SPI Mathematics · Quadratic Equations

Sum and Product of the Roots

For ax² + bx + c = 0 the roots add up to −b/a and multiply to c/a, so an equation can be built from its roots and its roots tested without solving.

All Quadratic Equations notes

Why this matters

Seven questions in the bank: three EASY, three MODERATE and one HARD. They form an equation from its roots, find a coefficient from roots in a ratio or reciprocal roots, or rebuild an equation two students copied wrongly. All of them need only the sum and the product, never the formula.

Concept 1 of 3: Sum and product of the roots from the coefficients

If the roots are α and β, the equation is

a(x - \alpha)(x - \beta) = 0

. Multiply it out and compare with

ax^2 + bx + c

: the sum and the product of the roots sit right there in the coefficients.

Definition

Sum: $\alpha + \beta = -\frac{b}{a}$ . Product: $\alpha\beta = \frac{c}{a}$ .
Roots in a ratio such as $1 : 2$ : call them $m$ and $2m$ , then use the sum and the product.
Reciprocal roots ( $\alpha$ and $\frac{1}{\alpha}$ ): the product is 1, so $c = a$ .
One root given: the product (or the sum) gives the other root at once.

Sum and product of roots

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

Worked example

Find the sum and the product of the roots of

2x^2 - 8x + 3 = 0

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Quadratic Equations · Sum and Product of the Roots

If the roots of the quadratic equation

x^2 + kx + 40 = 0

are in the ratio 2 : 5 then the value of k =

Forgetting the minus in the sum

The sum of the roots is

-\frac{b}{a}

. For

x^2 - 9x + 18 = 0

it is

+9

, and for

x^2 + 9x + 18 = 0

it is

-9

Not dividing by a

For

2x^2 - 8x + 3 = 0

the product of the roots is

\frac{3}{2}

, not 3. Divide by a unless a is 1.

Drill 1 more on sum and product of the roots from the coefficients

Concept 2 of 3: Forming a quadratic equation from its roots

Knowing the two roots is the same as knowing the sum and the product, and those two numbers are all an equation with

a = 1

needs.

Definition

The equation with roots α and β is $x^2 - (\text{sum})x + (\text{product}) = 0$ .
The middle term carries a minus sign in front of the sum.
If the sum or product is a fraction, multiply every term by the denominator to clear it.
Check by factorising: $x^2 - 10x + 24 = (x - 4)(x - 6)$ .

Equation from its roots

x^2 - (\alpha + \beta)x + \alpha\beta = 0

Worked example

Form the quadratic equation whose roots are 3 and

-5

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 2 · Quadratic Equations · Sum and Product of the Roots

The roots of the quadratic equation are

-2

and

\frac{11}{2}

then the equation is

Writing plus the sum

The equation is

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

. Roots 2 and 5 give

x^2 - 7x + 10 = 0

;

x^2 + 7x + 10 = 0

has the roots

-2

and

-5

Leaving fractions in the equation

Options are printed with whole-number coefficients.

x^2 - \frac{5}{2}x + 1 = 0

appears as

2x^2 - 5x + 2 = 0

, so multiply through before comparing.

Drill 2 more on forming a quadratic equation from its roots

Concept 3 of 3: Rebuilding an equation from wrongly copied coefficients

A student who copies one coefficient wrongly still gets the other relation right. If the constant term was wrong, the x term was right, so that student's sum of roots is correct. If the x term was wrong, the constant was right, so that student's product is correct.

Definition

Constant term copied wrongly: trust that student's sum of roots.
Coefficient of x copied wrongly: trust that student's product of roots.
Put the trusted sum and product together: $x^2 - (\text{sum})x + \text{product} = 0$ , then factorise if the roots are asked.

Correct equation from the two students' trusted values

x^2 - (\text{sum from the student with the right } x \text{ term})\,x + (\text{product from the student with the right constant}) = 0

Worked example

In solving

x^2 + bx + c = 0

, one student copies the constant term wrongly and gets the roots 1 and 7. Another copies the coefficient of x wrongly and gets the roots 2 and 5. Find the correct equation.

Practice this conceptself-check · 2 quick reps

The same idea in a practice question:

Practice questionHard

Example 3 · Quadratic Equations · Sum and Product of the Roots

One of the two students, while solving a quadratic equation in x, copied the constant term incorrectly and got the roots 3 and 2. The other copied the constant term and coefficient of

x^2

correctly as

-6

and 1 respectively. The correct roots are ____.

Trusting the wrong student's sum

A student who copied the coefficient of x wrongly has a wrong sum of roots. Take only the product from that student, and the sum from the one who got the x term right.

Drill 1 more on rebuilding an equation from wrongly copied coefficients

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Sum and product of the roots from the coefficients
Sum and product of roots
$\alpha + \beta = -\frac{b}{a} \qquad \alpha\beta = \frac{c}{a}$
Forming a quadratic equation from its roots
Equation from its roots
$x^2 - (\alpha + \beta)x + \alpha\beta = 0$
Rebuilding an equation from wrongly copied coefficients
Correct equation from the two students' trusted values
$x^2 - (\text{sum from the student with the right } x \text{ term})\,x + (\text{product from the student with the right constant}) = 0$

Watch out for (5)

Forgetting the minus in the sum
→ Sum and product of the roots from the coefficients
Not dividing by a
→ Sum and product of the roots from the coefficients
Writing plus the sum
→ Forming a quadratic equation from its roots
Leaving fractions in the equation
→ Forming a quadratic equation from its roots
Trusting the wrong student's sum
→ Rebuilding an equation from wrongly copied coefficients

Test yourself on a real paper

Sit a past SPI paper, timed and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Sit an SPI past paper Drill all 7 questions