College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Mid-Chapter Check Point - Page 165: 20


The solutions are $x=-5\pm2\sqrt{7}$

Work Step by Step

$x^{2}+10x-3=0$ Take the $3$ to the right side of the equation: $x^{2}+10x=3$ Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=10$ $x^{2}+10x+\Big(\dfrac{10}{2}\Big)^{2}=3+\Big(\dfrac{10}{2}\Big)^{2}$ $x^{2}+10x+25=3+25$ $x^{2}+10x+25=28$ Factor the left side of the equation, which is a perfect square trinomial: $(x+5)^{2}=28$ Take the square root of both sides: $\sqrt{(x+5)^{2}}=\pm\sqrt{28}$ $x+5=\pm2\sqrt{7}$ Solve for $x$: $x=-5\pm2\sqrt{7}$
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