Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Cumulative Review Exercises - Page 1127: 16


The value of the expression $6{{x}^{2}}-6<5x $ is $\left( -\frac{2}{3},\frac{3}{2} \right)$.

Work Step by Step

Rearrange the following expression as given below: $\begin{align} & 6{{x}^{2}}-6<5x \\ & 6{{x}^{2}}-6-5x<0 \\ & 6{{x}^{2}}-5x-6<0 \\ & 6{{x}^{2}}-9x+4x-6<0 \end{align}$ And simplify it further to get: $\begin{align} & 3x\left( 2x-3 \right)+2\left( 2x-3 \right)<0 \\ & \left( 2x-3 \right)\left( 3x+2 \right)<0 \end{align}$ Again, solve the inequality $\left( 2x-3 \right)\left( 3x+2 \right)<0$ as given below: $\left( 2x-3 \right)<0$ And $\left( 3x+2 \right)>0$ Or: $\left( 2x-3 \right)>0$ And $\left( 3x+2 \right)<0$ Then, from the first condition we get $ x-\frac{2}{3}$. From the second condition we get $ x>\frac{3}{2}$ and $ x-\frac{2}{3}$ and $ x
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