Question

how to prove quadratic equation formula
​

Answer 1

To prove

$\text{Quadratic equation formula : } \quad x = \dfrac{ - b \pm\sqrt{ b^{2} - 4ac} }{2a}$

Proof

Take the standard form of Quadratic equation and solve it by completing the square method.

$ax^2 + bx + c = 0$

Subtracting 'c' on both sides

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

In order to bring LHS as a perfect square, firstly we need to eliminate coefficient if x² i.e 'a'

Dividing throughout by 'a'

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

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

Now, coefficient of x² is eliminated. It is still not a perfect square.

To make LHS a perfect square it should be of the form x² + 2xy + y² [ as a² + 2xy + y² = (x + y)² ]

Here, we have x² + bx/a, comparing it with x² + 2xy we know only 'x' but not 'y'

Again comparing it with x² + 2xy we have, in the equation 2xy is there as bx/a

We have to bring bx/a of the form 2xy

Here only x is know. So, take out x from bx/a and value should remain same

bx/a can be written as

$\rightarrow x \bigg( \dfrac{b}{a} \bigg)$

Now we have take out 2 so that it will arrive in form of 2xy and and value should remain same.

$\rightarrow 2(x )\bigg( \dfrac{b}{2a} \bigg)$

So, now we got to know the values of x and y

Now, the equation will be

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

To make it perfect square add y² i.e [b/2a]² on both sides

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

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

[ Because x² + y² + 2xy = (x + y)² ]

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

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

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

Taking square root on both sides

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

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

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

Subtracting b/2a on both sides

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

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

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

Hence proved.

Answer 2

$\bold{\large{\underline{\underline{\sf{StEp\:by\:stEp\:explanation:}}}}}$

To prove:

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

Proof:

We will start with the the standard form of a quadratic equation,

$\bold\green{\implies ax^{2}+bx+c=0}$

Now, divide both sides of the equation by 'a' so you can complete the square,

$\sf{\implies x^{2}+\dfrac{b}{a}x+\dfrac{c}{a}=\dfrac{0}{a}}$

$\sf{\implies x^{2}+\dfrac{b}{a}x+\dfrac{c}{a}=0}$

Now, Subtract c/a from both sides.

$\sf{\implies x^{2}+\dfrac{b}{a}x+\dfrac{c}{a}-\dfrac{c}{a}=0-\dfrac{c}{a}}$

$\tt{\implies x^{2}+\dfrac{b}{a}x=-\dfrac{c}{a}}$

Now, by completing the square method.

The coefficient of the second term is b/a . Divide this coefficient by 2 and square the result to get (b/2a)², add (b/2a)² to both sides:

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

Since the left side of the equation right above is a perfect square, you can factor the left side by using the coefficient of the first term (x) and the base of the last term(b/2a).

Then, square the right side to get (b²)/(4a²).

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

Get the same denominator on the right side:

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

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

Now, take the square root of each side,

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

Now, Simplify the left side,

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

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

Rewrite the right sides,

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

Now, Subtract b/2a from both sides,

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

Adding the numerator and keeping the same denominator, we get the quadratic formula,

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

_________________(HENCE PROVED)