Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 10 - Radical Expressions and Equations - Get Ready: 17

Answer

$x = \frac{1}{2}$

Work Step by Step

First, we should look at the discriminant ($b^2 - 4ac$) to determine how many real number solutions we have. The rules are such: $(b^2 - 4ac) \gt 0$ then there are two real number solutions $(b^2 - 4ac) \lt 0$ then there is one real number solution $(b^2 - 4ac) = 0$ then there is no real number solution Evaluating the discriminant: Step 1: Write the equation in standard form. $4x^2 - 4x +1 = 0$ Step 2: Determine our a, b, and c values $a = 4, b = -4, c = 1$ Step 3: Evaluate $b^2 - 4ac$, using our values of a, b, and c $b^2 - 4ac$ = $(-4)^2 - 4(4)(1) = 16 - 16 = 0$ Based on our rules of the discriminant, since $b^2 - 4ac = 0$, there is only one real number solutions. Now, using the Quadratic Formula, we can find this real number solution. Step 1 and 2 are already complete from finding the discriminant. Step 1: Write the equation in standard form. $4x^2 - 4x +1 = 0$ Step 2: Determine our a, b, and c values $a = 4, b = -4, c = 1$ Step 3: Substitute our values of a, b, and c into the formula $x = \frac{-b \frac{+}{-} \sqrt(b^2 - 4ac)}{2a}$ $x = \frac{-(-4) \frac{+}{-} \sqrt((-4)^2 - 4(4)(1)}{2(4)}$ $x = \frac{4 \frac{+}{-} \sqrt(16 - 16)}{8}$ $x = \frac{4}{8}$ $x = \frac{1}{2}$ We have found our real number solution.
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