Question

Find the roots of $a{x}^2+bx+c=0(a\ne 0)$ by the method of completing the square.

Answer 1

$a{x}^2+bx+c=0$

$\Rightarrow a{x}^2+bx+{\left(\frac{b}{2\sqrt{a}}\right)}^2-{\left(\frac{b}{2\sqrt{a}}\right)}^2+c=0$

$\Rightarrow {(\sqrt{a}x+\frac{b}{2\sqrt{a}})}^2-\frac{b^2}{4a}+c=0$

$\Rightarrow {(\sqrt{a}+\frac{b}{2\sqrt{a}})}^2-{\left(\sqrt{\frac{b^2-ac}{4a}}\right)}^2=0$

$\Rightarrow (\sqrt{a}x+\frac{b}{2\sqrt{a}}+\sqrt{\frac{b^2-4ac}{4a}})(\sqrt{a}x+\frac{b}{2\sqrt{a}}-\sqrt{\frac{b^2-4ac}{2a}})=0$

$\sqrt{a}x+\frac{b}{2\sqrt{a}}+\sqrt{\frac{b^2-4ac}{4a}}=0$, $\sqrt{a}x+\frac{b}{2\sqrt{a}}-\sqrt{\frac{b^2-4ac}{2a}}=0$

$x=\frac{-b+\sqrt{b^2-4ac}}{2\sqrt{a}\cdot \sqrt{a}},x=\frac{-b-\sqrt{b^2-4ac}}{2\sqrt{a}\cdot \sqrt{a}}$

$\therefore x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.