Year 9 - Quadratics - Free practice

Year 9 Quadratics Worksheets & Practice Tests

Graphing parabolas, null factor law, solving quadratic equations.

Quick answer

How do you solve quadratic equations using the null factor law?

A quadratic equation is solved by the null factor law when you factorise it so that two brackets multiply to zero, then set each bracket equal to zero. For example, x² + x − 6 = 0 becomes (x + 3)(x − 2) = 0, giving x = −3 or x = 2. The solutions are also called the roots of the equation.

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Worked examples

Year 9 quadratics: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

Example 1 · Solving a quadratic by null factor law
Solve $x^2 - 5x + 6 = 0$ .
Show worked solution
Key idea: Factorise the quadratic into two brackets, then use the null factor law: if $a \times b = 0$ , then $a = 0$ or $b = 0$ .
1. Find two integers that multiply to $6$ and add to $-5$ : these are $-2$ and $-3$ .
2. Factorise: $(x - 2)(x - 3) = 0$
3. Apply the null factor law:
  $x - 2 = 0 \text{ or } x - 3 = 0$
4. Solve each equation:
  $x = 2 \text{ or } x = 3$
5. Check by substituting back: $(2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0$ ✓ and $(3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0$ ✓
Answer: $x = 2$ or $x = 3$
Example 2 · Describing transformations of a parabola
The graph of $y = x^2$ is transformed to give $y = -3x^2 + 4$ . Describe the transformations applied to the standard parabola, state whether the new parabola opens upward or downward, give its vertex, and find the $y$ -intercept.
Show worked solution
Key idea: In $y = ax^2 + c$ , the value of $a$ controls the stretch and reflection, and $c$ shifts the graph vertically.
1. Compare with the standard form $y = ax^2 + c$ : here $a = -3$ and $c = 4$ .
2. Reflection: Because $a = -3 < 0$ , the parabola is reflected in the $x$ -axis (opens downward).
3. Vertical stretch: $\lvert a \rvert = 3 > 1$ , so the parabola is narrower than $y = x^2$ by a factor of $3$ .
4. Vertical translation: $c = 4$ , so the graph is shifted $4$ units upward.
5. The vertex is at $(0,\ 4)$ and the $y$ -intercept is also $(0,\ 4)$ .
Answer: Reflected in the $x$ -axis, stretched vertically by factor $3$ , shifted $4$ units up; opens downward; vertex and $y$ -intercept at $(0, 4)$ .
Example 3 · Setting up and solving a quadratic from a worded problem
A rectangular garden has a length that is $5$ metres more than its width. The area of the garden is $36$ square metres. Find the dimensions of the garden.
Show worked solution
Key idea: Write an equation for the area of the rectangle, rearrange to equal zero, then factorise and apply the null factor law. Reject any negative solution since a length must be positive.
1. Let the width be $x$ metres. Then the length is $(x + 5)$ metres.
2. Set up the area equation:
  $x(x + 5) = 36$
3. Expand and rearrange:
  $x^2 + 5x - 36 = 0$
4. Factorise: find two integers that multiply to $-36$ and add to $5$ : these are $9$ and $-4$ .
  $(x + 9)(x - 4) = 0$
5. Apply the null factor law: $x = -9$ or $x = 4$ . Since width must be positive, $x = 4$ .
6. Width $= 4$ m, Length $= 4 + 5 = 9$ m. Check: $4 \times 9 = 36$ ✓
Answer: The garden is $4$ metres wide and $9$ metres long.

Common mistakes

Where students usually lose marks

Forgetting to set the equation equal to zero before factorising

$x^2 - x = 6 \Rightarrow x(x-1) = 6 \Rightarrow x = 6$ or $x - 1 = 6$

The null factor law only works when one side equals zero. Always rearrange first to get $x^2 - x - 6 = 0$ , then factorise to get $(x-3)(x+2) = 0$ .

Giving only one solution when there are two

$(x-4)(x+1) = 0 \Rightarrow x = 4$ (forgetting $x = -1$ )

Each bracket gives its own solution. Always write $x - 4 = 0$ or $x + 1 = 0$ and solve both equations separately.

Getting the sign wrong when reading roots from factorised form

$(x - 3)(x + 5) = 0 \Rightarrow x = 3$ or $x = 5$

Set each bracket equal to zero and solve: $x - 3 = 0$ gives $x = 3$ , and $x + 5 = 0$ gives $x = -5$ (not $+5$ ). The sign in the bracket flips when you solve.

Misreading the effect of a on the parabola's width

" $y = 5x^2$ is wider than $y = x^2$ because $5$ is a big number."

A larger positive value of $a$ makes the parabola narrower, not wider. The graph is steeper because $y$ values grow faster. A value of $a$ between $0$ and $1$ makes the parabola wider.

Try one yourself

Your turn

Solve $x^2 - 7x + 10 = 0$ using the null factor law. Enter the two solutions as a pair of numbers.

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About this worksheet

Why we built it

Year 9 is when students meet quadratics properly. These worksheets give your student varied practice across graphing parabolas, using the null factor law, and solving monic quadratics with integer roots - the exact skills tested at the Year 9 level.

What's covered

Sub-skills your student will practise

✓Recognising the standard parabola y = x^2
✓Effect of transformations on a parabola (y = ax^2 + c)
✓Solving quadratic equations by null factor law
✓Reading roots from a graph
✓Setting up a quadratic from a worded scenario

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Frequently asked

Questions parents ask about quadratics

What's a quadratic function?

A function where the highest power of x is 2 - so y = x^2, y = x^2 + 3x, y = 2x^2 - 5x + 1. Their graphs are always parabolas (U-shaped or upside-down U).

What's the null factor law?

If two things multiply to zero, at least one of them must be zero. So if (x - 3)(x + 5) = 0, then either x - 3 = 0 or x + 5 = 0. This is the main method Year 9 uses to solve quadratics.

How do students sketch a parabola?

Find the y-intercept (set x = 0), the x-intercepts if they exist (set y = 0 and solve), and the vertex (the turning point). Mark those points and draw a smooth curve.

What kind of quadratics does Year 9 solve?

Monic quadratics (where the x^2 coefficient is 1) with integer roots, using null factor law after factorising. Quadratic formula and completing the square arrive in Year 10.

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