Question

Find the real solutions of$4x^2-x-5=0$

Short answer

The real solution of the given equation is$\left(\frac{5}{4},-1\right)$.

Step-by-step solution

Step 1. Given Information

We have to find the real solutions of$4x^2-x-5=0$

Step 2. Now solving the equation.

In the given equation, the product of first and last term is 20 and the middle term is 3. So the possible factors are 4 and 5.

Step 3. Solving.

$$\begin{gathered} 4x^2-(5-4)x-5=0 \\ 4x^2-5+4x-5=0 \\ x(4x-5)+1(4x-5)=0 \\ (x+1)(4x-5)=0 \end{gathered}$$

Step 4. Now putting the value equal to zero.

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