Solutions for equations are found when an expression evaluates to zero. One method for solving a quadratic equation is by factorising. After factorising, find the value of `x` that makes each bracket zero.
Rearrange the equation so that one side of the equation is equal to zero. Next, factorise the expression. Then find the two values of `x` that will solve the equation. In some instances the two values may be the same - `x^2+6x+9` factorises to `(x+3)(x+3)` which gives you solutions of -3 and -3.
Note that not all quadratic expressions have a solution: for example, `x^2-3x+7` cannot be solved except by using advanced mathematics.
Solve `x^2-4x-5=0`
`x^2-4x-5` = 0
Factorise the expression:
`(x+1)(x-5) = 0`
Determine what values of `x` causes each set of brackets to equal zero:
`(x+1)=0` when `x=-1`, and `(x-5) =0` when `x=5`
The solutions are `x=-1 text( and ) x=5`
Check:
(-1)2 - 4(-1) - 5 =0✔
(502 - 4(5) - 5 = 0✔
Answer: `x=-1 text( and ) x=5`
Solve `2x^2-7x=4`
`2x^2-7x=4`
Rearrange the equation so that one side of the equation equals zero:
`2x^2 - 7x - 4 = 0`
Factorise the equation:
`(2x+1)(x-4)=0`
What values of `x` will make each bracket zero?
`(2x+1)=0` when `x=-0.5`
and `(x-4) =0` when `x=4`
The solutions are therefore `x=-0.5 text( and ) x=4`
Check:
2(-0.5)2 - 7(-0.5) = 4✔
2(4)2 - 7(4) = 4✔
Answer: `x=-frac(1)(2) text( and ) x=4`