Solving Equations with Surds

Linear equation with surd coefficients: collect x terms, factor x out of a surd bracket, divide, rationalise; answer p + q√3

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

Surds as coefficients

You can already simplify surds and rationalise a bottom like $3 - 2$ . AS questions now put surds in an equation: you must find $x$ , and give it as an exact surd.

Treat $3$ like any number. Get every $x$ term on one side, take $x$ out as a factor, divide, then rationalise.

Worked example

1
Solve $3 (2 x - 3) = x + 1$ . Give $x$ in the form $p + q 3$ , where $p$ and $q$ are fractions.
2
$23 x - 3 = x + 1$ — Expand the left side: $3 \times 3 = 3$ .
3
$23 x - x = 4$ — Move the $x$ terms left and the numbers right.
4
$x (23 - 1) = 4$ — Take $x$ out as a factor. The plain $x$ leaves a 1 behind.
5
$x = \frac{4}{2 3 - 1}$ — Divide by the bracket. A surd is on the bottom, so we are not finished.
6
$x = \frac{4 ( 2 3 + 1 )}{( 2 3 - 1 ) ( 2 3 + 1 )}$ — Multiply top and bottom by the bracket with the sign swapped.
7
$bottom = (23)^{2} - 1^{2} = 12 - 1 = 11$ — $(23)^{2} = 4 \times 3 = 12$ .
8
$x = \frac{8 3 + 4}{11} = \frac{4}{11} + \frac{8}{11} 3$ — Split into the form asked: $p = \frac{4}{11}$ , $q = \frac{8}{11}$ .

Memory trick

Gather, bag, divide, clean. Gather the $x$ terms, bag them in a bracket, divide by the bracket, then clean the surd off the bottom.

Rule

These questions say “show all stages of your working”. Method marks go to: collecting the $x$ terms, reaching $x = \dots$ , and the rationalising line. Jumping from $x (23 - 1) = 4$ to the answer looks like a calculator, and loses the last marks.

Common mistake

Writing $23 x - x$ as $x (23)$ . The lone $x$ is $1 x$ , so the bracket is $(23 - 1)$ .

Key idea

Why “surd”? Arab mathematicians called these roots asamm, “deaf”: numbers that cannot be spoken exactly. Latin translators wrote surdus, which also means deaf.

Key idea

A surd equation is a normal linear equation. The only extra job is rationalising at the end.

Simplify roots before collecting

Some questions mix roots like $2$ and $18$ . They look different, so you cannot collect them. Simplify every root first, then they match.

Worked example

1
Write $(2 x + 2)^{2} + (x - 18)^{2}$ in the form $a x^{2} + b x 2 + c$ .
2
$18 = 9 \times 2 = 32$ — Simplify first: 9 is the biggest square in 18.
3
$(2 x + 2)^{2} = 4 x^{2} + 42 x + 2$ — Every term times every term. $2 \times 2 = 2$ .
4
$(x - 32)^{2} = x^{2} - 62 x + 18$ — $(32)^{2} = 9 \times 2 = 18$ . The middle term is $2 \times x \times (- 32)$ .
5
$5 x^{2} - 22 x + 20$ — Collect: $4 x^{2} + x^{2}$ , then $42 x - 62 x$ , then $2 + 18$ . So $a = 5$ , $b = - 2$ , $c = 20$ .

The same habit works inside an equation.

Worked example

1
Solve $12 x + 5 = 3 x + 11$ .
2
$12 = 23$ — Simplify, so both $x$ terms use $3$ .
3
$23 x - 3 x = 6$ — Collect: $x$ terms left, numbers right.
4
$3 x = 6$ — $23 - 3 = 3$ , just like $2 a - a = a$ .
5
$x = \frac{6}{3} = \frac{6 3}{3} = 23$ — Divide, then rationalise.

Rule

A mark scheme can give a mark just for writing $18 = 32$ (or $8 = 22$ ). Write it as a line of its own.

Common mistake

Leaving $18$ as it is. Then $- 218 x$ and $42 x$ never combine, and the coefficient of $x$ comes out wrong.

Key idea

Before you collect, make every root as small as it can be.

Equations in √x

Where two graphs like $y = 3 x$ and $y = 6 - x$ meet, you get an equation in $x$ . Treat $x$ as the unknown, just as you treated $x$ . Square only at the very end.

Formula

3 x = 3 \times x

$3$ — the number, which comes out as its own root.
$x$ — the unknown, which stays as $x$ .

Worked example

1
Find the exact point where $y = 3 x$ meets $y = 6 - x$ .
2
$3 x = 6 - x$ — Set the two equal and split $3 x$ .
3
$3 x + x = 6$ — Collect the $x$ terms on one side.
4
$x (3 + 1) = 6$ — Take $x$ out as a factor.
5
$x = \frac{6 ( 3 - 1 )}{( 3 + 1 ) ( 3 - 1 )} = \frac{6 ( 3 - 1 )}{2}$ — Divide and rationalise: the bottom is $3 - 1 = 2$ .
6
$x = 33 - 3$ — This is $x$ , not $x$ . One more step.
7
$x = (33 - 3)^{2} = 27 - 183 + 9 = 36 - 183$ — Square the whole bracket: three terms, with $(33)^{2} = 27$ .
8
$y = 6 - x = 6 - (33 - 3) = 9 - 33$ — For $y$ , use the $x$ you already have. No new roots needed.

Rule

If the question gives the answer, as in “show that $x = 33 - 3$ ”, every line must be seen, including the rationalising. A missing line costs the final mark.

Common mistake

Squaring $33 - 3$ as $27 + 9$ . Squaring a bracket gives three terms: do not lose the middle one.

Key idea

Find $x$ first, as if it were $x$ . Square it last to get $x$ .

“Hence” uses part (a)

Many surd questions come in two parts. Part (a) simplifies a fraction. Part (b) says “hence solve”. That means: use your answer to (a), so you do not rationalise again.

Worked example

1
(a) Rationalise $\frac{1}{3 - 7}$ . (b) Hence solve $3 x = 7 x + 47$ .
2
$\frac{1}{3 - 7} \times \frac{3 + 7}{3 + 7} = \frac{3 + 7}{9 - 7} = \frac{3 + 7}{2}$ — (a) Swap the sign and multiply. The bottom is $9 - 7 = 2$ .
3
$3 x - 7 x = 47$ — (b) Collect the $x$ terms on the left.
4
$x (3 - 7) = 47$ — Take $x$ out. The bracket is the bottom from part (a).
5
$x = 47 \times \frac{1}{3 - 7}$ — Write $x$ as a number times the fraction from (a). This shows you used (a).
6
$x = 47 \times \frac{3 + 7}{2} = 27 (3 + 7)$ — Swap in your answer to (a). $4 \div 2 = 2$ .
7
$x = 67 + 14$ — $27 \times 7 = 2 \times 7 = 14$ .

Rule

“Hence” means use (a), so show the line $x = 47 \times (part a)$ . “Hence, or otherwise” lets you use another method, but “hence” is usually the shortest.

Common mistake

Ignoring part (a) and starting again. It costs time, and a fresh slip can cost marks that (a) had already won.

Key idea

Before part (b), look back at part (a). Its bracket is usually the one you need.

Which method? Read the question

The question tells you which tool to use.

You see	You do
$x$ terms with surd numbers in front	Collect, take $x$ out, divide, rationalise
Different roots, like $2$ and $8$	Simplify each root first, then collect
$x$ or $3 x$ in the equation	Solve for $x$ , then square the whole answer
“Hence” after a part (a) fraction	Write $x$ as a number times your part (a) answer
“Form $p + q 3$ ”	Split the fraction so the form shows
“Show all stages of your working”	Write the rationalising line in full

Key idea

End with no surd on the bottom, every root as small as possible, and the answer in the form asked.

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