18.02 Solve by factorisations (Extended)
Solve quadratic equations using factorisation
- Gather together all the terms on one side of the equation so that the other side is $0$0.
- Factorise these terms using any appropriate techniques.
- Use the null factor law to split the quadratic equation into two linear equations.
- Solve the linear equations to get the solutions to the original quadratic equation.
Worked example
Solve $2x^2+4x+2=x^2-4x-13$2x2+4x+2=x2−4x−13 for $x$x.
Think: Since there are square terms in the equation this is a quadratic equation, so we can use the method above.
Do:
| $2x^2+4x+2$2x2+4x+2 | $=$= | $x^2-4x-13$x2−4x−13 |
|
| $x^2+8x+15=0$x2+8x+15=0 | $=$= | $0$0 |
Gathering all of the terms on the left hand side and simplifying |
| $\left(x+5\right)\left(x+3\right)$(x+5)(x+3) | $=$= | $0$0 |
Factorising the quadratic expression on the left hand side Note that $5+3=8$5+3=8 and $5\times3=15$5×3=15 |
Now we use the null factor law to split this quadratic equation into two linear equations.
| $x+5$x+5 | $=$= | $0$0 |
Using one factor |
| $x$x | $=$= | $-5$−5 |
Solving for $x$x |
| $x+3$x+3 | $=$= | $0$0 |
Using the other factor |
| $x$x | $=$= | $-3$−3 |
Solving for $x$x |
So the solutions are $x=-5$x=−5 and $x=-3$x=−3.
We can solve many quadratic equations by using this method:
- Gather together all the terms on one side of the equation so that the other side is $0$0.
- Factorise these terms using any appropriate techniques.
- Use the null factor law to split the quadratic equation into two linear equations.
- Solve the linear equations to get the solutions to the original quadratic equation.
Practice questions
Question 1
The equation $2y-6y^2=0$2y−6y2=0 has two solutions. Solve for both values of $y$y, writing your answers on the same line separated by a comma.
Question 2
Solve $x^2=13x+114$x2=13x+114 for $x$x.
Enter each solution on the same line, separated by a comma.
Question 3
Solve $x^2+13x+42=0$x2+13x+42=0 for $x$x.
Enter each solution on the same line, separated by a comma.