2.04 The quadratic formula

There are four ways to solve a quadratic equation, that is, an equation of the form $ax^2+bx+c=0$ax2+bx+c=0. Three methods that have already been covered in this chapter include:

• The ± allows for the possibility of two solutions.
• The $b^2-4ac$b2−4ac under the square root sign is important as it will tell us how many solutions there are. This is known as the discriminant.

 

Practice questions

Question 1

Question 2

 

Properties of the discriminant

Quadratic functions can have two, one or no real solutions. Thinking about the graph of a quadratic, these solutions correspond to two, one or no $x$x-intercepts, respectively. This is because the solutions to a function $f(x)=0$f(x)=0 are the places where the function crosses the $x$x-axis.

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

• No real solutions: there are no $x$x-intercepts or real zeros.  
• One real solution: here, the two zeros are actually equal and correspond to the one $x$x-intercept where the quadratic just touches the $x$x-axis at the turning point.
• Two real solutions: these are the two zeros or $x$x-intercepts, where the quadratic passes through the $x$x-axis.

 

Sometimes, it gets complex....

If we were to study this area of mathematics further, we would eventually learn of a theorem about polynomials that states 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 will be talking only about real solutions - solutions that are real numbers. But there are other types of numbers that are not real, in fact, they are called imaginary numbers. (An unfortunately confusing name, 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 are ultimately the roots of polynomials. This is a very exciting area of mathematics, but we'll save that for later. For now, let's get back to reality!

 

Finding the number of solutions

Finding the number of solutions to a quadratic, or if there are any solutions at all, can be done without having to work through all the algebra required to solve the function. 

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 expression under the square root $\sqrt{b^2-4ac}$√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, the square root is $0$0 and the quadratic equation becomes $x=\frac{-b}{2a}$x=−b2a​. Does this look familiar? It 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,  $+$+$\sqrt{b^2-4ac}$√b2−4ac and $-\sqrt{b^2-4ac}$−√b2−4ac. The quadratic formula will then generate two distinct real roots.

If $b^2-4ac<0$b2−4ac<0, then the square root is negative, and it is impossible to take the square root of a negative number and get real solutions. The quadratic formula will then generate zero real roots.

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.

 

 

Practice question

Question 3

 

Solving quadratic problems with additional unknowns

Since the discriminant can predict the number of solutions for a quadratic, can it also be used to create a group of quadratics that are guaranteed to have a certain number of solutions? This is possible if an unknown constant is introduced and values found for this constant that lead to the discriminate $b^2-4ac$b2−4ac being either $<0$<0, $<0$<0 or $=0$=0, depending on the number of solutions needed.

 

Worked Example 

Example 1

Find the values for which the quadratic $x^2+kx+4=0$x2+kx+4=0 has two solutions.

Think: Consider the value that the discriminant must take for two solutions to be generated by a quadratic equation. 

Do:  The discriminant $b^2-4ac$b2−4ac for this quadratic would need to be $>0$>0 for two real solutions to exist. Here, $a=1$a=1, $b=k$b=k and $c=4$c=4

Substitute these values into the discriminant and set this to be greater than zero:

$k^2-4\times1\times4$k2−4×1×4	$>$>	$0$0
$k^2-16$k2−16	$>$>	$0$0
$k^2$k2	$>$>	$16$16

Solving this last line for $k$k gives the solutions $k>4$k>4 or $k<-4$k<−4. This means that the range of values that $k$k can take so that the quadratic $x^2+kx+4=0$x2+kx+4=0 has two solutions are any real values of $k$k such that $k<-4$k<−4 or $k>4$k>4.

 

Practice question

Question 4