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

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

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

From here it's straightforward to solve for $y:$

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

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

$$\begin{aligned} x_1+x_2&=-\dfrac{b}{a} \\ x_1x_2&=\dfrac{c}{a}. \end{aligned}$$

Square both sides of the first equation and multiply by 4 the second one:

$$\begin{aligned} x_1^2+2x_1x_2+x_2^2=\dfrac{b^2}{a^2} \\ 4x_1x_2=\dfrac{4c}{a}. \end{aligned}$$

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,

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