## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter R - Algebra Reference - R.4 Equations - R.4 Exercises - Page R-16: 17

#### Answer

$0$ and $4$

#### Work Step by Step

Divide the equation by 12, $12y^{2}-48y=0\qquad.../\div 12$ $y^{2}-4y=0\qquad$ ... factor out y $y(y-4)=0$ By the zero product principle, $y=0$ or $y-4=0$ $y=0$ or $y=4$ The solutions are $0$ and $4$.

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