Question

Solve the given quadratic equation by completing the square, $x^2-3x+1=0$

Answer 1

The given quadratic equation is $x^2-3x+1=0$.

Add and subtract $\frac{9}{4}$ in the equation $x^2-3x+1=0$ to make a perfect square as shown below:

$x^2-3x+1=0\Rightarrow x^2-3x+\frac{9}{4}-\frac{9}{4}+1=0\Rightarrow (x{)}^2-(2\times x\times \frac{3}{2})+{\left(\frac{3}{2}\right)}^2+(-\frac{9}{4}+1)=0\Rightarrow {(x-\frac{3}{2})}^2-\frac{5}{4}=0(\because (a-b{)}^2=a^2+b^2-2ab)\Rightarrow {(x-\frac{3}{2})}^2=\frac{5}{4}\Rightarrow x-\frac{3}{2}=\pm \sqrt{\frac{5}{4}}\Rightarrow x-\frac{3}{2}=\pm \frac{\sqrt{5}}{2}\Rightarrow x-\frac{3}{2}=-\frac{\sqrt{5}}{2},x-\frac{3}{2}=\frac{\sqrt{5}}{2}\Rightarrow x=-\frac{\sqrt{5}}{2}+\frac{3}{2},x=\frac{\sqrt{5}}{2}+\frac{3}{2}\Rightarrow x=\frac{3-\sqrt{5}}{2},x=\frac{3+\sqrt{5}}{2}$

Hence, $x=\frac{3-\sqrt{5}}{2}$ or $x=\frac{3+\sqrt{5}}{2}$.