Forming quadratic equations with new rootsEdexcel International A Level Further Maths: Revision notes

Section 1

The method

A quadratic with roots $p$ and $q$ is $x^2-(p+q)x+pq=0$ . To form an equation whose roots are functions of $\alpha$ and $\beta$ :

Find $\alpha+\beta=-\frac ba$ and $\alpha\beta=\frac ca$ for the original equation.
Work out the new sum $p+q$ and new product $pq$ in terms of these.
Write $x^2-(p+q)x+pq=0$ and, if asked for integer coefficients, multiply through to clear fractions.

You never need to find $\alpha$ and $\beta$ themselves.

Key termsnew sumnew product

Common mistake

Using the original sum and product in the new equation, or leaving the minus sign out of $x^2-(p+q)x+pq=0$ .

Section 2

Reciprocal roots

For roots $\frac1\alpha$ and $\frac1\beta$ : $\text{sum}=\frac{\alpha+\beta}{\alpha\beta},\qquad \text{product}=\frac{1}{\alpha\beta}.$ Example: $2x^2-7x+3=0$ has $\alpha+\beta=\frac72$ , $\alpha\beta=\frac32$ . New sum $=\frac73$ , new product $=\frac23$ , so $3x^2-7x+2=0$ . For roots $\frac{1}{2\alpha}$ and $\frac{1}{2\beta}$ include the factor of $2$ : the sum is $\frac{\alpha+\beta}{2\alpha\beta}$ and the product is $\frac{1}{4\alpha\beta}$ .

Key termsreciprocal roots

Exam tip

Check: reversing the coefficients of $ax^2+bx+c=0$ gives $cx^2+bx+a=0$ , whose roots are $\frac1\alpha$ and $\frac1\beta$ .

Section 3

Powers of the roots

For roots $\alpha^2,\beta^2$ : sum $=(\alpha+\beta)^2-2\alpha\beta$ , product $=(\alpha\beta)^2$ . For roots $\alpha^3,\beta^3$ : sum $=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)$ , product $=(\alpha\beta)^3$ . Example: $x^2-4x+2=0$ gives $\alpha^3+\beta^3=64-24=40$ and $(\alpha\beta)^3=8$ , so the new equation is $x^2-40x+8=0$ .

Key termscubed roots

Common mistake

Using $(\alpha+\beta)^3$ as the new sum, or $(\alpha+\beta)^3$ as the product. The product of the cubes is $(\alpha\beta)^3$ .

Section 4

Mixed roots such as $\alpha+\frac2\beta$

For roots $\alpha+\frac2\beta$ and $\beta+\frac2\alpha$ : $\text{sum}=(\alpha+\beta)+\frac{2(\alpha+\beta)}{\alpha\beta},$ $\text{product}=\alpha\beta+2+2+\frac{4}{\alpha\beta}.$ Example: $2x^2-6x+1=0$ has $\alpha+\beta=3$ , $\alpha\beta=\frac12$ . Sum $=3+12=15$ , product $=\frac12+4+8=\frac{25}{2}$ , so $2x^2-30x+25=0$ . Expand the product carefully: there are four terms and the two cross terms are each $2$ .

Key termsinteger coefficients

Exam tip

Shifted roots are easy too: for $\alpha+1,\beta+1$ the sum is $\alpha+\beta+2$ and the product is $\alpha\beta+\alpha+\beta+1$ .

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Forming quadratic equations with new roots

The roots of the equation $2x^2-7x+3=0$ are $\alpha$ and $\beta$ .
Find an equation, with integer coefficients, that has roots $\alpha+1$ and $\beta+1$ .2 marks
The roots of the equation $x^2-4x+2=0$ are $\alpha$ and $\beta$ .
Find an equation, with integer coefficients, that has roots $\frac1\alpha$ and $\frac1\beta$ .2 marks
The roots of the equation $x^2-5x+2=0$ are $\alpha$ and $\beta$ .
Find an equation, with integer coefficients, that has roots $\frac{1}{2\alpha}$ and $\frac{1}{2\beta}$ .3 marks

See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).