The three equations below are considered quadratic equations due to the highest index of $2$2 on any x term.
$x^2=25$x2=25
$2x^2-4x=4$2x2−4x=4
$x^2+5x-6=0$x2+5x−6=0
There are a number of ways to solve quadratic equations. The solutions to quadratic equations are also called the roots of the equation - we will cover this language in more detail in a later section.
The four methods we have seen are:
$x^2$x2 | $=$= | $49$49 |
$x$x | $=$= | $\pm7$±7 |
$x^2-2x-3$x2−2x−3 | $=$= | $0$0 |
$(x+1)(x-3)$(x+1)(x−3) | $=$= | $0$0 |
$x$x | $=$= | $-1,3$−1,3 |
When looking to solve a quadratic, we should start by checking the simplest options:
Some quadratics might look like they are factorised and ready to be solved using the null factor law, but have additional terms on the other side - such as the equation $\left(x-3\right)\left(x-4\right)=2$(x−3)(x−4)=2.
The null factor law only works if the other side of the equation is zero!
Solve for $x$x, expressing your answer in exact form.
$\left(x-5\right)^2-4=8$(x−5)2−4=8
Write all solutions on the same line, separated by commas.
Solve for $x$x:
$x^2=17x+60$x2=17x+60
Write all solutions on the same line, separated by commas.
Solve the equation $x\left(x+2\right)=48$x(x+2)=48.
Write all solutions on the same line, separated by commas.
If the quadratic is not easily factorisable, then we can either complete the square or use the quadratic formula.
The video in the following example demonstrates one method for completing the square.
Solve the quadratic equation $x^2+4x-5=0$x2+4x−5=0 by first completing the square.
Think: The following video shows how we can complete the square for the expression $x^2+4x-5$x2+4x−5.
Do: We can now solve this algebraically.
$x^2+4x-5$x2+4x−5 | $=$= | $0$0 |
$\left(x+2\right)^2-9$(x+2)2−9 | $=$= | $0$0 |
$\left(x+2\right)^2$(x+2)2 | $=$= | $9$9 |
$x+2$x+2 | $=$= | $\pm3$±3 |
$x$x | $=$= | $3-2,-3-2$3−2,−3−2 |
$x$x | $=$= | $1,-5$1,−5 |
We can also complete the square using an algebraic method. The steps involved in this are:
For this example:
$x^2+4x-5$x2+4x−5 | $=$= | $0$0 |
$x^2+4x$x2+4x | $=$= | $5$5 |
$x^2+4x+4$x2+4x+4 | $=$= | $5+4$5+4 |
$\left(x+2\right)^2$(x+2)2 | $=$= | $9$9 |
$x+2$x+2 | $=$= | $3-2,-3-2$3−2,−3−2 |
$x+2$x+2 | $=$= | $\pm3$±3 |
$x$x | $=$= | $3-2,-3-2$3−2,−3−2 |
$x$x | $=$= | $1,-5$1,−5 |
We might come across equations with fractions that don't look like quadratic equations at first glance. For example, the equations $x-4=-\frac{6}{x}$x−4=−6x and $\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2}$1x−1x+1=12 don't immediately look like quadratic equations. If we multiply each term by the lowest common denominator, however, we can see the true quadratic nature of the equation. For example, if we multiply both sides of $x-4=-\frac{6}{x}$x−4=−6x by $x$x we get:
$x^2-4x$x2−4x | $=$= | $-6$−6 |
$x^2-4x+6$x2−4x+6 | $=$= | $0$0 |
We can solve this using one of our current methods.
Solve the following quadratic equation by completing the square:
$x^2+18x+32=0$x2+18x+32=0
Solve the equation $4x^2+6x+2=0$4x2+6x+2=0 using the quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$x=−b±√b2−4ac2a.
Enter each solution on the same line, separated by a comma.
Solve the following equation:
$\frac{12+11x}{5x}=x$12+11x5x=x
Write all solutions on the same line, separated by commas.