Grade 9

Quadratic equations: practice problems

A quadratic equation $ax^2 + bx + c = 0$ can be solved by factorising, by square roots or with the quadratic formula. Pick the easiest method for the form you see.

Basic, Standard, Advanced: Grade 9 · Term 2

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Solving by factorising

If $AB = 0$ , then $A = 0$ or $B = 0$ .

x^2 - 5x + 6 = 0 \;\Rightarrow\; (x - 2)(x - 3) = 0 \;\Rightarrow\; x = 2,\ 3

Watch out

Do not divide $x^2 = 3x$ by $x$ ; you would lose $x = 0$ . Factorise $x(x - 3) = 0$ instead.

Square roots and completing the square

$(x + p)^2 = q$ gives $x = -p \pm \sqrt{q}$ . For $x^2 + 6x - 1 = 0$ , add $3^2$ to both sides of $x^2 + 6x = 1$ to get $(x + 3)^2 = 10$ .

The quadratic formula

Quadratic formula

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

Example: $2x^2 - 5x + 1 = 0$ gives $x = \dfrac{5 \pm \sqrt{17}}{4}$ .

Word problems

After solving, check which solutions fit the problem. Lengths and counts must be positive; sometimes both solutions are valid, for example in moving-point problems.

Q1BasicGr 9 · Term 2Quadratic equations

Solve the quadratic equation.

(x + 1)^2 = 4

Answer

x = -3,\ 1

Solution

Take square roots.
$x + 1 = \pm 2$
So
$x = -1 \pm 2 \quad\Rightarrow\quad x = -3,\ 1$

Q2StandardGr 9 · Term 2Quadratic equations

Solve the quadratic equation.

12x^2 - x - 1 = 0

Answer

x = -\frac{1}{4},\ \frac{1}{3}

Solution

By the quadratic formula,
$x = \frac{1 \pm \sqrt{49}}{24} = \frac{1 \pm 7}{24}$
So
$x = \frac{8}{24},\ \frac{-6}{24} \quad\Rightarrow\quad x = -\frac{1}{4},\ \frac{1}{3}$

Q3AdvancedGr 9 · Term 2Quadratic equations

A rectangle is 5 cm longer than it is wide. Its area is 176 cm $^2$ . Find its width.

Answer

11\text{ cm}

Solution

Let the width be $x$ cm:
$x(x + 5) = 176 \quad\Rightarrow\quad x^2 + 5x - 176 = 0$
Factorising,
$(x - 11)(x + 16) = 0 \quad\Rightarrow\quad x = 11,\ -16$
Since $x > 0$ ,
$x = 11$

Q1BasicGr 9 · Term 2Quadratic equations

Solve the quadratic equation.

3x^2 = -21x

Q2StandardGr 9 · Term 2Quadratic equations

One solution of $x^2 - 6x + a = 0$ is $7$ . Find $a$ and the other solution.

Q3AdvancedGr 9 · Term 2Quadratic equations

A rectangular plot is 11 m by 25 m. One path runs across it and one runs along it, both of the same width, and the rest is a flower bed of 207 m $^2$ . How wide are the paths?

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The function y = ax²