Answer
one real zero (multiplicity 2).
Work Step by Step
Given $4x^2-28x+49=0$, we have $a=4, b=-28, c=49$ and $b^2-4ac=(-28)^2-4(4)(49)=0$, thus there will be one real zero with a multiplicity of 2.
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 4 - Exponential and Logarithmic Functions - Section 4.5 Properties of Logarithms - 4.5 Assess Your Understanding - Page 332: 117
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