4.04 The discriminant and the quadratic formula
The quadratic formula
There are three main ways to solve a quadratic equation (i.e. an equation of the form $ax^2+bx+c=0$ax2+bx+c=0):
- By basic algebra - using inverse operations
- By factoring - using the zero product property
- By using the quadratic formula
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. The only downside to using it is that it gives no insight into the problem you are trying to solve, hence why teachers will often suggest other methods when solving quadratic equations.
Nevertheless, to see why using the quadratic formula is so efficient, refer to the following demonstration. It shows how any quadratic equation can be solved no matter what the values of $a$a, $b$b and $c$c are. It also allows you to see how the answer is affected when each of the values change. (To change the values, just move the sliders.)
The quadratic formula might seem quite complex when you first come across it, but it can be broken down into smaller parts:
- 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.
Worked example
Question 1
Find the $x$x-intercepts (zeros) of the graph $y=2x^2-3x+1$y=2x2−3x+1 using the quadratic formula.
Think: The $x$x-intercepts are found by substituting $y=0$y=0. This results in a quadratic equation $2x^2-3x+1=0$2x2−3x+1=0. Identify $a$a,$b$b and $c$c for the formula.
$a=2$a=2 and $b=-3$b=−3 and $c=1$c=1
Do: Once we have identified these values we can substitute them into the quadratic formula:
| $x$x | $=$= | $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$−b±√b2−4ac2a |
| $=$= | $\frac{-\left(-3\right)\pm\sqrt{\left(-3\right)^2-4\times2\times1}}{2\times2}$−(−3)±√(−3)2−4×2×12×2 | |
| $=$= | $\frac{3\pm\sqrt{9-8}}{4}$3±√9−84 | |
| $=$= | $\frac{3\pm\sqrt{1}}{4}$3±√14 | |
| $=$= | $\frac{3\pm1}{4}$3±14 |
The last line gives $x=\frac{4}{4}$x=44 and $x=\frac{2}{4}$x=24. So, for the quadratic $y=2x^2-3x+1$y=2x2−3x+1 the zeros are $x=1$x=1 and $x=\frac{1}{2}$x=12.
The discriminant
We have seen when solving quadratic equations that there can be two, one or zero 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 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 distinct real solutions: these are the two zeros or $x$x-intercepts where the quadratic passes through the $x$x-axis.
- One distinct real solution with a multiplicity of two: 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 used many techniques for 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±√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.
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 $b^2-4ac<0$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
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? This is because sometimes the solution are imaginary (an unfortunately confusing name, as they are just as real as real numbers! They just make up the other half of the set of complex numbers.)
Practice question
Question 2
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
ATwo real solutions
BOne real solution
C
Writing solutions in complex form
To be able to write some numbers in complex form a little algebraic manipulation may be necessary, mostly involving the fact that $\sqrt{-1}=i$√−1=i or that $-1=i^2$−1=i2.
When we combine this algebraic manipulation with the use of our quadratic manipulations using the Quadratic Formula, we can find both real and complex solutions to quadratic equations.
If the discriminant is negative, then we employ complex number simplification processes.
Worked example
Question 3
Solve $x^2-2x+3=0$x2−2x+3=0, using the quadratic formula
| $x$x | $=$= | $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$−b±√b2−4ac2a |
| $x$x | $=$= | $\frac{-\left(-2\right)\pm\sqrt{\left(-2\right)^2-4\left(1\right)\left(3\right)}}{2\left(1\right)}$−(−2)±√(−2)2−4(1)(3)2(1) |
| $x$x | $=$= | $\frac{2\pm\sqrt{4-12}}{2}$2±√4−122 |
| $x$x | $=$= | $\frac{2\pm\sqrt{-8}}{2}$2±√−82 |
| $x$x | $=$= | $\frac{2\pm2\sqrt{-2}}{2}$2±2√−22 |
| $x$x | $=$= | $1\pm\sqrt{-2}$1±√−2 |
| $x$x | $=$= | $1\pm\sqrt{-1}\sqrt{2}$1±√−1√2 |
| $x$x | $=$= | $1\pm i\sqrt{2}$1±i√2 |
| $x$x | $=$= | $1\pm\sqrt{2}i$1±√2i |
Practice questions
Question 4
Solve the following equation by using the quadratic formula.
$x\left(x-4\right)=5$x(x−4)=5
Question 5
Consider the equation $x\left(x+9\right)=-20$x(x+9)=−20.
Solve it by the method of factoring. Do not use the quadratic formula.
Write all solutions on the same line, separated by commas.
Check your solution by solving it using the quadratic formula.
QUESTION 6
Below is the result after using the quadratic formula to solve an equation.
$x=\frac{-3\pm\sqrt{9-4\cdot56}}{16}$x=−3±√9−4·5616
Which of the following is true?
The equation has one real solution.
AThe equation has two real solutions.
BThe equation has complex solutions.
C