Quadratic Equations

Hong Kong

Stage 4 - Stage 5

Lesson

A quadratic root is a value of $x$`x` that is a solution to the quadratic equation $ax^2+bx+c=0$`a``x`2+`b``x`+`c`=0. The roots of this quadratic are also the same as the zeros of the function $y=ax^2+bx+c$`y`=`a``x`2+`b``x`+`c`. Graphically, this where the quadratic crosses the $x$`x`-axis. But just how are the roots of equations related to the coefficients and constants of the equation itself?

We've already looked at how to find the roots of a factorised quadratic using the null factor law. Now we are going to learn how to find the roots of an expanded quadratic.

If we define the two roots of the quadratic to be the values $\alpha$`α` and $\beta$`β`, then the sum and difference of these roots can be found using the following formulae:

Sum of roots: $\alpha+\beta=\frac{-b}{a}$`α`+`β`=−`b``a`

Product of roots: $\alpha\beta=\frac{c}{a}$`α``β`=`c``a`

Where does this come from? If we let $\alpha$`α` and $\beta$`β` be the roots of the equation $ax^2+bx+c=0$`a``x`2+`b``x`+`c`=0, then

$x^2+\frac{b}{a}x+\frac{c}{a}$x2+bax+ca |
$=$= | $\left(x-\alpha\right)\left(x-\beta\right)$(x−α)(x−β) |

$=$= | $x^2-\left(\alpha+\beta\right)x+\alpha\beta$x2−(α+β)x+αβ |

By equating the coefficients on either side of the equation, we can see that:

$\alpha+\beta=\frac{-b}{a}$`α`+`β`=−`b``a` and $\alpha\beta=\frac{c}{a}$`α``β`=`c``a`

Interestingly, we can use the sum and the product of the roots of a quadratic equation to determine some characteristics of the quadratic *without finding the actual roots!*

$x=6$`x`=6 and $x=9$`x`=9 are the roots of a quadratic equation.

What is the equation in factored form?

What is the equation in expanded form?

Consider the equation $x^2+mx+18=0$`x`2+`m``x`+18=0

Find the sum of the roots in terms of $m$

`m`.Find the product of the roots.

Find the value of $m$

`m`when the product of the roots is equal to $6$6 times the sum.