Sum and product of rootsEdexcel International A Level Further Maths: Revision notes

Section 1

Sum and product of roots

If $\alpha$ and $\beta$ are the roots of $ax^2+bx+c=0$ , then $ax^2+bx+c=a(x-\alpha)(x-\beta)$ . Expanding and comparing coefficients gives $\alpha+\beta=-\frac ba,\qquad \alpha\beta=\frac ca.$ For $2x^2-7x+4=0$ : $\alpha+\beta=\frac72$ and $\alpha\beta=\frac42=2$ . You can find these without solving the equation, which is useful when the roots are awkward.

Key termssum of rootsproduct of roots

Common mistake

Forgetting to divide by $a$ , or losing the minus sign: the sum is $-\frac ba$ , not $\frac ba$ .

Section 2

Symmetric expressions

Any expression that stays the same when $\alpha$ and $\beta$ are swapped can be written using $\alpha+\beta$ and $\alpha\beta$ . Key results: $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$ . $\dfrac1\alpha+\dfrac1\beta=\dfrac{\alpha+\beta}{\alpha\beta}$ . $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$ . $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)$ . For $x^2+6x+7=0$ ( $\alpha+\beta=-6$ , $\alpha\beta=7$ ): $(\alpha-\beta)^2=36-28=8$ and $\alpha^3+\beta^3=-216-3(7)(-6)=-216+126=-90$ .

Key termssymmetric expression

Common mistake

Writing $\alpha^2+\beta^2=(\alpha+\beta)^2$ . The $-2\alpha\beta$ term is essential.

Exam tip

The identity for $\alpha^3+\beta^3$ comes from expanding $(\alpha+\beta)^3=\alpha^3+3\alpha^2\beta+3\alpha\beta^2+\beta^3$ .

Section 3

Forming an equation with new roots

To find an equation whose roots are related to $\alpha$ and $\beta$ : (1) write the sum $S$ and product $P$ of the new roots using $\alpha+\beta$ and $\alpha\beta$ ; (2) the equation is $x^2-Sx+P=0$ ; (3) clear any fractions. Example: new roots $\frac\alpha\beta$ and $\frac\beta\alpha$ for $3x^2-5x+1=0$ . $S=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{19/9}{1/3}=\frac{19}{3}$ and $P=1$ , so $x^2-\frac{19}{3}x+1=0$ , i.e. $3x^2-19x+3=0$ .

Key termsnew roots

Exam tip

Always write $x^2-Sx+P=0$ with the minus sign on $S$ .

Section 4

Roots shifted or scaled

For new roots $2\alpha$ and $2\beta$ : $S=2(\alpha+\beta)$ and $P=4\alpha\beta$ . For $x^2+6x+7=0$ : $S=-12$ , $P=28$ , giving $x^2+12x+28=0$ . For new roots $\alpha+2$ and $\beta+2$ with $\alpha+\beta=-7$ and $\alpha\beta=12$ : $S=-7+4=-3$ and $P=\alpha\beta+2(\alpha+\beta)+4=12-14+4=2$ , giving $x^2+3x+2=0$ . For new roots $\alpha^3$ and $\beta^3$ : $S=\alpha^3+\beta^3=-91$ and $P=(\alpha\beta)^3=1728$ .

Key termsscaled roots

Common mistake

Using $2\alpha\beta$ as the product of $2\alpha$ and $2\beta$ . It is $4\alpha\beta$ .

Section 5

Unknown coefficients

When a condition on the roots involves an unknown constant, express the condition using $\alpha+\beta$ and $\alpha\beta$ . For $x^2+kx+12=0$ with $\alpha^2+\beta^2=25$ : $\alpha+\beta=-k$ , $\alpha\beta=12$ , so $k^2-24=25$ and $k^2=49$ . With $k=7$ , $\alpha+\beta=-7$ and $\alpha^3+\beta^3=(-7)^3-3(12)(-7)=-343+252=-91$ . Check: the roots are $-3$ and $-4$ , and $(-3)^3+(-4)^3=-27-64=-91$ .

Key termsunknown coefficient

Exam tip

If the roots turn out to be nice numbers, check your answer by substituting them directly.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Sum and product of roots

The roots of the equation $2x^2-7x+4=0$ are $\alpha$ and $\beta$ .
Find the value of $\dfrac1\alpha+\dfrac1\beta$ .2 marks
The roots of the equation $x^2+6x+7=0$ are $\alpha$ and $\beta$ .
Find a quadratic equation, with integer coefficients, whose roots are $2\alpha$ and $2\beta$ .2 marks
The roots of the equation $3x^2-5x+1=0$ are $\alpha$ and $\beta$ .
Show that $\alpha^2+\beta^2=\frac{19}{9}$ .3 marks

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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).