Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises: 75

Answer

$$x = 2 \pm \frac{\sqrt7}{2}$$

Work Step by Step

$Find$ $all$ $real$ $solutions$ $of$ $the$ $quadratic$ $equation:$ $4x^2-16x+9=0$ Use the Quadratic Equation Formula: $x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}$ $4x^2-16x+9$ $a = 4, b = -16, c = 9$ $x = \frac{-(-16) \pm \sqrt {(-16)^2- 4(4 \times 9)}}{2(4)}$ $x = \frac{16 \pm \sqrt {256- 144}}{8}$ $x = \frac{16 \pm \sqrt {112}}{8}$ [Note: $\sqrt {112} = \sqrt {16\times7} = \sqrt {16}\times\sqrt7$ = $4\sqrt7$ $x = \frac{16 \pm 4\sqrt {7}}{8}$ $x = \frac{16}{8} \pm \frac{4\sqrt7}{8}$ $$x = 2 \pm \frac{\sqrt7}{2}$$
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