Question

Hlo...✌

⭐If ax^2 + bx + c = 0
Prove ➡ x = -b + √b^2-4ac /2a
➡ x= -b - √b^2 - 4ac /2a

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Answer 1

$ax^2+bx+c=0$

Multiply both sides by a :

$\implies a^2x^2+abx+ac=0$

Transpose ac to the other side :

$\implies a^2x^2+abx=-ac$

Add b²/4 on both sides :

$\implies a^2x^2+abx+\frac{b^2}{4}=\frac{b^2}{4}-ac$

Perfect square on the right hand side :

$\implies a^2x^2+2\times a\times \frac{b}{2}\times x+\frac{b^2}{4}=\frac{b^2}{4}-ac$

$\implies (ax+\frac{b}{2})^2=\frac{b^2-4ac}{4}$

Take square root both sides :

$\implies ( ax+\frac{b}{2})=\frac{\pm \sqrt{b^2-4ac}}{2}$

Transpose b/2 to the other side :

$\implies ax=\frac{-b}{2}+\frac{\pm\sqrt{b^2-4ac}}{2}$

$\implies ax=\frac{-b\pm\sqrt{b^2-4ac}}{2}$

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

Hence it is proved that :

$Either\:x=\frac{-b+\sqrt{b^2-4ac}}{2a}\\\\Or\:x=\frac{-b-\sqrt{b^2-4ac}}{2a}$

Answer 2

$\underline{\underline{\bold{Question:}}}$

$\bold{If\quad\:ax^2+bx+c=0\quad\:then}\\\\\\\bold{Prove\:that\qquad\:x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.}$

$\underline{\bold{Solution:}}$

Given :

$\bold{ax^2+bc+c=0}\\\\\\\underline{\bold{Dividing\:the\;whole\:equation\:by\:a.}}\\\\\\\implies{\bold{\dfrac{ax^2}{a}+\dfrac{bx}{a}+\dfrac{c}{a}=\dfrac{0}{a}}}\\\\\\\implies{\bold{x^2+\dfrac{bx}{a}+\dfrac{c}{a}=0}}\\\\\\\underline{\bold{Adding\:both\;side\:the\:square\:of\:half\:of\:coefficient\:of\:x.}}\\\\\\\implies{\bold{x^2+(\dfrac{b}{2a})^2+\dfrac{bx}{a}+\dfrac{c}{a}=(\dfrac{b}{2a})^2}}\\\\\\\implies{\bold{x^2+(\dfrac{b}{2a})^2+2*x*\dfrac{b}{2a}=\dfrac{b^2}{4a^2}-\dfrac{c}{a}}}$

$\boxed{\bold{(a+b)^2=a^2+2ab+b^2}}\\\\\\\implies{\bold{(x+\dfrac{b}{2a})^2=\dfrac{b^2}{4a^2}-\dfrac{c}{a}}}\\\\\\\implies{\bold{(x+\dfrac{b}{2a})^2=\dfrac{b^2-4ac}{4a^2}}}\\\\\\\implies{\bold{x+\dfrac{b}{2a}=\dfrac{\pm\:\sqrt{b^2-4ac}}{2a}}}\\\\\\\implies{\bold{x=-\dfrac{b}{2a}+\dfrac{\pm\:\sqrt{b^2-4ac}}{2a}}}\\\\\\\implies{\bold{x=\dfrac{-b\pm\:\sqrt{b^2-4ac}}{2a}}}\\\\\\\boxed{\boxed{\bold{x=\dfrac{-b\pm\:\sqrt{b^2-4ac}}{2a}}}}$