## Key Questions

• The quadratic formula is used to get the roots of a quadratic equation, if the roots exists at all.

We usually just perform factorization to get the roots of a quadratic equation. However, this is not always possible (especially when the roots are irrational)

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

Example 1:

$y = {x}^{2} - 3 x - 4$
$0 = {x}^{2} - 3 x - 4$

$\implies 0 = \left(x - 4\right) \left(x + 1\right)$
$\implies x = 4 , x = - 1$

Using the quadratic formula, let's try to solve the same equation

$x = \frac{- \left(- 3\right) \pm \sqrt[2]{{\left(- 3\right)}^{2} - 4 \cdot 1 \cdot \left(- 4\right)}}{2 \cdot 1}$
$\implies x = \frac{3 \pm \sqrt[2]{9 + 16}}{2}$
$\implies x = \frac{3 \pm \sqrt[2]{25}}{2}$
$\implies x = \frac{3 + 5}{2} , x = \frac{3 - 5}{2}$
$\implies x = 4 , x = - 1$

Example 2:

$y = 2 {x}^{2} - 3 x - 5$
$0 = 2 {x}^{2} - 3 x - 5$

Performing factorization is a little hard for this equation, so let's jump straight to using the quadratic formula

$x = \frac{- \left(- 3\right) \pm \sqrt[2]{{\left(- 3\right)}^{2} - 4 \cdot 2 \cdot \left(- 5\right)}}{2 \cdot 2}$

$x = \frac{3 \pm \sqrt[2]{9 + 40}}{4}$

$x = \frac{3 \pm \sqrt[2]{49}}{4}$

$x = \frac{3 + 7}{4} , x = \frac{3 - 7}{4}$

$x = \frac{5}{2} , x = - 1$

• Suppose that you have a function represented by $f \left(x\right) = A {x}^{2} + B x + C$.

We can use the quadratic formula to find the zeroes of this function, by setting $f \left(x\right) = A {x}^{2} + B x + C = 0$.

Technically we can also find complex roots for it, but typically one will be asked to work only with real roots. The quadratic formula is represented as:

$\frac{- B \pm \sqrt{{B}^{2} - 4 A C}}{2 A} = x$

... where x represents the x-coordinate of the zero.

If ${B}^{2} - 4 A C < 0$, we will be dealing with complex roots, and if ${B}^{2} - 4 A C \ge 0$, we will have real roots.

As an example, consider the function ${x}^{2} - 13 x + 12$. Here,

$A = 1 , B = - 13 , C = 12.$

Then for the quadratic formula we would have:

$x = \frac{13 \pm \sqrt{{\left(- 13\right)}^{2} - 4 \left(1\right) \left(12\right)}}{2 \left(1\right)}$ =

$\frac{13 \pm \sqrt{169 - 48}}{2} = \frac{13 \pm 11}{2}$

Thus, our roots are $x = 1$ and $x = 12$.

For an example with complex roots, we have the function $f \left(x\right) = {x}^{2} + 1$. Here $A = 1 , B = 0 , C = 1.$

$x = \frac{0 \pm \sqrt{{0}^{2} - 4 \left(1\right) \left(1\right)}}{2 \left(1\right)} = \pm \frac{\sqrt{- 4}}{2} = \pm i$

... where $i$ is the imaginary unit, defined by its property of ${i}^{2} = - 1$.

In the graph for this function on the real coordinate plane, we will see no zeroes, but the function will have these two imaginary roots.

• The discriminant is part of the quadratic formula.

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

Discriminant

${b}^{2} - 4 a c$

The discriminant tells you the number and types of solutions to a quadratic equation.

${b}^{2} - 4 a c = 0$, one real solution

${b}^{2} - 4 a c > 0$, two real solutions

${b}^{2} - 4 a c < 0$, two imaginary solutions

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
Negative b plus minus the square root of b squared minus 4*a*c over 2*a. To plug something into the quadratic formula the equation needs to be in standard form ($a {x}^{2} + b {x}^{2} + c$).