Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - Mid-Chapter Review - Mixed Review - Page 733: 16


Since the discriminant is a positive number that is a perfect square ($10^{2}=100$), there are $2$ rational solutions.

Work Step by Step

$ 3x^{2}=4x+7\qquad$....add $-4x-7$ to both sides. $ 3x^{2}-4x-7=0\qquad$.... $a=3,\ b=-4,\ c=-7$ $ b^{2}-4ac\qquad$....substitute $b$ for $-4,\ a$ for $3$ and $c$ for $-7$ $=(-4)^{2}-4\cdot(-7)\cdot 3$ $=16+84$ $=100$
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