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: 47


The real solutions of the equation $x^{2}-7x+12=0$ are $x=3$ and $x=4$.

Work Step by Step

1. Factor $x^{2}-7x+12$: The factors are (x−3)(x-4) because −3×-4 = 12 (the c value in $a^{2}+bx+c$) and -3+-4 =-7, the b value. 2. Set the factors equal to zero: $x-3=0$ and $x-4=0$ 3. Solve: $x=3$ and $x=4$
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