Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises - Page 121: 47


The solutions are $x=1\pm\dfrac{\sqrt{3}}{2}i$

Work Step by Step

$-4x^{2}+8x=7$ Divide the whole equation by $-4$: $-\dfrac{1}{4}(-4x^{2}+8x=7)$ $x^{2}-2x=-\dfrac{7}{4}$ Complete the square by adding $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation and simplifying. In this case, $b=-2$ $x^{2}-2x+\Big(-\dfrac{2}{2}\Big)^{2}=-\dfrac{7}{4}+\Big(-\dfrac{2}{2}\Big)^{2}$ $x^{2}-2x+1=-\dfrac{7}{4}+1$ $x^{2}-2x+1=-\dfrac{3}{4}$ Factor the left side of the equation, which is a perfect square trinomial: $(x-1)^{2}=-\dfrac{3}{4}$ Take the square root of both sides: $\sqrt{(x-1)^{2}}=\pm\sqrt{-\dfrac{3}{4}}$ $x-1=\pm\dfrac{\sqrt{3}}{2}i$ Solve for $x$: $x=1\pm\dfrac{\sqrt{3}}{2}i$
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