Question

Find the real solutions of the equation$x^2+3x-\frac{7}{4}=0$ by completing square.

Short answer

The real solution of the equation is $x=-\frac{7}{2}\text{and}x=\frac{1}{2}$.

Step-by-step solution

Step 1. Complete the square of given quadratic.

$$\begin{gathered} x^2+3x-\frac{7}{4}=0....\left(\text{Given}\right) \\ {x}^2+3x=\frac{7}{4}\mathrm{....}\left(\text{Add}\frac{7}{4}\text{bothsides}\right) \\ {x}^2+3x+\left(\frac{9}{4}\right)=\frac{7}{4}+\left(\frac{9}{4}\right)\mathrm{....}\left(\text{Add}{\left(\frac{3}{2}\right)}^2\text{\hspace{0.33em}bothsides}\right) \\ {\left(x+\frac{3}{2}\right)}^2=4....\left(1\right) \end{gathered}$$

Step 2. Further simplifying (1).

$$\begin{gathered} x+\frac{3}{2}=\pm \sqrt{4}....\left(\text{Squarerootof}\left(1\right)\right) \\ x=-\frac{3}{2}\pm 2 \\ \text{or,}x=-\frac{3}{2}-2,-\frac{3}{2}+2 \\ x=-\frac{7}{2}\text{or}x=\frac{1}{2} \end{gathered}$$

Step 3. DESCRIPTION OF STEPS.

From the obtained values, the real solution of the equation is $x=-\frac{7}{2}\text{and}x=\frac{1}{2}$.