2.05 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:
- Algebraic manipulation
- Factoring and using the null factor law
- Completing the square
The fourth & final method for finding solutions to a quadratic equation is to use the quadratic formula.
The quadratic formula can be derived by completing the square for the general equation $ax^2+bx+c=0$ax2+bx+c=0.
The advantage of using the quadratic formula is that it always works (unlike factoring) and it always follows the exact same process. However, other methods may be more efficient in many cases, depending on the quadratic.
The quadratic formula might seem quite complex, 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.
Before we can use the quadratic formula, we have to rearrange the quadratic equation into the form
$ax^2+bx+c=0$ax2+bx+c=0
where $a$a, $b$b, and $c$c are any number and $a\ne0$a≠0.
Once the equation is in this form, the solutions are given by the quadratic formula:
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$x=−b±√b2−4ac2a
Worked example
Solve $x^2+9x-7=0$x2+9x−7=0 for $x$x.
Think: There are no integers which add to give $9$9 and multiply to give $-7$−7 so we cannot factorise the left hand side.
However, we already have the equation in the right form to use the quadratic formula. Here, $a=1$a=1, $b=9$b=9 and $c=-7$c=−7.
Do: We can substitute these values into the quadratic formula.
| $x$x | $=$= | $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$−b±√b2−4ac2a |
The quadratic formula |
| $=$= | $\frac{-9\pm\sqrt{9^2-4\times1\times\left(-7\right)}}{2\times1}$−9±√92−4×1×(−7)2×1 |
Substituting $a=1$a=1, $b=9$b=9, and $c=-7$c=−7 |
|
| $=$= | $\frac{-9\pm\sqrt{109}}{2}$−9±√1092 |
Simplifying the expression |
So the solutions to the equation are $x=\frac{-9\pm\sqrt{109}}{2}$x=−9±√1092
Reflect: Notice that we ended up with a surd. This means that the solutions to this equation are irrational. This is why we could not solve the equation by factorising.
For a quadratic equation of the form
$ax^2+bx+c=0$ax2+bx+c=0
The solutions are given by the quadratic formula:
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$x=−b±√b2−4ac2a
Practice questions
Question 1
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.
Question 2
Solve the following equation: $-6-13x+5x^2=0$−6−13x+5x2=0.
Write all solutions on the same line, separated by commas.
Question 3
Solve $10-6m+2m^2=m^2+8m+9$10−6m+2m2=m2+8m+9 for $m$m by using the quadratic formula or otherwise.
Enter each solution as a surd on the same line, separated by a comma.