Quadratic equations

Q: What is a quadratic equation?

An equation whose highest power of the unknown is 2, in the form ax^2 + bx + c = 0.

Q: Which method should I use?

Try factorisation first when the numbers are friendly. Use the quadratic formula when factorising is not obvious, and completing the square when asked or when deriving.

Q: What does the discriminant tell you?

Its sign tells how many real roots there are: positive means two, zero means one repeated root, negative means none.

Q: Can both roots of a word problem be used?

Only if they make sense. A length or a count cannot be negative, so reject any negative root.

A quadratic equation has the form $a x^{2} + b x + c = 0$ with $a \neq = 0$ . It can have two roots, one repeated root or no real roots. You can solve it by factorising, completing the square or using the quadratic formula.

Formulas

Standard form

a x^{2} + b x + c = 0, a \neq = 0

Quadratic formula

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Discriminant

D = b^{2} - 4 a c : D > 0 two real roots, D = 0 equal roots, D < 0 no real roots

Worked examples

Example 1. Solve

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

by factorisation.

Find two numbers with product 6 and sum $- 5 : - 2$ and $- 3$ .
$(x - 2) (x - 3) = 0, so x = 2 or x = 3$ .

Answer:

x = 2 or x = 3

Example 2. Solve

2 x^{2} + 5 x - 3 = 0

using the quadratic formula.

$a = 2, b = 5, c = - 3, soD = 25 + 24 = 49$ .
$x = \frac{- 5 \pm 7}{4}$ , giving $x = \frac{1}{2}$ or $x = - 3$ .

Answer:

x = \frac{1}{2} or x = - 3

Example 3. Find the nature of the roots of

x^{2} + 2 x + 5 = 0

$D = 2^{2} - 4 (1) (5) = 4 - 20 = - 16$ .
$D < 0$ , so there are no real roots.

Answer: No real roots

Example 4. The product of two consecutive positive integers is 132. Find them.

Let them be $n$ and $n + 1 : n (n + 1) = 132$ , so $n^{2} + n - 132 = 0$ .
$(n + 12) (n - 11) = 0$ . Since $n$ is positive, $n = 11$ .

Answer: 11 and 12

Common mistakes

Not writing the equation in standard form (everything on one side, equal to 0) before factorising or using the formula.
Losing the sign of $b$ . In $- b$ , if $b = - 5$ then $- b = 5$ .
Forgetting the ± in the formula, which gives two roots, not one.

Practice questions with answers

1. Solve

x^{2} - 7 x + 10 = 0

A2 and 5
B1 and 10
C−2 and −5
D3 and 4

Show answer

Answer: A) 2 and 5

Product 10, sum 7 → numbers 2 and 5.

(x - 2) (x - 5) = 0

2. Solve by factorisation:

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

A $x = 2, 3$
B $x = - 2, - 3$
C $x = 1, 6$
D $x = - 1, - 6$

Show answer

Answer: A)

x = 2, 3

x^{2} - 5 x + 6 = (x - 2) (x - 3) = 0, so x = 2 or x = 3

3. The quadratic formula gives the roots of

a x^{2} + b x + c = 0

(a not 0). Which is the correct formula?

A $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$
B $x = \frac{b \pm b ^{2} - 4 a c}{2 a}$
C $x = \frac{- b \pm b ^{2} + 4 a c}{2 a}$
D $x = \frac{- b \pm b ^{2} - 4 a c}{a}$

Show answer

Answer: A)

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Completing the square on

a x^{2} + b x + c = 0

gives

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

4. For a quadratic equation

a x^{2} + b x + c = 0

, the discriminant D is given by:

A $b^{2} - 4 a c$
B $b^{2} + 4 a c$
C $4 a c - b^{2}$
D $2 b - 4 a c$

Show answer

Answer: A)

b^{2} - 4 a c

By definition the discriminant of

a x^{2} + b x + c = 0

D = b^{2} - 4 a c

5. Which of the following is the standard form of a quadratic equation?

Aax + b = 0
B $a x^{2} + b x + c = 0$ , a not equal to 0
C $a x^{3} + b x + c = 0$
D $a x^{2} + b x + c = 0, a = 0$

Show answer

Answer: B)

a x^{2} + b x + c = 0

, a not equal to 0

A quadratic equation in standard form is

a x^{2} + b x + c = 0

where a,

b, c

are real and a is not 0.

6. The equation

\frac{x - 1}{x + 1} + \frac{x + 1}{x - 1} = \frac{10}{3}

, where

x

≠ 1 and

x

≠ −1, reduces to which quadratic equation?

A $x^{2} + 4 = 0$
B $x^{2} - 4 = 0$
C $4 x^{2} - 1 = 0$
D $x^{2} - 2 x - 4 = 0$

Show answer

Answer: B)

x^{2} - 4 = 0

Put the left side over the common denominator

(x + 1) (x - 1)

: the numerator is

(x - 1)^{2} + (x + 1)^{2} = 2 x^{2} + 2

, because the

- 2 x

and +

2 x

cancel. So

\frac{2 x ^{2} + 2}{x ^{2} - 1} = \frac{10}{3}

→

3 (2 x^{2} + 2) = 10 (x^{2} - 1)

→

6 x^{2} + 6 = 10 x^{2} - 10

→

4 x^{2} = 16

→

x^{2} - 4 = 0

, giving

x

= ±2. Check

x = 2 : \frac{1}{3} + 3 = \frac{10}{3}

. Option 1 mis-signs the 10 as it crosses over and would have no real roots at all; option 3 divides

4 x^{2} = 16

by 16 instead of by 4; option 4 comes from missing the cancellation and expanding the numerator as

2 x^{2} - 2 x + 2

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Frequently asked questions

What is a quadratic equation?

An equation whose highest power of the unknown is 2, in the form $a x^{2} + b x + c = 0$ .

Which method should I use?

Try factorisation first when the numbers are friendly. Use the quadratic formula when factorising is not obvious, and completing the square when asked or when deriving.

What does the discriminant tell you?

Its sign tells how many real roots there are: positive means two, zero means one repeated root, negative means none.

Can both roots of a word problem be used?

Only if they make sense. A length or a count cannot be negative, so reject any negative root.

Quadratic equations

Formulas

Worked examples

Common mistakes

Practice questions with answers

Practice by class

Frequently asked questions

Related topics