Sec 3 · Additional Mathematics

Equations and inequalities

Practise equations and inequalities with step-by-step help. Free account.

1. Completing the square

a(x + h)² + k: take out a first

\displaystyle 2x^{2} - 12x + 5

e.g.

\displaystyle 2x^{2} - 12x + 5

\displaystyle = 2(x^{2} - 6x) + 5

\displaystyle = 2((x - 3)^{2} - 9) + 5

\displaystyle = 2(x - 3)^{2} - 13

Solving: divide by a first

\displaystyle 2x^{2} - 8x + 3 = 0

e.g.

\displaystyle 2x^{2} - 8x + 3 = 0

\displaystyle x^{2} - 4x + \dfrac{3}{2} = 0

\displaystyle (x - 2)^{2} = \dfrac{5}{2}

\displaystyle x = 2 \pm \sqrt{\dfrac{5}{2}}

\displaystyle x = 3.58 \text{ or } x = 0.42\text{ (2 d.p.)}

📖 Worked example

Express $3x^{2} - 3x + 10$ in the form $a(x + h)^2 + k$ .

Step 1 of 4

Take out the coefficient of x² from the x terms

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

Practise this →

2. Nature of roots

$b^{2} - 4ac$	Roots of $ax^{2} + bx + c = 0$	Example
$> 0$	two real, distinct roots	$x^{2} - 5x + 4 = 0$ has $b^{2} - 4ac = 9$
$= 0$	two real, equal roots	$x^{2} - 6x + 9 = 0$ has $b^{2} - 4ac = 0$
$< 0$	no real roots	$x^{2} + 2x + 5 = 0$ has $b^{2} - 4ac = -16$
$\geq 0$	real roots

Equal roots: find k

\displaystyle x^{2} + kx + 36 = 0

e.g.

\displaystyle k^{2} - 4(1)(36) = 0

\displaystyle k = -12 \text{ or } k = 12

No real roots: range of k

\displaystyle x^{2} - 4x + k = 0

$(-4)^{2} - 4(1)(k) < 0$
$16 - 4k < 0$ , so $k > 4$

📖 Worked example

Find the range of values of $k$ for which $x^2 - 4x + k = 0$ has no real roots.

Step 1 of 3

Which condition on the discriminant gives no real roots?

b^2 - 4ac < 0

k in more than one coefficient: b² − 4ac is a quadratic in k

\displaystyle kx^{2} + (k + 3)x + k = 0

$b^{2} - 4ac = (k + 3)^{2} - 4(k)(k)$
$= -3k^{2} + 6k + 9$
Equal roots: $-3(k - 3)(k + 1) = 0$ , so $k = 3$ or $k = -1$
Reject a $k$ that makes the $x^{2}$ coefficient $0$ : then it is not a quadratic

Range of k: solve a quadratic inequality in k

\displaystyle x^{2} + (k - 6)x + k - 3 = 0

No real roots: $(k - 6)^{2} - 4(k - 3) < 0$
$k^{2} - 16k + 48 < 0$ , so $(k - 4)(k - 12) < 0$
$4 < k < 12$ (between the roots)

📖 Worked example

The equation $kx^2 + (k + 3)x + k = 0$ , where $k$ is a constant, has two equal roots. Find the possible values of $k$ .

Step 1 of 6

Which condition on the discriminant gives two equal roots?

b^2 - 4ac = 0

📖 Worked example

Find the range of values of $k$ for which the equation $x^2 + (k - 6)x + k - 3 = 0$ has no real roots.

Step 1 of 7

Which condition on the discriminant gives no real roots?

b^2 - 4ac < 0

Practise this →

3. Where a line meets a curve

Three graphs of a U-shaped curve and a line: the line cuts the curve at two points, touches it at one point (a tangent), or misses it.

The line	$b^{2} - 4ac$ of the combined quadratic
meets (intersects) the curve	$\geq 0$
meets it at two distinct points	$> 0$
is a tangent to the curve	$= 0$
does not meet the curve	$< 0$

Make the y's equal, then use b² − 4ac

\displaystyle x^{2} - 3x + 2 = -x + k

e.g.

\displaystyle x^{2} - 3x + 2 = -x + k

\displaystyle x^{2} - 2x + 2 - k = 0

Tangent: b² − 4ac = 0

\displaystyle (-2)^{2} - 4(1)(2 - k) = 0

e.g.

\displaystyle (-2)^{2} - 4(1)(2 - k) = 0

\displaystyle 4k - 4 = 0

\displaystyle k = 1

📖 Worked example

Find the $x$ -coordinates of the points where the line $y = 2x + 1$ meets the curve $y = x^2 - 2x + 1$ .

Step 1 of 2

Where they meet, the y-values are equal: make one quadratic

x^2 - 2x + 1

=

2x + 1

x^2 - 4x = 0

Curves with xy or y² terms: substitute the line

\displaystyle xy = 12,\ y = x - 4

e.g.

\displaystyle x(x - 4) = 12

\displaystyle x^{2} - 4x - 12 = 0

\displaystyle (x - 6)(x + 2) = 0

\displaystyle x = 6 \text{ or } x = -2

📖 Worked example

The line $y = x - 4$ meets the curve $xy = 12$ at two points. Find the $x$ -coordinates of these points.

Step 1 of 3

Substitute the line into the curve: replace every y

x(x - 4) = 12

Practise this →

4. Quadratic inequalities

The curve y = (x − 1)(x − 4) is below the x-axis between 1 and 4, and above it for x < 1 or x > 4.

Inequality ( $a < b$ )	Solution
$(x - a)(x - b) < 0$	$a < x < b$ (between)
$(x - a)(x - b) > 0$	$x < a$ or $x > b$ (outside)
$(x - a)(x - b) \leq 0$	$a \leq x \leq b$ (the roots included)

Factorise, then sketch

\displaystyle x^{2} - 5x + 4 < 0

e.g.

\displaystyle x^{2} - 5x + 4 < 0

\displaystyle (x - 1)(x - 4) < 0

\displaystyle 1 < x < 4

Outside the roots

\displaystyle x^{2} - 5x + 4 > 0

e.g.

\displaystyle (x - 1)(x - 4) > 0

\displaystyle x < 1 \text{ or } x > 4

📖 Worked example

Solve $x^2 - 6x > 0$ .

Step 1 of 4

Factorise

x(x - 6) > 0

Practise this →

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Sec 3 · Additional Mathematics

Equations and inequalities

Practise equations and inequalities with step-by-step help. Free account.

Start free Worksheet (PDF)

1. Completing the square

a(x + h)² + k: take out a first

\displaystyle 2x^{2} - 12x + 5

e.g.

\displaystyle 2x^{2} - 12x + 5

\displaystyle = 2(x^{2} - 6x) + 5

\displaystyle = 2((x - 3)^{2} - 9) + 5

\displaystyle = 2(x - 3)^{2} - 13

Solving: divide by a first

\displaystyle 2x^{2} - 8x + 3 = 0

e.g.

\displaystyle 2x^{2} - 8x + 3 = 0

\displaystyle x^{2} - 4x + \dfrac{3}{2} = 0

\displaystyle (x - 2)^{2} = \dfrac{5}{2}

\displaystyle x = 2 \pm \sqrt{\dfrac{5}{2}}

\displaystyle x = 3.58 \text{ or } x = 0.42\text{ (2 d.p.)}

📖 Worked example

Express $3x^{2} - 3x + 10$ in the form $a(x + h)^2 + k$ .

Step 1 of 4

Take out the coefficient of x² from the x terms

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

Practise this →

2. Nature of roots

$b^{2} - 4ac$	Roots of $ax^{2} + bx + c = 0$	Example
$> 0$	two real, distinct roots	$x^{2} - 5x + 4 = 0$ has $b^{2} - 4ac = 9$
$= 0$	two real, equal roots	$x^{2} - 6x + 9 = 0$ has $b^{2} - 4ac = 0$
$< 0$	no real roots	$x^{2} + 2x + 5 = 0$ has $b^{2} - 4ac = -16$
$\geq 0$	real roots

Equal roots: find k

\displaystyle x^{2} + kx + 36 = 0

e.g.

\displaystyle k^{2} - 4(1)(36) = 0

\displaystyle k = -12 \text{ or } k = 12

No real roots: range of k

\displaystyle x^{2} - 4x + k = 0

$(-4)^{2} - 4(1)(k) < 0$
$16 - 4k < 0$ , so $k > 4$

📖 Worked example

Find the range of values of $k$ for which $x^2 - 4x + k = 0$ has no real roots.

Step 1 of 3

Which condition on the discriminant gives no real roots?

b^2 - 4ac < 0

k in more than one coefficient: b² − 4ac is a quadratic in k

\displaystyle kx^{2} + (k + 3)x + k = 0

$b^{2} - 4ac = (k + 3)^{2} - 4(k)(k)$
$= -3k^{2} + 6k + 9$
Equal roots: $-3(k - 3)(k + 1) = 0$ , so $k = 3$ or $k = -1$
Reject a $k$ that makes the $x^{2}$ coefficient $0$ : then it is not a quadratic

Range of k: solve a quadratic inequality in k

\displaystyle x^{2} + (k - 6)x + k - 3 = 0

No real roots: $(k - 6)^{2} - 4(k - 3) < 0$
$k^{2} - 16k + 48 < 0$ , so $(k - 4)(k - 12) < 0$
$4 < k < 12$ (between the roots)

📖 Worked example

The equation $kx^2 + (k + 3)x + k = 0$ , where $k$ is a constant, has two equal roots. Find the possible values of $k$ .

Step 1 of 6

Which condition on the discriminant gives two equal roots?

b^2 - 4ac = 0

📖 Worked example

Find the range of values of $k$ for which the equation $x^2 + (k - 6)x + k - 3 = 0$ has no real roots.

Step 1 of 7

Which condition on the discriminant gives no real roots?

b^2 - 4ac < 0

Practise this →

3. Where a line meets a curve

The line	$b^{2} - 4ac$ of the combined quadratic
meets (intersects) the curve	$\geq 0$
meets it at two distinct points	$> 0$
is a tangent to the curve	$= 0$
does not meet the curve	$< 0$

Make the y's equal, then use b² − 4ac

\displaystyle x^{2} - 3x + 2 = -x + k

e.g.

\displaystyle x^{2} - 3x + 2 = -x + k

\displaystyle x^{2} - 2x + 2 - k = 0

Tangent: b² − 4ac = 0

\displaystyle (-2)^{2} - 4(1)(2 - k) = 0

e.g.

\displaystyle (-2)^{2} - 4(1)(2 - k) = 0

\displaystyle 4k - 4 = 0

\displaystyle k = 1

📖 Worked example

Find the $x$ -coordinates of the points where the line $y = 2x + 1$ meets the curve $y = x^2 - 2x + 1$ .

Step 1 of 2

Where they meet, the y-values are equal: make one quadratic

x^2 - 2x + 1

=

2x + 1

x^2 - 4x = 0

Curves with xy or y² terms: substitute the line

\displaystyle xy = 12,\ y = x - 4

e.g.

\displaystyle x(x - 4) = 12

\displaystyle x^{2} - 4x - 12 = 0

\displaystyle (x - 6)(x + 2) = 0

\displaystyle x = 6 \text{ or } x = -2

📖 Worked example

The line $y = x - 4$ meets the curve $xy = 12$ at two points. Find the $x$ -coordinates of these points.

Step 1 of 3

Substitute the line into the curve: replace every y

x(x - 4) = 12

Practise this →

4. Quadratic inequalities

Inequality ( $a < b$ )	Solution
$(x - a)(x - b) < 0$	$a < x < b$ (between)
$(x - a)(x - b) > 0$	$x < a$ or $x > b$ (outside)
$(x - a)(x - b) \leq 0$	$a \leq x \leq b$ (the roots included)

Factorise, then sketch

\displaystyle x^{2} - 5x + 4 < 0

e.g.

\displaystyle x^{2} - 5x + 4 < 0

\displaystyle (x - 1)(x - 4) < 0

\displaystyle 1 < x < 4

Outside the roots

\displaystyle x^{2} - 5x + 4 > 0

e.g.

\displaystyle (x - 1)(x - 4) > 0

\displaystyle x < 1 \text{ or } x > 4

📖 Worked example

Solve $x^2 - 6x > 0$ .

Step 1 of 4

Factorise

x(x - 6) > 0

Practise this →

Try a question: Equations and inequalities

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