Using the Quadratic Formula

The quadratic formula, is of the form \( x = \frac { - b \pm \sqrt{ b^2 - 4ac } } { 2a} .\) It is also known as Shreedhara Acharya's formula, named after the ancient Indian mathematician who derived it.

If we have a quadratic polynomial in the form \( ax^2 + bx + c ,\) then we can use the formula \( x = \frac { - b \pm \sqrt{ b^2 - 4ac } } { 2a} \) to find when it equals zero. (Note the plus-or-minus means there are two solutions, not just one.)

Suppose you have the polynomial function \( f(x) = 3x^2 - 2x + 3 .\) For what values of \(x\) does it equal 0?

We can let \( a = 3 ,\) \(b = -2 ,\) and \( c = -3 .\) Applying this to the quadratic formula, \( x = \frac { -2 \pm \sqrt{ 4 - 4(3)(3) } } { 2(3)} = \frac { -2 \pm \sqrt{ 40 } } { 6 } = \frac{-1}{3} \pm \frac{\sqrt{10}}{3}.\)

This can also aid in factorization, especially in cases where the roots of a polynomial are not rational:

Factorize \(x^2 + x - 1 = 0.\)

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Using our formula, we obtain \[\phi= \frac{-b+ \sqrt{b^2 - 4ac}}{2a} = \frac{-1 +\sqrt{5}}{2}, \Phi= \frac{-b - \sqrt{b^2 - 4ac}}{2a} = \frac{-1 - \sqrt{5}}{2} .\] Our quadratic is thus \[k(x-\phi)(x-\Phi) = 0\] for some constant \(k\). However, since the leading coefficient of our first term is 1, we know that the factorization must be \[(x-\phi)(x - \Phi) = \left(x - \frac{-1 +\sqrt{5}}{2} \right)\left(x- \frac{-1 - \sqrt{5}}{2}\right). \ _\square\]

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

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

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 \[\begin{align} (2ax+b)^{2}-b^{2}+4ac&=0 \\ (2ax+b)^{2}&=b^{2}-4ac \\ \sqrt{(2ax+b)^{2}}&=\pm \sqrt{b^{2}-4ac}\\ 2ax+b&=\pm \sqrt{b^{2}-4ac} \\ 2ax&=-b \pm \sqrt{b^{2}-4ac} \\ x&=\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}. \ _\square \end{align}\]

Make the substitution \(x=y-\dfrac{b}{2a}:\)

\[\begin{align} a\left(y-\dfrac{b}{2a}\right)^2+b\left(y-\dfrac{b}{2a}\right)+c&=0\\ ay^2-by+\dfrac{b^2}{4a}+by-\dfrac{b^2}{2a}+c&=0. \end{align}\]

From here it's straightforward to solve for \(y:\)

\[\begin{align} ay^2-\dfrac{b^2}{4a}+c&=0 \\ y^2&=\dfrac{b^2-4ac}{4a^2} \\ y&=\pm \dfrac{\sqrt{b^2-4ac}}{2a}. \end{align}\]

Finally undo the substitution:

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

Let \(ax^2+bx+c=0\) have roots \(x_1\) and \(x_2\). 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,

\[\begin{align} x_1&=\dfrac{-b+\sqrt{b^2-4ac}}{2a} \\\\ x_2&=\dfrac{-b-\sqrt{b^2-4ac}}{2a}. \ _ \square \end{align}\]