Method 1
Expand the brackets (by multiplying all terms in the first bracket by all terms in the second)

$\sqrt{a} \times \sqrt{a} + \sqrt{a} \times (- 2 \sqrt{b}) + \sqrt{4 b} \times \sqrt{a} + \sqrt{4 b} \times (- 2 \sqrt{b}) = \sqrt{a} \sqrt{a} - 2 \sqrt{a} \sqrt{b} + \sqrt{4 b} \sqrt{a} - 2 \sqrt{4 b} \sqrt{b}$

Simplify $\sqrt{4 b}$ using the surd rule that $\sqrt{x y} = \sqrt{x} \sqrt{y}$

$\sqrt{4 b} = \sqrt{4} \sqrt{b} = 2 \sqrt{b}$

[1]

Replace $\sqrt{4 b}$ with $2 \sqrt{b}$ in the expansion above and cancel the middle terms

$\sqrt{a} \sqrt{a} - 2 \sqrt{a} \sqrt{b} + 2 \sqrt{b} \sqrt{a} - 4 \sqrt{b} \sqrt{b} = \sqrt{a} \sqrt{a} - 4 \sqrt{b} \sqrt{b}$

Simplify $\sqrt{a} \sqrt{a}$ using the surd rule that $\sqrt{x} \sqrt{x} = {(\sqrt{x})}^{2} = x$

$\sqrt{a} \sqrt{a} = a$

[1]

Replace $\sqrt{a} \sqrt{a}$ with $a$ and $\sqrt{b} \sqrt{b}$ with $b$ in the expansion

$a - 4 b$ [1]

Method 2

Simplify $\sqrt{4 b}$ using the surd rule that $\sqrt{x y} = \sqrt{x} \sqrt{y}$

$\sqrt{4 b} = \sqrt{4} \sqrt{b} = 2 \sqrt{b}$

[1]

Replace $\sqrt{4 b}$ with $2 \sqrt{b}$ in the question

$(\sqrt{a} + 2 \sqrt{b}) (\sqrt{a} - 2 \sqrt{b})$

This is the "difference of two squares", (p + q)(p - q) = p² - q²
Use the "difference of two squares" to expand the brackets

${(\sqrt{a})}^{2} - {(2 \sqrt{b})}^{2}$

Square the terms

$a - 2^{2} b$

[1]

$a - 4 b$ [1]

Expanding Brackets (OCR GCSE Maths)

Exam Questions