Solving Quadratic equation by Completing the square method

A mathematical approach of solving a quadratic equation by completely transforming quadratic expression as a square of a binomial is called completing the square method.

Method

$ax^2+bx+c = 0$ is a representation of a quadratic equation in standard algebraic form. It can be solved by converting the quadratic equation as a square of a binomial by some acceptable mathematical adjustments.

01

The quadratic equation can be converted as a square of a binomial and it helps us to find the roots of the equation easily.

$\implies ax^2 + bx = -c$

$\implies$ $\dfrac{ax^2 + bx}{a} = -\dfrac{c}{a}$

$\implies$ $\Bigg(\dfrac{a}{a}\Bigg)x^2 + \Bigg(\dfrac{b}{a}\Bigg)x = -\dfrac{c}{a}$

$\implies$ $\require{cancel} \Bigg(\dfrac{\cancel{a}}{\cancel{a}}\Bigg)x^2 + \Bigg(\dfrac{b}{a}\Bigg)x = -\dfrac{c}{a}$

$\implies$ $x^2 + \Bigg(\dfrac{b}{a}\Bigg)x = -\dfrac{c}{a}$

$\implies$ $x^2 + 1 \times \Bigg(\dfrac{b}{a}\Bigg)x$ $\,=\,$ $-\dfrac{c}{a}$

$\implies$ $x^2 + \Bigg(\dfrac{2}{2}\Bigg) \times \Bigg(\dfrac{b}{a}\Bigg)x$ $\,=\,$ $-\dfrac{c}{a}$

$\implies$ $x^2 + 2\Bigg(\dfrac{b}{2a}\Bigg)x$ $\,=\,$ $-\dfrac{c}{a}$

$\implies$ $x^2 + 2\Bigg(\dfrac{b}{2a}\Bigg)x + \Bigg(\dfrac{b}{2a}\Bigg)^2 -\Bigg(\dfrac{b}{2a}\Bigg)^2$ $\,=\,$ $-\dfrac{c}{a}$

$\implies$ $x^2 + 2\Bigg(\dfrac{b}{2a}\Bigg)x + \Bigg(\dfrac{b}{2a}\Bigg)^2$ $\,=\,$ $\Bigg(\dfrac{b}{2a}\Bigg)^2 -\dfrac{c}{a}$

$\implies$ $x^2 + \Bigg(\dfrac{b}{2a}\Bigg)^2 + 2x\Bigg(\dfrac{b}{2a}\Bigg)$ $\,=\,$ $\Bigg(\dfrac{b}{2a}\Bigg)^2 -\dfrac{c}{a}$

$\implies {\Bigg(x + \dfrac{b}{2a}\Bigg)}^2$ $\,=\,$ $\dfrac{b^2}{4a^2} -\dfrac{c}{a}$

The quadratic equation in standard form is successfully transformed as a square of a binomial.

$\implies {\Bigg(x + \dfrac{b}{2a}\Bigg)}^2$ $\,=\,$ $\dfrac{b^2 -4ac}{4a^2}$

02

Finding the Roots of the Equation

It is used to find the values of the roots of the quadratic equation.

$\implies$ $x + \dfrac{b}{2a}$ $\,=\,$ $\pm \sqrt{\dfrac{b^2 -4ac}{4a^2}}$

$\implies$ $x + \dfrac{b}{2a}$ $\,=\,$ $\pm \sqrt{\dfrac{b^2 -4ac}{{(2a)}^2}}$

$\implies$ $x + \dfrac{b}{2a}$ $\,=\,$ $\pm \dfrac{\sqrt{b^2 -4ac}}{2a}$

$\implies$ $x = -\dfrac{b}{2a} \pm \dfrac{\sqrt{b^2 -4ac}}{2a}$

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

Example

$2x^2 -7x + 3 = 0$ is a quadratic equation and understand the procedure of completing the square method.

Make some basic adjustment to transform the quadratic equation as a square of a binomial.

$\implies 2x^2 -7x = -3$

$\implies \dfrac{2x^2 -7x}{2} = -\dfrac{3}{2}$

$\implies \dfrac{2x^2}{2} -\dfrac{7x}{2} = -\dfrac{3}{2}$

$\implies \Bigg(\dfrac{2}{2}\Bigg)x^2 -\Bigg(\dfrac{7}{2}\Bigg)x = -\dfrac{3}{2}$

$\implies \require{cancel} \Bigg(\dfrac{\cancel{2}}{\cancel{2}}\Bigg)x^2 -\Bigg(\dfrac{7}{2}\Bigg)x = -\dfrac{3}{2}$

$\implies x^2 -\Bigg(\dfrac{7}{2}\Bigg)x = -\dfrac{3}{2}$

$\implies x^2 -x\Bigg(\dfrac{7}{2}\Bigg) = -\dfrac{3}{2}$

$\implies x^2 -\Bigg(\dfrac{2}{2}\Bigg)x\Bigg(\dfrac{7}{2}\Bigg) = -\dfrac{3}{2}$

$\implies x^2 -2x\Bigg(\dfrac{7}{4}\Bigg) = -\dfrac{3}{2}$

$\implies x^2 -2x\Bigg(\dfrac{7}{4}\Bigg) + {\Bigg(\dfrac{7}{4}\Bigg)}^2 -{\Bigg(\dfrac{7}{4}\Bigg)}^2 = -\dfrac{3}{2}$

$\implies x^2 -2x\Bigg(\dfrac{7}{4}\Bigg) + {\Bigg(\dfrac{7}{4}\Bigg)}^2 -\dfrac{49}{16} = -\dfrac{3}{2}$

$\implies x^2 -2x\Bigg(\dfrac{7}{4}\Bigg) + {\Bigg(\dfrac{7}{4}\Bigg)}^2 = \dfrac{49}{16} -\dfrac{3}{2}$

$\implies x^2 -2x\Bigg(\dfrac{7}{4}\Bigg) + {\Bigg(\dfrac{7}{4}\Bigg)}^2 = \dfrac{49 -24}{16}$

$\implies {\Bigg(x -\dfrac{7}{4}\Bigg)}^2 = \dfrac{25}{16}$

Now, find the values of $x$ by solving it.

$\implies \Bigg(x -\dfrac{7}{4}\Bigg) = \pm \sqrt{\dfrac{25}{16}}$

$\implies x -\dfrac{7}{4} = \pm \dfrac{5}{4}$

$\implies x = \dfrac{7}{4} \pm \dfrac{5}{4}$

$\implies x = \dfrac{7 \pm 5}{4}$

$\implies x = \dfrac{7 + 5}{4}$ and $x = \dfrac{7 -5}{4}$

$\implies x = \dfrac{12}{4}$ and $x = \dfrac{2}{4}$

$\,\,\, \therefore \,\,\,\,\,\, x = 3$ and $x = \dfrac{1}{2}$

Therefore, the roots of quadratic equation $2x^2 -7x + 3 = 0$ by using completing the square method are $3$ and $\dfrac{1}{2}$.