2}-4x+3x-3=0$ $4x(x-1)+3(x-1)=0$ $(x-1)(4x+3)=0$ $(x-1)=0$ or $(4x+3)=0$ $x=1$ or $x=-\frac{3}{4}$ Check: Substituting $x=1$ in the equation, $4(1)^{2}-(1)-3=0$ $4-1-3=0$ $4-4=0$ $0=0$ Substituting $x=-\frac{3}{4}$ in the equation, $4(-\frac{3}{4})^{2}-(-\frac{3}{4})-3=0$ $\frac{36}{16}+\frac{3}{4}-3=0$ $\frac{9}{4}+\frac{3}{4}-3=0$ $\frac{12}{4}-3=0$ $3-3=0$ $0=0$ Both values check and the solution set is {$-\frac{3}{4},1$}
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