4.06 Discriminant and parabolas

Quadratic equations and parabolas

We have seen when solving quadratic equations that there can be two, one or no real solutions. Let's think about the graphs of quadratics:

No solutions	One solution	Two solutions

 

As shown above there are only three possibilities, with respect to $x$x intercepts, when graphing quadratics. These possibilities are as follows:

• No real solutions. This means there are no $x$x-intercepts or real zeros.  
• One real solution. This is where the two zeros are actually equal, and the parabola has one $x$x-intercept where it just touches the $x$x-axis at the turning point.
• Two real solutions. These are the two distinct zeros or $x$x-intercepts, where the quadratic passes through the $x$x-axis.

The solutions to a quadratic equation correspond to the $x$x values that occur when $y=0$y=0 in a quadratic function, and these are the places where a function crosses the $x$x-axis.

 

The discriminant

We have revised a range of algebraic techniques to solve quadratic equations, and obviously, if we are able to find these actual solutions, we can answer the question of how many solutions or roots a quadratic has. But there is a faster way!

Let's look again at the quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$x=−b±√b2−4ac2a​

Specifically, let's look at what happens if the square root part $\sqrt{b^2-4ac}$√b2−4ac takes on different values...

$b^2-4ac<0$b2−4ac<0  or  $b^2-4ac=0$b2−4ac=0  or  $b^2-4ac>0$b2−4ac>0 

• If $b^2-4ac<0$b2−4ac<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.
• If $b^2-4ac=0$b2−4ac=0, then the square root is $0$0 and then the quadratic equation becomes just $x=\frac{-b}{2a}$x=−b2a​. and we have a single root. This is actually the equation for the axis of symmetry.
• If $b^2-4ac>0$b2−4ac>0, then the square root will have two values, one for $+\sqrt{b^2-4ac}$+√b2−4ac and one for $-\sqrt{b^2-4ac}$−√b2−4ac. The quadratic formula will then generate for us two distinct real roots. If this value is a square number, we will have two rational roots, seeing as the square root of a square number will lead us with a rational result. For any other positive number, we will end up with two roots with surds in them. 

This expression $b^2-4ac$b2−4ac within the quadratic formula is called the discriminant, and it determines the number of real solutions a quadratic function will have. The symbol $\Delta$Δ is used as a shortcut for$b^2-4ac$b2−4ac .

 

 

Remember that every quadratic function can be sketched, even if it has no real roots. Every quadratic function is a parabola and has a vertex, which means that it has either a maximum or minimum value which can be found. Knowing the value of the discriminant, and the value of $a$a in $y=ax^2+bx+c$y=ax2+bx+c which determines its concavity, can give us enough information for a rough sketch of the parabola.

 

Practice questions

Question 1

Question 2

Question 3

Question 4

Simultaneous equations involving quadratic functions

Having revised our algebraic techniques for simultaneous equations, we will now think about how this connects with our graphical understanding of functions. As with all algebraic expressions, simultaneous equations can also be shown as graphs on a number plane. Each graph of a function represents all of the possible solutions of that equation. So, when we are finding a solution that solves both of our simultaneous equations, we are looking for the point of intersection of the two graphs.

When we have two linear equations (or straight lines), they will only have one point of intersection. However, when we have a quadratic (parabola) and a linear equation (straight line), there can be:

• two points of intersection

 

• one point of intersection (So the line is a tangent to the parabola. A tangent is a straight line that touches a curve only once.)

• or even no points of intersection (meaning there are no solutions)

 

If we were to solve two quadratic equations simultaneously, we could similarly have $0$0, $1$1 or $2$2 solutions. Can you visualise these three scenarios too?

 

Worked example

example 1

Show that the line $y=4x-8$y=4x−8 is a tangent to the parabola $y=x^2-4$y=x2−4.

Think: To show that the line is a tangent to the parabola, we must show that the functions only intersect once.

Do: To find the point of intersection we can use the substitution method to substitute the first equation into the second to get: 

$4x-8$4x−8	$=$=	$x^2-4$x2−4
$0$0	$=$=	$x^2-4x+4$x2−4x+4
$x^2-4x+4$x2−4x+4	$=$=	$0$0

 

Now we can either solve this quadratic equation to show that there is only one point of intersection. Or we can use the discriminant to show that this quadratic only has one solution. For this lesson, we will use the discriminant:

$\triangle$△	$=$=	$b^2-4ac$b2−4ac
	$=$=	$(-4)^2-4\times1\times(4)$(−4)2−4×1×(4)
	$=$=	$16-16$16−16
	$=$=	$0$0

 

Since the discriminant equals zero, we know that the quadratic equation $x^2-4x+4=0$x2−4x+4=0 has only one solution. Since that equation came from solving $y=4x-8$y=4x−8 and $y=x^2-4$y=x2−4 simultaneously, we know the simultaneous equations only have one solution, and therefore the corresponding functions only have one point of intersection. Therefore the line $y=4x-8$y=4x−8 is a tangent to the parabola $y=x^2-4$y=x2−4. 

 

Practice questions

Question 5

Question 6

Question 7