# Quadratic Formula

The quadratic formula, also known as Shreedhara Acharya's formula, named after the ancient Indian mathematician who derived it, states that the roots of the equation \( ax^2 + bx + c = 0 \) are of the form

\[ x = \frac { - b \pm \sqrt{ b^2 - 4ac } } { 2a} .\]

#### Contents

## Proof 1: (Completing the square)

We start with \(a{ x }^{ 2 }+bx+c=0.\)

We then subtract \(c\) from both sides to get \(a{ x }^{ 2 }+bx=-c.\)

Divide both sides by \(a\) to get \(\displaystyle { x }^{ 2 }+\frac { bx }{ a } =\frac { -c }{ a }.\)

We can then add \(\dfrac { { b }^{ 2 } }{ 4{ a }^{ 2 } } \) to both sides to get \[{ x }^{ 2 }+\frac { bx }{ a }+\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } =\frac { -c }{ a }+\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } .\]

We can then complete the square on the left side to get \[\left(x+\frac { b }{ 2a } \right)^2=\frac { -c }{ a }+\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } }.\]

Simplify the right side to get \(\displaystyle \left(x+\frac { b }{ 2a } \right)^{ 2 }=\frac { { b }^{ 2 }-4ac }{ 4{ a }^{ 2 } }.\)

Take the square root of both sides and get \[ x+\frac { b }{ 2a } =\frac { \pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }.\]
Then, subtract \(\dfrac { b }{ 2a } \) from both sides to get \[ x = \dfrac { - b \pm \sqrt{ b^2 - 4ac }}{2a}. \ _\square\]

## Proof 2:

We again start with \(ax^{2}+bx+c=0.\)

We multiply both sides by \(4a\) to get \(4a^{2}x^{2}+4abx+4ac=0\).

Now notice that \(4a^{2}x^{2}+4abx\) is almost the square of \(2ax+b,\) or more precisely, \((2ax+b)^{2}-b^{2}\). Our equation then becomes
\[\begin{align}
(2ax+b)^{2}-b^{2}+4ac&=0 \\
(2ax+b)^{2}&=b^{2}-4ac \\
\sqrt{(2ax+b)^{2}}&=\pm \sqrt{b^{2}-4ac}\\
2ax+b&=\pm \sqrt{b^{2}-4ac} \\
2ax&=-b \pm \sqrt{b^{2}-4ac} \\
x&=\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}. \ _\square
\end{align}\]

## Proof 3: (by substitution)

Make the substitution \(x=y-\dfrac{b}{2a}\): \[a\left(y-\dfrac{b}{2a}\right)^2+b\left(y-\dfrac{b}{2a}\right)+c=0\\ ay^2-by+\dfrac{b^2}{4a}+by-\dfrac{b^2}{2a}+c=0\]

From here it's straightforward to solve for \(y\): \[ay^2-\dfrac{b^2}{4a}+c=0 \implies y^2=\dfrac{b^2-4ac}{4a^2} \implies y=\pm \dfrac{\sqrt{b^2-4ac}}{2a}\]

Finally undo the substitution: \[x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}. \ _\square\]

## Proof 4: (by Vieta's formula)

Let \(ax^2+bx+c=0\) have roots \(x_1,x_2\). Then, by vieta's formula we have: \[\begin{align} x_1+x_2&=-\dfrac{b}{a} \\ x_1x_2&=\dfrac{c}{a} \end{align}\]

Square both sides of the first equation and multiply by 4 the second one: \[\begin{align} x_1^2+2x_1x_2+x_2^2=\dfrac{b^2}{a^2} \\ 4x_1x_2=\dfrac{4c}{a} \end{align}\]

Then subtract them: \[x_1^2-2x_1x_2+x_2^2=\dfrac{b^2-4ac}{a^2}\]

We got a perfect square on the left side: \[(x_1-x_2)^2=\dfrac{b^2-4ac}{a^2} \implies x_1-x_2=\dfrac{\sqrt{b^2-4ac}}{a}\]

Finally, we have the sum and the difference of the roots. So: \[\begin{align} x_1&=\dfrac{-b+\sqrt{b^2-4ac}}{2a} \\ x_2&=\dfrac{-b-\sqrt{b^2-4ac}}{2a}. \ _ \square \end{align}\]

**Cite as:**Quadratic Formula.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/quadratic-formula/