Sum and product of roots | Year 11

01 / LEARN

Understand the idea.

Compare coefficients

If a quadratic has roots $r$ and $s$ , write it as $a (x - r) (x - s)$ . Expansion gives $a x^{2} - a (r + s) x + a r s$ . Comparing this with $a x^{2} + b x + c$ yields the root sum $r + s = - \frac{b}{a}$ and product $r s = \frac{c}{a}$ .

Both formulas divide by the leading coefficient, and the sum contains a minus sign. They describe the pair without requiring separate root calculations, which is especially useful when the roots contain surds or parameters.

r + s = - \frac{b}{a}, r s = \frac{c}{a}

Build or transform an equation

To construct a monic quadratic with root sum $S$ and product $P$ , use $x^{2} - S x + P = 0$ . Any nonzero constant multiple gives the same roots, so integer coefficients can be obtained by clearing denominators when necessary.

For transformed roots, calculate the new sum and product before constructing the equation. Doubling both roots doubles their sum but quadruples their product. If real roots are required, also verify the discriminant of the resulting quadratic.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Read without solving

Find the sum and product of roots of $3 x^{2} - 7 x + 2 = 0$ .

Worked solution

1
Identify the coefficients with their signs.
$a = 3, b = - 7, c = 2$
The formulas use coefficients from the equation with all terms on one side.
2
Calculate the root sum.
$r + s = - \frac{b}{a} = - \frac{- 7}{3} = \frac{7}{3}$
The negative of the middle coefficient is divided by the leading coefficient.
3
Calculate the root product.
$r s = \frac{c}{a} = \frac{2}{3}$
The constant coefficient is divided by the same leading coefficient.
4
Verify by a factorisation.
$3 x^{2} - 7 x + 2 = (3 x - 1) (x - 2)$
The roots one third and two have sum seven thirds and product two thirds.

Answer

Sum $7 / 3$ and product $2 / 3$ .

EXAMPLE 02

Build the equation

Roots have sum 5 and product -6. Form a monic equation.

Worked solution

1
Use the sum-product form of a monic equation.
$x^{2} - S x + P = 0$
Expanding two root factors gives middle coefficient negative S and constant P.
2
Substitute the given sum and product.
$x^{2} - 5 x - 6 = 0$
The required sum is five and the required product is negative six.
3
Check the roots against both conditions.
$\begin{aligned} x^{2} - 5 x - 6 & = (x - 6) (x + 1) \\ 6 + (- 1) & = 5 \\ 6 (- 1) & = - 6 \end{aligned}$
The constructed equation has real roots six and negative one with exactly the required sum and product.

Answer

$x^{2} - 5 x - 6 = 0$

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the root product for $2 x^{2} + 9 x - 5 = 0$ .

Show answer & reasoning

$- 5 / 2$

The product is $\frac{c}{a} = \frac{- 5}{2} = - \frac{5}{2}$ . Include the sign of the constant coefficient; the middle coefficient is not needed for this calculation.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Find root sum/product for $x^{2} - 8 x + 10 = 0$ .
2 marks
Find root sum/product for $4 x^{2} + 3 x - 2 = 0$ .
2 marks
Form a monic equation with root sum -3 and product 2.
2 marks
One root of $x^{2} - 7 x + 12 = 0$ is 3. Find the other without full factorisation.
2 marks
Roots of $x^{2} - 4 x + 1 = 0$ are $r$ , $s$ . Form an equation with roots $2 r, 2 s$ .
3 marks
One positive root of $x^{2} - k x + 15 = 0$ is three times the other. Find $k$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find root sum/product for $x^{2} - 8 x + 10 = 0$ .
2 marks
Apply $\frac{- b}{a}$ and $\frac{c}{a}$ .
Sum 8; product 10.
Find root sum/product for $4 x^{2} + 3 x - 2 = 0$ .
2 marks
Divide by leading coefficient 4.
Sum $\frac{- 3}{4}$ ; product $\frac{- 1}{2}$ .
Form a monic equation with root sum -3 and product 2.
2 marks
Substitute into $x^{2} - S x + P = 0$ .
$x^{2} + 3 x + 2 = 0$
One root of $x^{2} - 7 x + 12 = 0$ is 3. Find the other without full factorisation.
2 marks
The sum is 7, so the other is 4.
Product $3 (4) = 12$ checks it.
4
Roots of $x^{2} - 4 x + 1 = 0$ are $r$ , $s$ . Form an equation with roots $2 r, 2 s$ .
3 marks
New sum is 8 and new product is 4.
Use the monic form.
$x^{2} - 8 x + 4 = 0$
One positive root of $x^{2} - k x + 15 = 0$ is three times the other. Find $k$ .
3 marks
Let roots $t, 3 t$ , giving $3 t^{2} = 15$ .
Both are positive, so $t = \sqrt{5}$ . Add roots to find $k$ .
$k = 4 \sqrt{5}$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topics 2.1–2.2: quadratic relationships, cubic and quartic polynomials. Original Ascent teaching sequence.

Official curriculum source ↗

In this lesson

Understand the idea.

Compare coefficients

Build or transform an equation

See the method in action.

Read without solving

Identify the coefficients with their signs.

Calculate the root sum.

Calculate the root product.

Verify by a factorisation.

Build the equation

Use the sum-product form of a monic equation.

Substitute the given sum and product.

Check the roots against both conditions.

Try it yourself.

Check your understanding

Keep practising