Question

Solve: $x^2+3x+1=0$

• A. $x=\frac{3\pm \sqrt{5}}{2}$
• B. $x=\frac{3+\sqrt{5}}{2},\frac{-3+\sqrt{5}}{2}$
• C. $x=\frac{\pm 3-\sqrt{5}}{2}$
• D. $x=\frac{-3+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}$

Answer 1

The correct option is D $x=\frac{-3+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}$

We have $x^2+3x+1=0$
Add and subtract ${\left(\frac{1}{2}\text{coefficient of x}\right)}^2$ in LHS and get $x^2+3x+1+{\left(\frac{3}{2}\right)}^2-{\left(\frac{3}{2}\right)}^2=0$

$\Rightarrow x^2+2\left(\frac{3}{2}\right)x+{\left(\frac{3}{2}\right)}^2-{\left(\frac{3}{2}\right)}^2+1=0$

$\Rightarrow {(x+\frac{3}{2})}^2-\frac{5}{4}=0$

$\Rightarrow {(x+\frac{3}{2})}^2={\left(\frac{\sqrt{5}}{2}\right)}^2$

$\Rightarrow x+\frac{3}{2}=\pm \frac{\sqrt{5}}{2}$

This gives $x=\frac{-3+\sqrt{5}}{2}$ or $x=\frac{-3-\sqrt{5}}{2}$

Therefore, $x=\frac{-3+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}$ are the solutions of the given equation