Question

Solve the following quadratic equation by completing the square: $x^2+3x+1=0$

• A. $\{\frac{-3+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}\}$
• B. $\{\frac{-3-\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}\}$
• C. $\{\frac{-3+\sqrt{5}}{2},\frac{-3+\sqrt{5}}{2}\}$
• D. None of these

Answer 1

Answer: (A)

$\{\frac{-3+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}\}$

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

The 3 will have to be divided by $2$ then the result should be squared and the final result is $\frac{9}{4}$. This number will be added and subtracted in the equation on one side.

$x^2+3x+\frac{9}{4}-\frac{9}{4}+1=0$

$(x^2+3x+\frac{9}{4})-\frac{9}{4}+1=0$

${(x+\frac{3}{2})}^2-\frac{9}{4}+1=0$

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

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

Now taking square root on both sides we get,

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

$\therefore$ $x=-\frac{3}{2}\pm \frac{\sqrt{5}}{2}$

$\therefore$ $x=-\frac{3}{2}+\frac{\sqrt{5}}{2}$ and $x=-\frac{3}{2}-\frac{\sqrt{5}}{2}$

$\Rightarrow$ $x=(\frac{-3+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2})$