2C - Sums and products of roots

The coefficients of a quadratic reveal the sum and product of its roots, even when the roots themselves are awkward. You will use these links to calculate expressions in the roots without solving the equation, then work backwards to form a quadratic from a known sum and product.

Roots and coefficients

Can you find the sum and product of the roots of $x^2-6x+2=0$ without finding either root?

Start with an equation whose roots you can see. The equation $(x-2)(x-5)=0$ has roots $2$ and $5$ . Expanding gives

(x-2)(x-5)=x^2-7x+10.

The coefficient $-7$ is the negative of the sum $2+5$ , and the constant $10$ is the product $2\times5$ . These links come from the expansion, so they work even when the roots are not integers.

Roots and their sum and product

The roots of an equation are the values that satisfy it. We use the Greek letters $\alpha$ and $\beta$ for the two roots of a quadratic, and write

S=\alpha+\beta,\qquad P=\alpha\beta.

A repeated root counts twice: if both roots are $3$ , then $S=6$ and $P=9$ .

For a quadratic $ax^2+bx+c=0$ , the leading coefficient $a$ must be nonzero. If the roots are $\alpha$ and $\beta$ , its factorised form is

\begin{aligned} a(x-\alpha)(x-\beta)&=0\\ ax^2-a(\alpha+\beta)x+a\alpha\beta&=0 &&\text{expand the factors} \end{aligned}

Comparing the coefficient of $x$ gives $-a(\alpha+\beta)=b$ ; comparing the constant terms gives $a\alpha\beta=c$ . Divide by $a$ to obtain the two links.

Sum and product of roots

If $ax^2+bx+c=0$ , with $a\ne0$ , has roots $\alpha$ and $\beta$ , then

\alpha+\beta=-\frac{b}{a},\qquad \alpha\beta=\frac{c}{a}.

Not on the formulae sheet: learn them.

Read the coefficients from an equation with zero on one side. Keep each coefficient's sign.

Worked example 1: reading the two links

Question. The roots of $x^2-6x+2=0$ are $\alpha$ and $\beta$ . Find $\alpha+\beta$ and $\alpha\beta$ without solving the equation.

Plan. Identify $a$ , $b$ and $c$ , then use the coefficient links.

\begin{aligned} a&=1,\quad b=-6,\quad c=2\\ \alpha+\beta&=-\frac{-6}{1}=6 &&\text{negate the signed coefficient }b\\ \alpha\beta&=\frac{2}{1}=2 &&\text{divide the constant by }a \end{aligned}

Answer. The sum is $6$ and the product is $2$ .

Check. For a separate check, the quadratic formula gives roots $3+\sqrt{7}$ and $3-\sqrt{7}$ . Their sum is $6$ , and their product is $9-7=2$ . Solving was unnecessary for the answer.

Now change only the leading coefficient. The sum and product must both be divided by that coefficient.

Worked example 2: a leading coefficient other than one

Question. The roots of $3x^2-6x+2=0$ are $\alpha$ and $\beta$ . Find their sum and product without solving.

Plan. Use $a=3$ while retaining $b=-6$ and $c=2$ .

\begin{aligned} \alpha+\beta&=-\frac{-6}{3}=2\\ \alpha\beta&=\frac{2}{3} \end{aligned}

Answer. $\alpha+\beta=2$ and $\alpha\beta=\frac{2}{3}$ .

Check. The roots are $1+\frac{\sqrt{3}}{3}$ and $1-\frac{\sqrt{3}}{3}$ . Adding gives $2$ ; multiplying gives $1-\frac{1}{3}=\frac{2}{3}$ .

The minus sign in the sum formula is easy to lose when $b$ is already negative.

Watch out

For $3x^2-6x+2=0$ , the wrong line $\alpha+\beta=-6$ ignores both the minus sign in $-b$ and the division by $a$ . Write $-(-6)/3$ before simplifying.

Use the same two links for this fresh equation.

Expressions in the roots

An expression that treats the two roots in the same way can often be rewritten using only their sum and product.

For example, roots $1$ and $4$ have sum $5$ , but the sum of their squares is $1+16=17$ , not $25$ . Squaring their sum also includes two cross-products:

(\alpha+\beta)^2=\alpha^2+2\alpha\beta+\beta^2.

Subtract $2\alpha\beta$ to isolate the squares. For cubes, expand one stage further:

\begin{aligned} (\alpha+\beta)^3 &=\alpha^3+3\alpha^2\beta+3\alpha\beta^2+\beta^3\\ &=\alpha^3+\beta^3+3\alpha\beta(\alpha+\beta) \end{aligned}

The two middle terms have the common factor $3\alpha\beta$ . This is why subtracting $3PS$ leaves just the sum of the cubes.

Square and cube identities

Writing $S=\alpha+\beta$ and $P=\alpha\beta$ ,

\alpha^2+\beta^2=S^2-2P,\qquad \alpha^3+\beta^3=S^3-3PS.

These identities are not on the formulae sheet; you can recover them by expansion.

Reciprocal sums use addition of fractions rather than a power expansion:

\frac{1}{\alpha}+\frac{1}{\beta} =\frac{\beta+\alpha}{\alpha\beta} =\frac{S}{P}.

This requires $\alpha\ne0$ and $\beta\ne0$ , equivalently $P\ne0$ . For a quadratic with $a\ne0$ , checking $c\ne0$ establishes this immediately because $P=c/a$ . If $c=0$ , at least one root is zero and this reciprocal sum is undefined. A zero sum $S$ is allowed; it is the denominator $P$ that matters.

Worked example 3: three expressions from two values

Question. The roots of $x^2-5x+4=0$ are $\alpha$ and $\beta$ . Without solving, find $\alpha^2+\beta^2$ , $\alpha^3+\beta^3$ and $\frac{1}{\alpha}+\frac{1}{\beta}$ .

Plan. Read $S$ and $P$ , then substitute into the identities; check $P\ne0$ before taking reciprocals.

\begin{aligned} S&=5,\quad P=4\\ \alpha^2+\beta^2&=5^2-2(4)=17\\ \alpha^3+\beta^3&=5^3-3(4)(5)=65\\ \frac{1}{\alpha}+\frac{1}{\beta}&=\frac{5}{4} &&\text{valid because }P=4\ne0 \end{aligned}

Answer. $17$ , $65$ and $\frac{5}{4}$ , respectively.

Check. The equation factors as $(x-1)(x-4)=0$ . Direct calculation gives $1^2+4^2=17$ , $1^3+4^3=65$ and $1+\frac{1}{4}=\frac{5}{4}$ .

Change only the sign of the constant term. The sum stays the same, but the product becomes negative, so the subtraction signs in the identities need care.

Worked example 4: a negative product

Question. The roots of $x^2-5x-4=0$ are $\alpha$ and $\beta$ . Without solving, find the same three expressions.

Plan. Use $S=5$ and $P=-4$ , keeping brackets around the negative product.

\begin{aligned} \alpha^2+\beta^2&=5^2-2(-4)=33\\ \alpha^3+\beta^3&=5^3-3(-4)(5)=185\\ \frac{1}{\alpha}+\frac{1}{\beta}&=\frac{5}{-4}=-\frac{5}{4} &&\text{since }P\ne0 \end{aligned}

Answer. $33$ , $185$ and $-\frac{5}{4}$ , respectively.

Check. The roots are approximately $5.701562$ and $-0.701562$ . Substituting these into the original expressions gives approximately $33$ , $185$ and $-1.25$ . The exact identity calculations give the answers without rounding.

In the cube calculation, the exponent belongs to the whole sum.

Watch out

$\alpha^3+\beta^3\ne(\alpha+\beta)^3$ in general. For roots $1$ and $4$ , these values are $65$ and $125$ . The correction is $-3\alpha\beta(\alpha+\beta)$ , including its sign.

Use all three identities on the next equation.

Forming a quadratic

A known sum and product let you rebuild an equation even when you have not been given its roots separately.

The expansion $(x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha\beta$ already supplies the construction. An equation with leading coefficient $1$ is called monic.

An equation from a sum and product

If two roots have sum $S$ and product $P$ , an equation with those roots is

x^2-Sx+P=0.

The coefficient of $x$ is the negative of the sum; the constant is the product. Multiplying the entire equation by a nonzero constant gives an equivalent equation.

Not on the formulae sheet: learn it.

If fractions occur, multiply every term by a common denominator to obtain integer coefficients. If the required roots are real, their sum and product necessarily satisfy $S^2-4P\ge0$ , since this is the discriminant of the equation you construct.

Worked example 5: integer sum and product

Question. Two roots have sum $7$ and product $10$ . Form a quadratic equation with these roots.

Plan. Substitute the sum and product into $x^2-Sx+P=0$ .

\begin{aligned} x^2-Sx+P&=0\\ x^2-7x+10&=0 &&\text{use }S=7,\ P=10 \end{aligned}

Answer. $x^2-7x+10=0$ .

Check. This factors as $(x-2)(x-5)=0$ . Its roots add to $7$ and multiply to $10$ .

Now allow a fractional sum. The construction stays the same; only the final clearing of fractions is new.

Worked example 6: clearing a denominator

Question. Two roots have sum $\frac{7}{2}$ and product $3$ . Form a quadratic equation with integer coefficients.

Plan. Form the monic equation first, then multiply the whole equation by $2$ .

\begin{aligned} x^2-\frac{7}{2}x+3&=0 &&\text{insert the sum and product}\\ 2x^2-7x+6&=0 &&\text{multiply every term by }2 \end{aligned}

Answer. $2x^2-7x+6=0$ .

Check. $(2x-3)(x-2)=0$ gives roots $\frac{3}{2}$ and $2$ . Their sum is $\frac{7}{2}$ and their product is $3$ , as required.

An expression does not yet state which values of $x$ are the roots.

Exam tip

When asked to form an equation, finish with $=0$ . When integer coefficients are requested, clear all denominators by multiplying the whole equation by the same nonzero number.

The opening problem is now reversible: $x^2-6x+2=0$ gives $S=6$ , $P=2$ , and those two numbers rebuild that equation. Next you can change the roots and construct an equation for the changed pair.

Quick Recap

For $ax^2+bx+c=0$ , $a\ne0$ , read $S=-b/a$ and $P=c/a$ .
Use $S^2-2P$ for the sum of squares and $S^3-3PS$ for the sum of cubes.
The reciprocal sum is $S/P$ , provided $P\ne0$ .
Rebuild an equation as $x^2-Sx+P=0$ , then clear fractions if needed.

The roots of $x^2-6x+2=0$ add to $6$ and multiply to $2$ , without needing to find either one.

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2C - Sums and products of roots

Roots and coefficients

Expressions in the roots

Forming a quadratic

Quick Recap

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