Hidden Quadratics

Spot a quadratic in disguise: in √x, x², x³, x^(1/3) or a^x; let u = …

In the exam for Edexcel International AS Maths · 5 real past-paper questions below

Spot the disguise

You can already solve any quadratic, by factorising, completing the square or the formula. AS questions hide a quadratic inside something that does not look like one.

Look for two powers where one is double the other. Then the bigger one is the smaller one squared.

Formula

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

$u$ — the smaller power: $x$ , $x^{\frac{1}{3}}$ , $x^{2}$ , $x^{3}$ or $2^{x}$ . Its square is the bigger power.

You see	Let
$x$ and $x$	$u = x$ , so $u^{2} = x$
$x^{\frac{2}{3}}$ and $x^{\frac{1}{3}}$	$u = x^{\frac{1}{3}}$ , so $u^{2} = x^{\frac{2}{3}}$
$x^{4}$ and $x^{2}$	$u = x^{2}$
$x^{6}$ and $x^{3}$	$u = x^{3}$
$4^{x}$ and $2^{x}$	$u = 2^{x}$ , since $4^{x} = (2^{x})^{2}$

Worked example

1
Solve $2 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 2 = 0$ .
2
$u = x^{\frac{1}{3}}, u^{2} = x^{\frac{2}{3}}$ — The powers are $\frac{1}{3}$ and $\frac{2}{3}$ . One is double the other.
3
$2 u^{2} + 3 u - 2 = 0$ — Write the quadratic in $u$ . Show this line: it earns the first method mark.
4
$(2 u - 1) (u + 2) = 0$ — Factorise. Show the brackets, not just the answers.
5
$u = \frac{1}{2} or u = - 2$ — These are values of $u$ , not of $x$ . Not finished yet.
6
$x^{\frac{1}{3}} = \frac{1}{2} \Rightarrow x = (\frac{1}{2})^{3} = \frac{1}{8}$ — Swap back. Undo a cube root by cubing.
7
$x^{\frac{1}{3}} = - 2 \Rightarrow x = (- 2)^{3} = - 8$ — A cube root can be negative, so keep this one too.

Memory trick

Swap, solve, swap back. Like a parcel: unwrap it, sort what is inside, then wrap it up again.

Common mistake

Stopping at $u = \frac{1}{2}$ and $u = - 2$ . The question asked for $x$ . Always swap back.

Key idea

Around 820, al-Khwarizmi in Baghdad wrote a book showing how to turn any equation into one of a few standard types. Its title gave us the word algebra, from al-jabr. A hidden quadratic is the same idea: change it into a type you know.

Key idea

If one power is double the other, it is a quadratic in the smaller power.

Square roots: keep or reject

With $x$ and $x$ , let $u = x$ . Then you must check each $u$ : a square root is never negative.

Worked example

1
Solve $2 x - 5 x - 12 = 0$ .
2
$u = x, 2 u^{2} - 5 u - 12 = 0$ — $x$ is $(x)^{2}$ , so $x = u^{2}$ .
3
$(2 u + 3) (u - 4) = 0$ — Factorise. $2 \times (- 12) = - 24$ , and $3 - 8 = - 5$ .
4
$u = 4 or u = - \frac{3}{2}$ — Two values of $u$ .
5
$x = - \frac{3}{2} is impossible$ — Reject it, and say why: $x$ cannot be negative.
6
$x = 4 \Rightarrow x = 16$ — Square to swap back. Check: $32 - 20 - 12 = 0$ .

Common mistake

Squaring $- \frac{3}{2}$ to get $x = \frac{9}{4}$ . Put it back in: $4.5 - 7.5 - 12 \neq = 0$ . It is not a solution. If you find it, cross it out with a reason.

When the quadratic in $u$ does not factorise, the answers are surds. Keep them exact.

Worked example

1
Solve $x - 6 x + 2 = 0$ . Give your answers in the form $a + b c$ .
2
$u^{2} - 6 u + 2 = 0$ — Let $u = x$ .
3
$(u - 3)^{2} - 9 + 2 = 0 \Rightarrow (u - 3)^{2} = 7$ — Complete the square. Half of 6 is 3.
4
$u = 3 \pm 7$ — Both are positive, since $7 \approx 2.65 < 3$ . Keep both.
5
$x = (3 + 7)^{2} = 9 + 67 + 7 = 16 + 67$ — Swap back by squaring the whole bracket. There are three terms.
6
$x = (3 - 7)^{2} = 16 - 67$ — Same steps with a minus sign in the middle term.

Common mistake

Squaring term by term: $(3 + 7)^{2} = 9 + 7$ . You lose the middle term $2 \times 3 \times 7 = 67$ , and the mark.

Rule

These questions say “you must show all stages of your working”. Write $u = x$ , the quadratic in $u$ , and the factorised form, completed square or formula with numbers in. Roots straight from a calculator score nothing.

Key idea

Square root: reject any negative $u$ , with a reason. Then square the whole $u$ , brackets and all.

Powers like 2ˣ

When $x$ is in the power, look for $2^{2 x}$ or $4^{x}$ beside $2^{x}$ . Use index laws to write it as $(2^{x})^{2}$ .

Formula

2^{2 x + 1} = 2^{1} \times 2^{2 x} = 2 (2^{x})^{2}

$2^{x}$ — this becomes $u$ . A $+ 1$ in the power splits off as a $\times 2$ in front.

Worked example

1
Solve $2^{2 x + 1} - 7 \times 2^{x} - 4 = 0$ .
2
$2^{2 x + 1} = 2 \times 2^{2 x} = 2 (2^{x})^{2}$ — Write the index law out. In a “show that” part, this line is required.
3
$u = 2^{x}, 2 u^{2} - 7 u - 4 = 0$ — Now it is a quadratic in $u$ .
4
$(2 u + 1) (u - 4) = 0$ — Factorise. $1 - 8 = - 7$ .
5
$u = 4 or u = - \frac{1}{2}$ — Two values of $u$ .
6
$2^{x} = - \frac{1}{2} is impossible$ — Reject it: a power of 2 is always positive.
7
$2^{x} = 4 = 2^{2} \Rightarrow x = 2$ — Write 4 as a power of 2, then match the powers.

A positive $u$ that is not a whole power still gives an exact answer. For example $2^{x} = \frac{1}{8} = 2^{- 3}$ gives $x = - 3$ , and $9^{x} = 3 = 9^{\frac{1}{2}}$ gives $x = \frac{1}{2}$ .

Exam questions often come in two parts. Part (a): use $u = 3^{x}$ to show that the equation becomes $3 u^{2} - 10 u + 3 = 0$ . Part (b): “Hence solve”. Solve that quadratic: $(3 u - 1) (u - 3) = 0$ , so $u = \frac{1}{3}$ or $u = 3$ . Swap back: $x = - 1$ or $x = 1$ .

Common mistake

Writing $2^{2 x + 1} = 2^{2 x} + 2$ . Adding powers means multiplying: $2^{2 x + 1} = 2^{2 x} \times 2$ .

Common mistake

Writing $2^{2 x} = 4 x$ . It is $(2^{x})^{2}$ , which you can also write as $4^{x}$ .

Key idea

Split the power with index laws until you see $(a^{x})^{2}$ . Any $a^{x} \leq 0$ is rejected.

Simultaneous equations to a quartic

Sometimes substituting one equation into the other gives $x^{4}$ and $x^{2}$ only. That is a quadratic in $x^{2}$ .

Worked example

1
Solve $x y + x^{2} = 24$ and $y = x^{3} - 3 x$ . Give exact answers.
2
$x (x^{3} - 3 x) + x^{2} = 24$ — Replace $y$ . Keep the bracket.
3
$x^{4} - 3 x^{2} + x^{2} = 24$ — Multiply out. Write both $x^{2}$ terms before you collect them.
4
$x^{4} - 2 x^{2} - 24 = 0$ — Collect. A “show that” would stop here.
5
$(x^{2} - 6) (x^{2} + 4) = 0$ — A quadratic in $x^{2}$ : $- 6 \times 4 = - 24$ and $- 6 + 4 = - 2$ .
6
$x^{2} = 6 (x^{2} = - 4 rejected)$ — A square cannot be negative, so $x^{2} = - 4$ has no real roots.
7
$x = \pm 6$ — Square root, and keep both signs.
8
$y = (6)^{3} - 36 = 66 - 36 = 36$ — Back-substitute. $(6)^{3} = 66$ .
9
$(6, 36) and (- 6, - 36)$ — Each $x$ gives its own $y$ . Write them as pairs.

A follow-on part may ask for the exact distance between the two points. Here it is $(26)^{2} + (66)^{2} = 24 + 216 = 240 = 415$ .

Common mistake

Stopping at $x^{2} = 6$ or writing only $x = 6$ . Find both $x$ values, then a $y$ for each.

Common mistake

Listing $x = \pm 6$ , $y = \pm 36$ with no pairing. The last accuracy mark needs clear pairs.

Key idea

Substitute, collect to $a x^{4} + b x^{2} + c = 0$ , solve for $x^{2}$ , reject a negative $x^{2}$ , then $\pm$ and back-substitute.

Which method? Read the question

The powers tell you what $u$ is. The meaning of $u$ tells you what to reject.

You see	You do
$x$ and $x$	$u = x$ ; reject negative $u$ ; square to finish
$x^{\frac{2}{3}}$ and $x^{\frac{1}{3}}$ , or $x^{6}$ and $x^{3}$	$u = x^{\frac{1}{3}}$ or $u = x^{3}$ ; keep negative $u$ ; cube or cube root to finish
$4^{x}$ , $2^{2 x + 1}$ and $2^{x}$	Index laws, $u = 2^{x}$ ; reject $u \leq 0$
Two equations giving $x^{4}$ and $x^{2}$	$u = x^{2}$ ; reject negative $x^{2}$ ; $\pm$ ; find each $y$
“Hence solve” after a cubic or quadratic	Set the disguised part equal to each answer you already have
“In the form $a + b c$ ”	Complete the square or use the formula; square the whole bracket
The unknown is a length	Reject a negative root: a length cannot be negative
“Show all stages of your working”	Write $u = \dots$ , the quadratic, and your solving line

Common mistake

Crossing out a root with no reason. Write the reason: “ $x \geq 0$ ”, “ $2^{x} > 0$ ”, “ $x^{2}$ cannot be negative” or “a length cannot be negative”.

Key idea

Spot the double power, swap to $u$ , solve, reject what is impossible, swap back.

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Real past-paper questions on hidden quadratics

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