Hidden Quadratics and Harder Simultaneous Equations

Spot a quadratic in disguise, substitute to solve it and reject impossible values, and solve a linear and a non-linear equation simultaneously.

In the exam for Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606 · 14 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

Quadratics in disguise

Look at $x^{4} - 13 x^{2} + 36 = 0$ . It has a power 4, so it does not look like a quadratic. But $x^{4}$ is just $(x^{2})^{2}$ . Call $x^{2}$ by a new letter, u, and it becomes $u^{2} - 13 u + 36 = 0$ : an ordinary quadratic in disguise.

Formula

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

$u$ — the piece that appears twice: once squared, once on its own. It might be $x^{2}$ , $x^{\frac{1}{3}}$ , $x$ or $2^{x}$ .
$a, b, c$ — ordinary numbers, exactly as in any quadratic.

Rule

The test: one power is exactly double the other. $x^{4}$ and $x^{2}$ ; $x^{\frac{2}{3}}$ and $x^{\frac{1}{3}}$ ; $x$ and $x$ ; $4^{x}$ and $2^{x}$ .

Steps

1
1. Spot the repeated piece and call it u.
2. Rewrite the equation as $a u^{2} + b u + c = 0$ .
3. Solve for u: factorise, or use the quadratic formula (printed on 0606 and 4037 papers; learn it for 4PM1).
4. Swap back and solve for x.
5. Reject any u-value the piece can never take.

0606 and 4037 test these equations directly. 4PM1 uses the same trick inside log and trig questions, and tests the simultaneous equations later in this lesson.

Key idea

Around the year 820, Muhammad ibn Musa al-Khwarizmi wrote a book in Baghdad on solving equations, including quadratics. Its title contains the word al-jabr, which became the word algebra.

Powers of x

Powers of x are the most common disguise. Choose u as the smaller power, then check that the bigger power is $u^{2}$ .

Worked example

1
Solve $x^{4} - 13 x^{2} + 36 = 0$ .
2
$u = x^{2}$ — x⁴ = (x²)², so let u stand for x².
3
$u^{2} - 13 u + 36 = 0$ — Rewrite using u.
4
$(u - 4) (u - 9) = 0$ — −4 and −9 multiply to 36 and add to −13.
5
$x^{2} = 4 or x^{2} = 9$ — Swap back: u is x².
6
$x = \pm 2, x = \pm 3$ — Each square has a positive and a negative root: four answers.

Worked example

1
Solve $x^{\frac{2}{3}} - x^{\frac{1}{3}} - 6 = 0$ .
2
$u = x^{\frac{1}{3}}$ — Two thirds is double one third, so x to the power two thirds is u².
3
$u^{2} - u - 6 = 0$ — Rewrite using u.
4
$(u - 3) (u + 2) = 0$ — −3 and 2 multiply to −6 and add to −1.
5
$x^{\frac{1}{3}} = 3 \Rightarrow x = 27$ — Cube both sides: 3³ = 27.
6
$x^{\frac{1}{3}} = - 2 \Rightarrow x = - 8$ — Cube both sides: (−2)³ = −8. Keep it: a cube root can be negative.

Rule

If the piece is under a fraction, multiply through first. For example, $x^{\frac{1}{3}} + 2 = \frac{15}{x ^{\frac{1}{3}}}$ becomes $u^{2} + 2 u - 15 = 0$ after multiplying every term by $x^{\frac{1}{3}}$ .

Mark schemes give a method mark for a 3-term quadratic in u, another for solving it, and accuracy marks for the x-values. Throwing away a correct answer, like $x = - 8$ , loses a mark.

Common mistake

Stopping at the u-values. The question asks for x, so always swap back.

Common mistake

Forgetting the negative root: $x^{2} = 4$ gives $x = 2$ and $x = - 2$ .

Key idea

Find u, then undo whatever u stands for. Square, cube or square root: do the opposite.

Values you must reject

Some disguises can never be negative. Then one of your u-values may be impossible, and you must say so and reject it.

Rule

$x$ is never negative, and neither is any even root such as $x^{\frac{1}{4}}$ . A power like $2^{x}$ is always positive. Cube roots can be negative.

Worked example

1
Solve $2 x + x - 6 = 0$ .
2
$u = x, x = u^{2}$ — x is double the power of √x.
3
$2 u^{2} + u - 6 = 0$ — Rewrite using u.
4
$(2 u - 3) (u + 2) = 0$ — Check: 2u² + 4u − 3u − 6.
5
$x = \frac{3}{2} \Rightarrow x = \frac{9}{4}$ — Square both sides.
6
$x = - 2$ — Reject: a square root is never negative.

Worked example

1
Solve $4^{x} - 2^{x + 2} - 32 = 0$ .
2
$4^{x} = (2^{2})^{x} = (2^{x})^{2}$ — Write 4 as 2², then swap the order of the powers.
3
$2^{x + 2} = 2^{x} \times 2^{2} = 4 \times 2^{x}$ — Adding powers means multiplying: split off the 2².
4
$u^{2} - 4 u - 32 = 0$ — Let u = 2ˣ.
5
$(u - 8) (u + 4) = 0$ — −8 and 4 multiply to −32 and add to −4.
6
$2^{x} = 8 \Rightarrow x = 3$ — 8 = 2³.
7
$2^{x} = - 4$ — Reject: a power of 2 is always positive.

If $2^{x}$ equals a number that is not a neat power of 2, such as 6, you need logarithms to finish. That is done in the exponential equations lesson.

Common mistake

Keeping an impossible value. Write the reason when you reject it, for example “ $x = - 2$ has no solution”.

Key idea

Before you swap back, ask: can this piece really equal that number?

A line and a curve

Now two equations: one straight line and one curve with $x^{2}$ , $x y$ or $y^{2}$ terms. Use the line to replace one letter in the curve. What is left is a quadratic in one letter.

Steps

1
1. Make x or y the subject of the linear equation. Pick the one with no fractions.
2. Substitute it into the other equation.
3. Expand carefully and collect everything into a quadratic equal to 0.
4. Solve the quadratic.
5. Put each answer into the linear equation to find its partner.

Worked example

1
Solve $x - y = 3$ and $x^{2} - 2 x y + 3 y^{2} = 11$ .
2
$y = x - 3$ — Rearrange the linear equation.
3
$x^{2} - 2 x (x - 3) + 3 (x - 3)^{2} = 11$ — Replace every y in the curve's equation.
4
$x^{2} - 2 x^{2} + 6 x + 3 x^{2} - 18 x + 27 = 11$ — Expand: (x − 3)² = x² − 6x + 9, then times 3.
5
$2 x^{2} - 12 x + 16 = 0$ — Collect like terms and subtract 11.
6
$x^{2} - 6 x + 8 = 0, (x - 2) (x - 4) = 0$ — Divide by 2, then factorise.
7
$x = 2, y = - 1 x = 4, y = 1$ — Use y = x − 3 for each x.

Mark schemes give a method mark for an equation in one letter only, an accuracy mark for the correct quadratic, a method mark for solving, and marks for the pairs.

Common mistake

Finding y from the curve's equation. That can give extra pairs that do not fit the line. Always use the linear equation.

Key idea

Line into curve, then quadratic, then pair up. Every answer is an (x, y) pair.

Shared chunks and fractions

Two common twists, set by 0606 and 4037. (4PM1 sets one linear and one quadratic equation, so it uses only the method in the last step.) First, both equations may share a whole chunk, such as $x y$ . Swap the chunk in first, and a line appears.

Worked example

1
Solve $x y = 20$ and $x y + 2 x + y = 33$ .
2
$20 + 2 x + y = 33$ — Replace xy by 20 in the second equation.
3
$y = 13 - 2 x$ — Now it is a straight line.
4
$x (13 - 2 x) = 20$ — Substitute into xy = 20.
5
$2 x^{2} - 13 x + 20 = 0$ — Expand to 13x − 2x² = 20, then move everything to one side.
6
$(2 x - 5) (x - 4) = 0$ — Check: 2x² − 8x − 5x + 20.
7
$x = 4, y = 5 x = \frac{5}{2}, y = 8$ — Use y = 13 − 2x for each x.

Second, an equation may have fractions with letters underneath. Multiply every term by both denominators first.

Worked example

1
Solve $y = x + 1$ and $\frac{x}{y} + \frac{y}{x} = \frac{5}{2}$ .
2
$2 x^{2} + 2 y^{2} = 5 x y$ — Multiply every term by 2xy to clear all the fractions.
3
$2 x^{2} + 2 (x + 1)^{2} = 5 x (x + 1)$ — Substitute y = x + 1.
4
$4 x^{2} + 4 x + 2 = 5 x^{2} + 5 x$ — Expand both sides.
5
$x^{2} + x - 2 = 0, (x + 2) (x - 1) = 0$ — Collect on one side, then factorise.
6
$x = 1, y = 2 x = - 2, y = - 1$ — Use y = x + 1 for each x.

Common mistake

Multiplying only some terms. When you clear fractions, every term on both sides is multiplied, including the $\frac{5}{2}$ .

Key idea

Simplify first: swap in a shared chunk, or clear the fractions. Then it is the usual line-into-curve method.

Which method? Read the question

Read the equations first. Their shape tells you which move to make.

You see	You do
$x^{4}$ and $x^{2}$	Let $u = x^{2}$ ; finish with ± square roots
$x^{\frac{2}{3}}$ and $x^{\frac{1}{3}}$	Let $u = x^{\frac{1}{3}}$ ; cube to finish; negatives are fine
$x$ and $x$	Let $u = x$ ; reject negative u
$4^{x}$ and $2^{x}$	Let $u = 2^{x}$ ; reject negative u
The piece under a fraction	Multiply every term through first
A line and a curve	Line into curve, solve, pair each x with its y
Both equations share $x y$	Swap the shared chunk in first

Common mistake

Giving x-values only in a simultaneous question. The answer is a set of pairs, each x with its own y.

Key idea

Make it a quadratic, solve it, then go back carefully. Most lost marks come from the going back.

Practise hidden quadratics and harder simultaneous equations now

Interactive visual aids, then real past-paper questions marked on your working like the exam. Free to start.

Practise free on Lemmara

Real past-paper questions on hidden quadratics and harder simultaneous equations

Each with a full worked solution.

All 14 questions on this topic →

Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education or Pearson Edexcel. The lessons and worked examples are Lemmara’s own.