Question

Find the real solutions of the equation$x^2+2x-5=0$by completing square

Short answer

The real solution of the equation is $x=-1+\sqrt{6}\text{and}x=-1-\sqrt{6}$.

Step-by-step solution

Step 1. Complete the square of given quadratic.

$$\begin{gathered} x^2+2x-5=0....\left(\text{Given}\right) \\ {x}^2+2x=5....\left(\text{Add5bothsides}\right) \\ {x}^2+2x+{\left(\frac{2}{2}\right)}^2=5+{\left(\frac{2}{2}\right)}^2\mathrm{....}\left(\text{Add}{\left(\frac{2}{2}\right)}^2\text{\hspace{0.33em}bothsides}\right) \\ {x}^2+2x+1=5+1 \\ {\left(x+1\right)}^2=6....\left(1\right) \end{gathered}$$

Step 2. Further simplifying (1).

$$\begin{gathered} x+1=\pm \sqrt{6}\mathrm{....}\left(\text{Squarerootof}\left(1\right)\right) \\ \text{or,}x=-1\pm \sqrt{6} \\ x=-1+\sqrt{6}\text{or}x=-1-\sqrt{6} \end{gathered}$$

Step 3. Description of steps.

From the obtained values, the real solution of the equation is $x=-1+\sqrt{6}\text{and}x=-1-\sqrt{6}$.