## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Section 1.5 - Equations - 1.5 Exercises: 47

#### Answer

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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