Solving quadratic equations by factorisation
Solving quadratic equations by factorisation
A quadratic equation can be written as \(ax^2+bx+c=0\), where \(a\ne0\). Its solutions are called roots. To solve it, factorise the left side and use the zero-product principle: if \(PQ=0\), then \(P=0\) or \(Q=0\).
Why: if two numbers multiply to give \(0\), at least one of them must be \(0\); for example \(5\times0=0\), but \(5\times2\ne0\). So each factor gives one equation that is easy to solve. The principle needs a product equal to \(0\); \(PQ=6\) tells us nothing about \(P\) or \(Q\) on its own.
| Equation | Factorised | Roots |
|---|---|---|
| \(x^2-5x-14=0\) | \((x-7)(x+2)=0\) | \(x=7\) or \(x=-2\) |
| \(2x^2-6x=0\) | \(2x(x-3)=0\) | \(x=0\) or \(x=3\) |
| \(4x^2-9=0\) | \((2x+3)(2x-3)=0\) | \(x=-\dfrac{3}{2}\) or \(x=\dfrac{3}{2}\) |
| \(x^2+10x+25=0\) | \((x+5)^2=0\) | \(x=-5\) (one repeated root) |
Method
- Make sure one side of the equation is \(0\).
- Factorise the other side completely.
- Set each factor that contains \(x\) equal to \(0\) and solve.
- Check each root by substituting it into the original equation.
Example
Solve \(x^2-5x-14=0\).
\(\begin{aligned} x^2-5x-14 &= 0 \\ (x-7)(x+2) &= 0 \\ x-7=0 \quad &\text{or}\quad x+2=0 \\ x=7 \quad &\text{or}\quad x=-2 \end{aligned}\)
Check: \(7^2-5(7)-14=49-35-14=0\) and \((-2)^2-5(-2)-14=4+10-14=0\). A common error is to divide \(2x^2-6x=0\) by \(x\); that gives only \(x=3\) and loses the root \(x=0\). Factorise instead.
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