Manipulating expressions in the rootsEdexcel International A Level Further Maths: Revision notes

For $ax^2+bx+c=0$ with roots $\alpha$ and $\beta$, factorising $a(x-\alpha)(x-\beta)=0$ and comparing coefficients gives $\alpha+\beta=-\frac{b}{a},\qquad \alpha\beta=\frac{c}{a}.$ These two values are all you need: any symmetric expression in $\alpha$ and $\beta$ (unchanged when $\alpha$ and $\beta$ are swapped) can be found without solving the equation. For $2x^2+5x-3=0$: $\alpha+\beta=-\frac52$ and $\alpha\beta=-\frac32$.

Expand $(\alpha+\beta)^2$ to rearrange: $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta.$ Similarly $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$. For fractions, combine over a common denominator: $\frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta},\qquad \frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}.$ Example: for $x^2-6x+4=0$, $\alpha^2+\beta^2=36-8=28$ and $\frac1\alpha+\frac1\beta=\frac64=\frac32$.

Expand $(\alpha+\beta)^3=\alpha^3+\beta^3+3\alpha\beta(\alpha+\beta)$ to obtain $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta).$ Higher powers are built in stages. For fourth powers, first find $\alpha^2+\beta^2$ and then $\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2(\alpha\beta)^2.$ Example: $x^2-5x+2=0$ gives $\alpha^3+\beta^3=125-3(2)(5)=95$, and $\alpha^2+\beta^2=21$, so $\alpha^4+\beta^4=441-8=433$.

Rewrite the target using $\alpha+\beta$ and $\alpha\beta$ only.

• $\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}=\frac{\alpha^3+\beta^3}{\alpha\beta}$.
• $(\alpha+1)(\beta+1)=\alpha\beta+(\alpha+\beta)+1$.
• $\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{\alpha^2+\beta^2}{(\alpha\beta)^2}$.
• $|\alpha-\beta|=\sqrt{(\alpha+\beta)^2-4\alpha\beta}$, taking the positive root.

If a question gives a condition such as $\alpha^2+\beta^2=40$, form an equation in the unknown constant first. For $x^2-8x+k=0$: $64-2k=40$, so $k=12$, and then $\alpha^3+\beta^3=512-288=224$.