Quadratic Equations

Hong Kong

Stage 4 - Stage 5

Lesson

We have seen when solving quadratic equations that there can be two, one or no solutions. If we think about the graphs of quadratics, this means that there can be two, one or no $x$`x`-intercepts. This is because the solutions to a quadratic equation correspond to the values of $x$`x` that we find when we set $y=0$`y`=0 in an equation, and these are the places where a function crosses the $x$`x`-axis.

Looking at the image above, we can see that a quadratic equation can have either:

- Two real solutions: these are the two zeros or $x$
`x`-intercepts where the quadratic passes through the $x$`x`-axis. - One real solution: where the two zeros are actually equal, i.e. the one $x$
`x`-intercept where the quadratic just touches the $x$`x`-axis at the turning point. - No real solutions: meaning there are no $x$
`x`-intercepts or real zeros.

We have revised many techniques fos solving quadratic equations, and obviously, if we have found the actual solutions we can answer the question of how many roots exist. But there is a quicker way to answer the question without working through all of the algebra required!

Let's look again at the quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$`x`=−`b`±√`b`2−4`a``c`2`a`

Specifically, let's look at what happens if the square root part $\sqrt{b^2-4ac}$√`b`2−4`a``c` takes on different values...

$b^2-4ac<0$`b`2−4`a``c`<0 or $b^2-4ac=0$`b`2−4`a``c`=0 or $b^2-4ac>0$`b`2−4`a``c`>0

If $b^2-4ac=0$`b`2−4`a``c`=0, then the square root is $0$0 and then the quadratic equation becomes just $x=\frac{-b}{2a}$`x`=−`b`2`a`. Does this look familiar? It is actually the equation for the axis of symmetry.

If $b^2-4ac>0$`b`2−4`a``c`>0, then the square root will have two values, one for $+$+$\sqrt{b^2-4ac}$√`b`2−4`a``c` and one for $-\sqrt{b^2-4ac}$−√`b`2−4`a``c`. The quadratic formula will then generate for us two distinct real roots.

If $b^2-4ac<0$`b`2−4`a``c`<0, then the square root is negative, and we know that we cannot take the square root of a negative number and get real solutions. This is the case where we have zero real roots.

This expression $b^2-4ac$`b`2−4`a``c` within the quadratic formula is called the discriminant, and it** **determines the number of real solutions a quadratic function will have.

Discriminant of a Quadratic

$b^2-4ac=0$`b`2−4`a``c`=0, $1$1 real solution, $2$2 equal real roots, the quadratic just touches the $x$`x`-axis (it looks like it bounced off)

$b^2-4ac>0$`b`2−4`a``c`>0, $2$2 real solutions, $2$2 distinct real roots, the quadratic passes through two different points on the $x$`x`-axis

$b^2-4ac<0$`b`2−4`a``c`<0, $0$0 real solutions, $2$2 complex roots, the quadratic has no $x$`x`-intercepts

In our introduction to polynomials we saw that for a polynomial of degree $n$`n`, there will be $n$`n` solutions. Since quadratics are polynomials of degree $2$2, we should always be expecting two solutions. So how is it that we can have one or even zero real solutions?

The key here is the *type* of solutions we care about. At this stage we can be content talking only about solutions that are real numbers. But there are other types of numbers that are not real, called imaginary numbers. (An unfortunately confusing name, as they are just as *real* as real numbers!)

The set of real numbers and the set of imaginary numbers combine to make up the set of complex numbers, and it is the complex numbers that will ensure we have $n$`n` roots for any polynomial degree $n$`n`. This is a very exciting area of maths, but we'll save that for later. For now let's get back to *real*ity!

How many real solutions are there to the equation $-\left(x-10\right)^2=5$−(`x`−10)2=5?

More than $2$2 real solutions

ANo real solutions

B$1$1 real solution

C$2$2 real solutions

D

Consider the equation $x^2+22x+121=0$`x`2+22`x`+121=0.

Find the value of the discriminant.

Using your answer from the previous part, determine whether the solutions to the equation are rational or irrational.

Irrational

ARational

B

Consider the equation $x^2+18x+k+7=0$`x`2+18`x`+`k`+7=0.

Find the values of $k$

`k`for which the equation has no real solutions.If the equation has no real solutions, what is the smallest integer value that $k$

`k`can have?

Consider the equation $2x^2-2x=x-1$2`x`2−2`x`=`x`−1.

Find the value of the discriminant.

Using your answer from the previous part, determine the number of real solutions the equation has.

No real solutions

ATwo real solutions

BOne real solution

C