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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 5 (page 464)

Find the real solutions of ${(2 x - 1)}^{2} - 3 (2 x - 1) - 4 = 0$

Short Answer

Expert verified

The solution set is $\{0, \frac{5}{2}\}$

Step by step solution

01

Step 1. Given Information

We have to find the real solutions of ${(2 x - 1)}^{2} - 3 (2 x - 1) - 4 = 0$

02

Step 2. For this equation, let u=2x-1, Then u2=(2x-1)2and the original equation, (2x-1)2-3(2x-1)-4=0 becomes u2-3u-4=0.

Now factor the equation.

In the equation, the multiply of first and last term is $4$ and the middle factor is $3 .$

03

Step 3. The possible factors are 1 and 4.

$u^{2} - (4 - 1) u - 4 = 0 u^{2} - 4 u + u - 4 = 0 u (u - 4) + 1 (u - 4) = 0 (u + 1) (u - 4) = 0$

04

Step 4. Now putting the value equal to zero.

$u + 1 = 0 u - 4 = 0 u + 1 - 1 = 0 - 1 u - 4 + 4 = 0 + 4 u = - 1 u = 4$

05

Step 5. But we want to solve for x. Because u=2x-1, we have

$2 x - 1 = - 1 2 x - 1 = 4 2 x - 1 + 1 = - 1 + 1 2 x - 1 + 1 = 4 + 1 2 x = 0 2 x = 5 x = 0 x = \frac{5}{2}$

The solution set is $\{0, \frac{5}{2}\} .$

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