Question

Derive this formula,
$\sf a{x}^{2} + bx + c = 0$
$\rm \implies x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$
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Answer 1

Required Answer:-

Given to derive:

• The formula for finding out the roots of a quadratic equation.

Derivation:

We have derive the quadratic formula.

Here is the derivation.

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

$\rm \implies a \bigg({x}^{2} + \dfrac{b}{a} x + \dfrac{c}{a} \bigg) = 0$

$\rm \implies \bigg({x}^{2} + \dfrac{b}{a} x + \dfrac{c}{a} \bigg) = 0$

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

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

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

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

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

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

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

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

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

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

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

Thus,

$\rm \implies x_{1,2} = \dfrac{ - b \pm\sqrt{ {b}^{2} - 4ac} }{2a}$

Hence Derived.