Roots α and β of a Quadratic

Use α + β = −b/a and αβ = c/a to find expressions in the roots without solving, form new equations from given roots, and find unknown coefficients.

In the exam for Edexcel Further Pure Maths 4PM1 · 22 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

Sum and product of roots

The roots of $2 x^{2} - 6 x + 3 = 0$ are messy surds: $\frac{3 \pm 3}{2}$ . Yet you can find their sum and their product in two seconds, with no surds at all. This lesson shows you how.

Call the two roots $α$ (alpha) and $β$ (beta). If $α$ and $β$ are the roots, the quadratic must factorise as $a (x - α) (x - β)$ . Expand it and compare:

$a (x - α) (x - β)$ $= a x^{2} - a (α + β) x + a α β$ — expand the brackets.

$a x^{2} + b x + c$ $= a x^{2} - a (α + β) x + a α β$ — this is the same quadratic, so the parts match.

$b = - a (α + β)$ — match the x terms.

$c = a α β$ — match the numbers on their own.

Formula

α + β = - \frac{b}{a}

$α$ — one root of $a x^{2} + b x + c = 0$ .
$β$ — the other root.
$- \frac{b}{a}$ — minus the x-number, divided by the $x^{2}$ -number.

Formula

α β = \frac{c}{a}

$α β$ — the two roots multiplied together.
$\frac{c}{a}$ — the number on its own, divided by the $x^{2}$ -number.

Rule

4PM1 does not print these on the formula sheet. Learn both, together with the quadratic formula.

Key idea

These two rules are often called Vieta's formulas, after François Viète, a French lawyer who did mathematics in the late 1500s. He was one of the first to write algebra with letters, the way you do today.

Reading and reversing

First practise reading the sum and product. The equation must equal 0, and the signs matter.

Worked example

1
The roots of $2 x^{2} - 6 x + 3 = 0$ are $α$ and $β$ . Write down $α + β$ and $α β$ .
2
$a = 2, b = - 6, c = 3$ — Read the numbers with their signs.
3
$α + β = - \frac{- 6}{2} = 3$ — Minus b over a. Two minus signs make a plus.
4
$α β = \frac{3}{2}$ — c over a.

You can also go backwards: from two roots to the equation. Put the sum and product into this form:

Formula

x^{2} - (sum) x + product = 0

$sum$ — the two roots added: $α + β$ .
$product$ — the two roots multiplied: $α β$ .

Rule

4PM1 lists this form as a formula to learn. If the question asks for integer coefficients, multiply every term by the same number to clear the fractions.

Worked example

1
Form a quadratic equation with integer coefficients whose roots are $\frac{2}{3}$ and $- \frac{1}{2}$ .
2
$sum = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}$ — Add the roots. The common denominator is 6.
3
$product = \frac{2}{3} \times (- \frac{1}{2}) = - \frac{1}{3}$ — Multiply the roots.
4
$x^{2} - \frac{1}{6} x - \frac{1}{3} = 0$ — Minus the sum, plus the product.
5
$6 x^{2} - x - 2 = 0$ — Multiply every term by 6. Keep the = 0.

Common mistake

Getting the sign of the sum wrong. $α + β = - \frac{b}{a}$ , so $x^{2} - 4 x + 1 = 0$ has $α + β = 4$ , not $- 4$ .

Common mistake

Writing $x^{2} + (sum) x + product$ . The sum goes in with a minus sign in front.

Key idea

Sum is minus b over a. Product is c over a. From roots back to an equation: x squared, minus the sum x, plus the product.

Squares and fractions

Exam questions ask for things like $α^{2} + β^{2}$ without solving. The trick: rewrite each one using only $α + β$ and $α β$ , which you already know.

Start from $(α + β)^{2} = α^{2} + 2 α β + β^{2}$ . Take away the $2 α β$ and you are left with the squares.

Formula

α^{2} + β^{2} = (α + β)^{2} - 2 α β

$2 α β$ — the middle term of $(α + β)^{2}$ , taken away.

Formula

(α - β)^{2} = (α + β)^{2} - 4 α β

$4 α β$ — $(α - β)^{2} = α^{2} + β^{2} - 2 α β$ , and $α^{2} + β^{2}$ already has $- 2 α β$ . Together that is $- 4 α β$ .

Formula

\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β}

$α + β$ — the top, after you use the common denominator.
$α β$ — the common denominator.

Worked example

1
The roots of $2 x^{2} - 5 x + 1 = 0$ are $α$ and $β$ . Find $α^{2} + β^{2}$ , $(α - β)^{2}$ and $\frac{1}{α} + \frac{1}{β}$ .
2
$α + β = \frac{5}{2}, α β = \frac{1}{2}$ — Minus b over a, and c over a.
3
$α^{2} + β^{2} = (\frac{5}{2})^{2} - 2 (\frac{1}{2}) = \frac{21}{4}$ — $\frac{25}{4} - 1 = \frac{21}{4}$ .
4
$(α - β)^{2} = \frac{25}{4} - 4 (\frac{1}{2}) = \frac{17}{4}$ — $\frac{25}{4} - 2 = \frac{17}{4}$ .
5
$\frac{1}{α} + \frac{1}{β} = \frac{5}{2} \div \frac{1}{2} = 5$ — Sum over product.

Common mistake

Writing $α^{2} + β^{2} = (α + β)^{2}$ . The square of a sum has a middle term, $2 α β$ , that you must take away.

Common mistake

Writing $\frac{1}{α} + \frac{1}{β} = \frac{1}{α + β}$ . Add fractions with a common denominator: the answer is $\frac{α + β}{α β}$ .

Key idea

Every symmetric expression can be built from the sum and the product. Your job is to find the identity that does it.

Cubes and fourth powers

Higher powers use the same idea. Expand a bracket you know, then take away the parts you do not want.

Formula

α^{3} + β^{3} = (α + β)^{3} - 3 α β (α + β)

$3 α β (α + β)$ — $(α + β)^{3}$ $= α^{3} + 3 α^{2} β + 3 α β^{2}$ $+ β^{3}$ . The two middle terms make $3 α β (α + β)$ .

Formula

α^{4} + β^{4} = (α^{2} + β^{2})^{2} - 2 (α β)^{2}

$2 (α β)^{2}$ — the middle term of $(α^{2} + β^{2})^{2}$ , which is $2 α^{2} β^{2}$ .

Formula

α^{2} - β^{2} = (α + β) (α - β)

$α - β$ — find it from $(α - β)^{2}$ , then square root. It can be + or −.

Worked example

1
For the same equation, $2 x^{2} - 5 x + 1 = 0$ , find $α^{3} + β^{3}$ and $α^{4} + β^{4}$ .
2
$α^{3} + β^{3} = (\frac{5}{2})^{3} - 3 (\frac{1}{2}) (\frac{5}{2})$ — Use the cube identity.
3
$= \frac{125}{8} - \frac{15}{4} = \frac{95}{8}$ — Change $\frac{15}{4}$ to $\frac{30}{8}$ and subtract.
4
$α^{4} + β^{4} = (\frac{21}{4})^{2} - 2 (\frac{1}{2})^{2}$ — Use $α^{2} + β^{2} = \frac{21}{4}$ from before.
5
$= \frac{441}{16} - \frac{8}{16} = \frac{433}{16}$ — $2 \times \frac{1}{4} = \frac{1}{2} = \frac{8}{16}$ .

Rule

When a question says “show that” an identity is true, expand the right-hand side fully and show every line. The expanding is where the marks are.

Common mistake

Squaring $α β$ wrongly in the fourth-power identity. $(α β)^{2}$ with $α β = \frac{1}{2}$ is $\frac{1}{4}$ , not $\frac{1}{2}$ .

Key idea

Reuse earlier answers. The fourth-power identity needs $α^{2} + β^{2}$ , which you usually found in an earlier part.

New roots, new equation

The favourite long question: “form an equation whose roots are” something new, like $α^{2}$ and $β^{2}$ . You never find $α$ or $β$ . You find the new sum and the new product.

Steps

1
1. Write down $α + β$ and $α β$ from the given equation.
2. New sum: add the two new roots, then rewrite using the identities.
3. New product: multiply the two new roots, then simplify.
4. Write $x^{2} - (new sum) x + new product = 0$ , then clear fractions.

Worked example

1
The roots of $2 x^{2} - 5 x + 1 = 0$ are $α$ and $β$ . Form an equation with integer coefficients whose roots are $α^{2}$ and $β^{2}$ .
2
$sum = α^{2} + β^{2} = \frac{21}{4}$ — Found earlier from the identity.
3
$product = α^{2} β^{2} = (\frac{1}{2})^{2} = \frac{1}{4}$ — $α^{2} β^{2}$ is $(α β)^{2}$ .
4
$x^{2} - \frac{21}{4} x + \frac{1}{4} = 0$ — Minus the sum, plus the product.
5
$4 x^{2} - 21 x + 1 = 0$ — Multiply every term by 4.

Worked example

1
With the same $α$ and $β$ , form an equation with integer coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$ .
2
$\frac{α}{β} + \frac{β}{α} = \frac{α ^{2} + β ^{2}}{α β}$ — Common denominator $α β$ .
3
$= \frac{21}{4} \div \frac{1}{2} = \frac{21}{2}$ — New sum.
4
$\frac{α}{β} \times \frac{β}{α} = 1$ — New product: everything cancels.
5
$2 x^{2} - 21 x + 2 = 0$ — From $x^{2} - \frac{21}{2} x + 1 = 0$ , times 2.

Common mistake

Dropping the “= 0” at the end. An expression is not an equation, and the final mark needs the whole equation.

Key idea

New roots, new equation: find the new sum, find the new product, then x squared minus sum x plus product equals 0.

Find the unknown

Sometimes the equation has an unknown coefficient, and you are told something about the roots. Turn that fact into an equation using the sum and product.

Worked example

1
The roots of $x^{2} + p x + 12 = 0$ are $α$ and $β$ , and $α^{2} + β^{2} = 25$ . Find the possible values of p.
2
$α + β = - p, α β = 12$ — Minus b over a, and c over a, with a = 1.
3
$(- p)^{2} - 2 (12) = 25$ — Use $α^{2} + β^{2} = (α + β)^{2} - 2 α β$ .
4
$p^{2} = 49$ — Add 24 to both sides.
5
$p = 7 or p = - 7$ — Keep both square roots.

Worked example

1
One root of $x^{2} - 8 x + k = 0$ is three times the other. Find k.
2
$α = 3 β$ — Write the fact as an equation.
3
$3 β + β = 8 \Rightarrow β = 2$ — The sum of the roots is 8.
4
$α = 6$ — Three times 2.
5
$k = α β = 12$ — The product of the roots is c over a, which is k.

Common mistake

Taking only the positive square root. $p^{2} = 49$ gives $p = \pm 7$ . Check both, and keep both unless the question rules one out.

Key idea

Whatever you are told about the roots, write it with $α + β$ and $α β$ . Then it is just an equation to solve.

Which tool? Read the question

Read the question, then pick the tool.

You see	You do
“Write down $α + β$ and $α β$ ”	$- \frac{b}{a}$ and $\frac{c}{a}$ , after making the equation = 0
$α^{2} + β^{2}$ , $(α - β)^{2}$ , $α^{3} + β^{3}$	Use the identity, then substitute the sum and product
$\frac{1}{α} + \frac{1}{β}$ or $\frac{α}{β} + \frac{β}{α}$	Common denominator $α β$
“Form an equation with roots …”	New sum and new product, then $x^{2} - (sum) x + product = 0$
“Integer coefficients”	Multiply through to clear fractions; keep = 0
A fact about the roots and an unknown	Write the fact using the sum and product, then solve

Rule

“Without solving the equation” means you get no marks for finding $α$ and $β$ with the formula. Use the sum and product.

Key idea

Two numbers hold everything: the sum and the product. Every question in this lesson is about rewriting things in terms of those two.

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