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 - Page 56: 53

Answer

The real solutions to the equation $2x^{2}=8$ are $x=2$ and $x=-2$

Work Step by Step

1. $2x^{2}=8$ 2. Divide both sides by 2: $x^{2}=4$ 3. Subtract 4 from both sides: $x^{2}-4=0$ 4. Factor the equation using the difference of two squares formula: $a^{2}-b^{2}=(a+b)(a-b)$: $(x+2)(x-2)=0$ 5. Set both factors equal to zero: $x+2=0$ and $x-2=0$ 6. Solve: $x=-2$ and $x=2$
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