Question

Solve the following quadratic equations by completing square.
x² + 3x + 1 = 0

Answer 1

Given: x² + 3x + 1 = 0
=> x² + 3x = –1…(1)
We make L.H.S a perfect square by using

Third term = ${\left(\frac{1}{2}\mathrm{coefficient}\mathrm{of}x\right)}^2={\left(\frac{1}{2}\times 3\right)}^2={\left(\frac{3}{2}\right)}^2=\frac{9}{4}$

On adding $\frac{9}{4}$ to both sides of equation (1), we get:
x² + 3x + $\frac{9}{4}$ = –1 + $\frac{9}{4}$
=> x² + 2 × x ×$\frac{3}{2}$ +${\left(\frac{3}{2}\right)}^2$ =$\frac{5}{4}$

=> ${\left(x+\frac{3}{2}\right)}^2={\left(\frac{\sqrt{5}}{2}\right)}^2$
On taking square root of both sides, we get:
$x+\frac{3}{2}=\pm \frac{\sqrt{5}}{2}$
Thus,
x = $\left(\frac{\sqrt{5}}{2}-\frac{3}{2}\right),\left(-\frac{\sqrt{5}}{2}-\frac{3}{2}\right)$