1.05 Quadratic formula

There are four ways to solve a quadratic equation (i.e. an equation of the form $ax^2+bx+c=0$ax2+bx+c=0): 

• 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

As we have seen with quadratic equations of the form  $ax^2+bx+c=0$ax2+bx+c=0 there can be two, one or no real 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 function in the form $y=ax^2+bx+c$y=ax2+bx+c are the places where the function crosses the $x$x-axis, hence the $y$y value is $0$0.

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 graph of the quadratic function 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 graph of the quadratic function just touches the $x$x-axis at the turning point.
• No real solutions, meaning the graph of the quadratic function has no $x$x-intercepts or real zeros.  

 

An aside

If we were to study this area of mathematics further, we would eventually learn of a theorem about polynomials that says 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 maths, but we'll save that for later. For now let's get back to reality!

 

The discriminant - 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 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 $0$0 and then the quadratic equation becomes just $x=\frac{-b}{2a}$x=−b2a​. Does this look familiar? It is actually the equation for the axis of symmetry. It is just one number so there is just one real root (solution).

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 for us two distinct real roots.

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.

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.

 

 

Rational or irrational solutions?

Remember that square roots of perfect squares like $\sqrt{9}$√9 are rational because they can be written as fractions such as $\frac{3}{1}$31​. Numbers like $\sqrt{7}$√7 are irrational because they cannot be written as a fraction of whole numbers. Therefore if the discriminant is equal to a perfect square we know the quadratic equation has rational solutions. If it is not equal to a perfect square then the solutions must be irrational. 

 

Practice questions

Question 3

Question 4

Question 5