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):
If $ax^2+bx+c=0$ax2+bx+c=0, then:
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$x=−b±√b2−4ac2a
The advantage of using the quadratic formula is that it always works (unlike factoring) and it always follows the exact same process. However, the other methods can be more efficient in many cases.
The quadratic formula might seem quite complex when you first come across it, but it can be broken down into smaller parts.
Solve the equation $x^2-5x+6=0$x2−5x+6=0 by using the quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$x=−b±√b2−4ac2a.
Write each solution on the same line, separated by a comma.
Solve the equation $25x^2+30x=-9$25x2+30x=−9 using the quadratic formula.
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:
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!
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.
$b^2-4ac=0$b2−4ac=0, $1$1 real solution, $2$2 equal real roots, the quadratic just touches the $x$x-axis (it looks like it bounced off)
$b^2-4ac>0$b2−4ac>0, $2$2 real solutions, $2$2 distinct real roots, the quadratic passes through two different points on the $x$x-axis
$b^2-4ac<0$b2−4ac<0, $0$0 real solutions, $2$2 complex roots, the quadratic has no $x$x-intercepts
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.
Consider the equation $2x^2-2x=x-1$2x2−2x=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
Two real solutions
One real solution
Consider the equation $x^2+22x+121=0$x2+22x+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
Rational
Consider the equation $x^2+18x+k+7=0$x2+18x+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?