Quadratic equations - Year 10 Mathematics

What you will learn

solve quadratic equations by factorisation and the null factor law,
solve quadratic equations by completing the square,
apply the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ ,
calculate the discriminant $\Delta = b^2 - 4ac$ and use it to determine the number of real solutions,
connect the discriminant to the graph of the corresponding parabola.

Worked example 0 Real-world example: dimensions of a garden bed

A rectangular garden bed has length $3$ m more than its width. Its area is $54$ m $^2$ . Find the dimensions.

Let the width be $x$ m. Then the length is $(x + 3)$ m.
Area: $x(x + 3) = 54$ , so $x^2 + 3x - 54 = 0$ .
Factorise: $(x + 9)(x - 6) = 0$ , giving $x = -9$ or $x = 6$ .
Since width must be positive, $x = 6$ . The bed is $6$ m by $9$ m.

Key idea: always check that solutions make sense in the real-world context.

1. Solving by factorisation

If a quadratic can be written as a product of two linear factors, apply the null factor law: if $AB = 0$ then $A = 0$ or $B = 0$ .

Worked example 1 Factorisation with the null factor law

Solve $2x^2 - 5x - 3 = 0$ .

Find two numbers that multiply to $2 \times (-3) = -6$ and add to $-5$ : $-6$ and $1$ .
Split: $2x^2 - 6x + x - 3 = 0$ .
Group: $2x(x - 3) + 1(x - 3) = 0$ .
Factorise: $(x - 3)(2x + 1) = 0$ .
Solutions: $x = 3$ or $x = -\dfrac{1}{2}$ .

Worked example 2 Equations not in standard form

Solve $x(x + 4) = 12$ .

Expand: $x^2 + 4x = 12$ .
Rearrange: $x^2 + 4x - 12 = 0$ .
Factorise: $(x + 6)(x - 2) = 0$ .
Solutions: $x = -6$ or $x = 2$ .

Key idea: always rearrange to $ax^2 + bx + c = 0$ before factorising — never split a product equal to a non-zero number.

2. Completing the square

Completing the square rewrites $ax^2 + bx + c$ in the form $a(x - h)^2 + k$ , where $(h, k)$ is the turning point of the parabola.

Formula reference

x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \frac{b^2}{4} + c

Worked example 3 Solving by completing the square

Solve $x^2 + 6x + 2 = 0$ .

Move the constant: $x^2 + 6x = -2$ .
Half of $6$ is $3$ ; add $3^2 = 9$ to both sides: $x^2 + 6x + 9 = 7$ .
Factor the left side: $(x + 3)^2 = 7$ .
Square root: $x + 3 = \pm\sqrt{7}$ .
Solutions: $x = -3 + \sqrt{7}$ or $x = -3 - \sqrt{7}$ .

Worked example 4 Completing the square with a leading coefficient

Solve $2x^2 - 8x + 3 = 0$ .

Divide by $2$ : $x^2 - 4x + \dfrac{3}{2} = 0$ .
Move constant: $x^2 - 4x = -\dfrac{3}{2}$ .
Half of $-4$ is $-2$ ; add $4$ : $x^2 - 4x + 4 = -\dfrac{3}{2} + 4 = \dfrac{5}{2}$ .
$(x - 2)^2 = \dfrac{5}{2}$ .
$x = 2 \pm \sqrt{\dfrac{5}{2}} = 2 \pm \dfrac{\sqrt{10}}{2}$ .

3. The quadratic formula

Formula reference

\text{If } ax^2 + bx + c = 0 \text{ then } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Worked example 5 Using the quadratic formula

Solve $3x^2 + 2x - 4 = 0$ .

$a = 3$ , $b = 2$ , $c = -4$ .
$x = \dfrac{-2 \pm \sqrt{4 - 4(3)(-4)}}{2(3)} = \dfrac{-2 \pm \sqrt{4 + 48}}{6} = \dfrac{-2 \pm \sqrt{52}}{6}$ .
$\sqrt{52} = 2\sqrt{13}$ .
$x = \dfrac{-2 \pm 2\sqrt{13}}{6} = \dfrac{-1 \pm \sqrt{13}}{3}$ .
As decimals: $x \approx 0.869$ or $x \approx -1.535$ .

4. The discriminant

The expression under the square root, $\Delta = b^2 - 4ac$ , is called the discriminant. It determines how many real solutions the equation has.

Formula reference

\Delta = b^2 - 4ac

Discriminant	Number of solutions	Graph
$\Delta > 0$	Two distinct real solutions	Parabola crosses $x$ -axis at two points
$\Delta = 0$	One repeated real solution	Parabola touches $x$ -axis at one point
$\Delta < 0$	No real solutions	Parabola does not cross $x$ -axis

Worked example 6 Using the discriminant

Determine the number of real solutions of $5x^2 - 3x + 1 = 0$ .

$\Delta = (-3)^2 - 4(5)(1) = 9 - 20 = -11$ .
$\Delta < 0$ , so the equation has no real solutions.
The parabola $y = 5x^2 - 3x + 1$ sits entirely above the $x$ -axis.

Worked example 7 Finding values that give a specific number of solutions

For what values of $k$ does $x^2 + kx + 9 = 0$ have exactly one solution?

One solution means $\Delta = 0$ : $k^2 - 4(1)(9) = 0$ .
$k^2 = 36$ .
$k = 6$ or $k = -6$ .

Worked example 8 Choosing the right method

Solve $x^2 - 10x + 7 = 0$ , giving exact answers.

Try to factorise: we need two numbers that multiply to $7$ and add to $-10$ . No integer pair works.
Use the quadratic formula: $x = \dfrac{10 \pm \sqrt{100 - 28}}{2} = \dfrac{10 \pm \sqrt{72}}{2} = \dfrac{10 \pm 6\sqrt{2}}{2} = 5 \pm 3\sqrt{2}$ .

Practice

Fluency

Tier 1: solve by factorisation and formula

Solve $x^2 - 7x + 10 = 0$ by factorisation.
Solve $x^2 + 2x - 15 = 0$ by factorisation.
Solve $3x^2 - 12x = 0$ .
Solve $x^2 - 25 = 0$ .
Use the quadratic formula to solve $x^2 + 4x + 1 = 0$ . Give exact answers.
Use the quadratic formula to solve $2x^2 - 3x - 1 = 0$ . Give exact answers.
Find the discriminant of $x^2 + 6x + 9 = 0$ and state the number of solutions.
Find the discriminant of $2x^2 - x + 3 = 0$ and state the number of solutions.
Solve $x^2 + 8x + 12 = 0$ by completing the square.
Solve $(x - 1)(x + 5) = 7$ .

Reasoning

Tier 2: applications and analysis

A rectangle has length $(x + 4)$ cm and width $(x - 2)$ cm. Its area is $32$ cm $^2$ . Find $x$ .
Solve $4x^2 - 12x + 9 = 0$ and explain why there is only one solution.
Solve $x^2 - 6x + 3 = 0$ by (a) completing the square and (b) the quadratic formula. Verify the answers agree.
For what values of $k$ does $x^2 + 4x + k = 0$ have two distinct real solutions?
The height of a ball is $h = 25t - 5t^2$ metres after $t$ seconds. When is the ball at a height of $20$ m?
Solve $\dfrac{6}{x} + x = 5$ by first multiplying through by $x$ .
The sum of a number and its reciprocal is $\dfrac{10}{3}$ . Find the number.
Show that $x^2 + 2x + 5 = 0$ has no real solutions using (a) the discriminant and (b) completing the square.

Reasoning

Tier 3: extended reasoning

Prove that the quadratic formula follows from completing the square on $ax^2 + bx + c = 0$ .
A farmer encloses a rectangular area using $80$ m of fencing against a river (three sides). Find the maximum area and the dimensions that achieve it.
The parabola $y = 2x^2 + bx + c$ passes through $(1, 3)$ and $(3, 7)$ . Find $b$ and $c$ .
For what values of $m$ does the line $y = mx + 1$ intersect the parabola $y = x^2$ at exactly one point?

Challenge

Reasoning

Harder reasoning

Solve $\dfrac{x}{x+1} + \dfrac{x+1}{x} = \dfrac{5}{2}$ by letting $u = \dfrac{x}{x+1}$ .
The roots of $x^2 - 5x + 3 = 0$ are $\alpha$ and $\beta$ . Without finding $\alpha$ and $\beta$ , evaluate $\alpha^2 + \beta^2$ and $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$ .
Find all values of $k$ such that $kx^2 + (k+2)x + 1 = 0$ has a repeated root. State any restrictions on $k$ .
A ball is thrown upward from a $15$ m platform with initial velocity $20$ m/s. Its height is $h = 15 + 20t - 5t^2$ . Find the times when it is at $30$ m height, and the time when it hits the ground. Give exact answers.

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

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Year 10 core - answers

Fluency

Tier 1: solve by factorisation and formula

$(x - 2)(x - 5) = 0$ : $x = 2$ or $x = 5$ .
$(x + 5)(x - 3) = 0$ : $x = -5$ or $x = 3$ .
$3x(x - 4) = 0$ : $x = 0$ or $x = 4$ .
$(x - 5)(x + 5) = 0$ : $x = 5$ or $x = -5$ .
$x = \dfrac{-4 \pm \sqrt{16 - 4}}{2} = \dfrac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}$ .
$x = \dfrac{3 \pm \sqrt{9 + 8}}{4} = \dfrac{3 \pm \sqrt{17}}{4}$ .
$\Delta = 36 - 36 = 0$ . One repeated solution: $x = -3$ .
$\Delta = 1 - 24 = -23 < 0$ . No real solutions.
$x^2 + 8x = -12$ . $(x + 4)^2 = 16 - 12 = 4$ . $x + 4 = \pm 2$ . $x = -2$ or $x = -6$ .
$x^2 + 4x - 5 = 7$ , so $x^2 + 4x - 12 = 0$ . $(x + 6)(x - 2) = 0$ : $x = -6$ or $x = 2$ .

Reasoning

Tier 2: applications and analysis

$(x + 4)(x - 2) = 32$ , $x^2 + 2x - 8 = 32$ , $x^2 + 2x - 40 = 0$ . $x = \dfrac{-2 \pm \sqrt{4 + 160}}{2} = \dfrac{-2 \pm \sqrt{164}}{2} = -1 \pm \sqrt{41}$ . Since $x > 2$ : $x = -1 + \sqrt{41} \approx 5.40$ cm.
$\Delta = 144 - 144 = 0$ . $(2x - 3)^2 = 0$ , so $x = \dfrac{3}{2}$ . There is one repeated solution because the parabola touches the $x$ -axis at exactly one point.
(a) $x^2 - 6x = -3$ . $(x - 3)^2 = 9 - 3 = 6$ . $x = 3 \pm \sqrt{6}$ . (b) $x = \dfrac{6 \pm \sqrt{36 - 12}}{2} = \dfrac{6 \pm \sqrt{24}}{2} = 3 \pm \sqrt{6}$ . Both methods give the same answer.
Two distinct real solutions requires $\Delta > 0$ : $16 - 4k > 0$ , so $k < 4$ .
$25t - 5t^2 = 20$ . $5t^2 - 25t + 20 = 0$ . $t^2 - 5t + 4 = 0$ . $(t - 1)(t - 4) = 0$ : $t = 1$ s or $t = 4$ s.
$\dfrac{6}{x} + x = 5$ . Multiply by $x$ : $6 + x^2 = 5x$ . $x^2 - 5x + 6 = 0$ . $(x - 2)(x - 3) = 0$ : $x = 2$ or $x = 3$ .
Let the number be $x$ . $x + \dfrac{1}{x} = \dfrac{10}{3}$ . Multiply by $3x$ : $3x^2 + 3 = 10x$ . $3x^2 - 10x + 3 = 0$ . $(3x - 1)(x - 3) = 0$ : $x = \dfrac{1}{3}$ or $x = 3$ .
(a) $\Delta = 4 - 20 = -16 < 0$ , so no real solutions. (b) $x^2 + 2x + 5 = (x + 1)^2 + 4$ . Since $(x+1)^2 \geq 0$ , the expression $\geq 4 > 0$ , so it can never equal zero.

Reasoning

Tier 3: extended reasoning

$ax^2 + bx + c = 0$ . Divide by $a$ : $x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0$ . Complete the square: $\left(x + \dfrac{b}{2a}\right)^2 = \dfrac{b^2}{4a^2} - \dfrac{c}{a} = \dfrac{b^2 - 4ac}{4a^2}$ . Square root: $x + \dfrac{b}{2a} = \pm\dfrac{\sqrt{b^2 - 4ac}}{2a}$ . Therefore $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .
Width $= x$ , length $= 80 - 2x$ . Area $= x(80 - 2x) = 80x - 2x^2$ . Axis of symmetry: $x = \dfrac{80}{4} = 20$ . Maximum area: $80(20) - 2(400) = 1600 - 800 = 800$ m $^2$ . Dimensions: $20$ m by $40$ m.
$(1, 3)$ : $2 + b + c = 3$ , so $b + c = 1$ . $(3, 7)$ : $18 + 3b + c = 7$ , so $3b + c = -11$ . Subtract: $2b = -12$ , $b = -6$ . $c = 1 - (-6) = 7$ .
$x^2 = mx + 1$ , so $x^2 - mx - 1 = 0$ . Exactly one intersection: $\Delta = 0$ . $m^2 + 4 = 0$ . Since $m^2 \geq 0$ , $m^2 + 4 \geq 4 > 0$ for all real $m$ . So $\Delta > 0$ always, meaning the line always intersects the parabola at two points — there is no value of $m$ giving exactly one intersection.

Reasoning

Challenge

Let $u = \dfrac{x}{x+1}$ , so $\dfrac{x+1}{x} = \dfrac{1}{u}$ . The equation becomes $u + \dfrac{1}{u} = \dfrac{5}{2}$ . Multiply by $2u$ : $2u^2 + 2 = 5u$ , $2u^2 - 5u + 2 = 0$ , $(2u - 1)(u - 2) = 0$ , $u = \dfrac{1}{2}$ or $u = 2$ . If $u = \dfrac{1}{2}$ : $\dfrac{x}{x+1} = \dfrac{1}{2}$ , $2x = x + 1$ , $x = 1$ . If $u = 2$ : $\dfrac{x}{x+1} = 2$ , $x = 2x + 2$ , $x = -2$ . Solutions: $x = 1$ or $x = -2$ .
By Vieta’s formulas: $\alpha + \beta = 5$ and $\alpha\beta = 3$ . $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 25 - 6 = 19$ . $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{5}{3}$ .
$\Delta = (k+2)^2 - 4k = k^2 + 4k + 4 - 4k = k^2 + 4$ . For a repeated root: $\Delta = 0$ , so $k^2 + 4 = 0$ . Since $k^2 \geq 0$ , $k^2 + 4 \geq 4 > 0$ for all real $k$ . There is no real value of $k$ that gives a repeated root (restriction: $k \neq 0$ since the equation must be quadratic).
$15 + 20t - 5t^2 = 30$ : $5t^2 - 20t + 15 = 0$ , $t^2 - 4t + 3 = 0$ , $(t-1)(t-3) = 0$ : $t = 1$ or $t = 3$ s. Hits ground: $15 + 20t - 5t^2 = 0$ , $t^2 - 4t - 3 = 0$ , $t = \dfrac{4 \pm \sqrt{16 + 12}}{2} = 2 \pm \sqrt{7}$ . Taking the positive root: $t = 2 + \sqrt{7} \approx 4.65$ s.

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