Question

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

Short answer

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

Step-by-step solution

Step 1. Given Information

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

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$3.$

Step 3. The possible factors are 1 and 4.

$$\begin{gathered} u^2-(4-1)u-4=0 \\ {u}^2-4u+u-4=0 \\ u(u-4)+1(u-4)=0 \\ (u+1)(u-4)=0 \end{gathered}$$

Step 4. Now putting the value equal to zero.

$$\begin{gathered} u+1=0u-4=0 \\ u+1-1=0-1u-4+4=0+4 \\ u=-1u=4 \end{gathered}$$

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

$$\begin{gathered} 2x-1=-12x-1=4 \\ 2x-1+1=-1+12x-1+1=4+1 \\ 2x=02x=5 \\ x=0x=\frac{5}{2} \end{gathered}$$

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