Math Doubts

Proof of Quadratic formula

$ax^2+bx+c = 0$ is a quadratic equation in standard algebraic form. The solution of the quadratic equation in algebraic formulation is called quadratic formula.

Quadratic formula

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


The quadratic formula is derived mathematically by transforming the quadratic equation in square form completely.

Basic simplification for the Perfect square conversion

Move the constant term $c$ to right hand side of the equation.

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

Divide both sides by the literal coefficient of the $x^2$.

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

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

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

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

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

Transforming the Equation as a square of a binomial

The left hand side expression can be transformed as a square of a binomial by some adjustments.

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

The multiplying factor $1$ can be written as the quotient of $2$ by $2$ because the division of $2$ by $2$ is $1$.

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

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

There is a chance to transform the left hand side expression as a square of sum of two terms. So, add $\Bigg(\dfrac{b}{2a}\Bigg)^2$ and subtract it by the same term.

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

Move the negative term to right hand side.

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

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

The left hand side expression represents the expansion of the square of the sum of the two terms. It can be simplified by the a+b whole square formula.

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

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

Obtaining solution of the equation

The square of the left hand side expression becomes square root to right hand side expression.

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

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

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

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

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

This is called quadratic formula and thus it is derived in step by step procedure from the quadratic equation.

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