Question

a = v
V7+ √cand b = √63+√d where cand d are positive
integers.
Given that c:d=1:9
find, in its simplest form, the ratio a : b
(3 marks)

Answer 1

Answer:

1 : 3

Step-by-step explanation:

Given:

$a = \sqrt{7}+\sqrt{c}$

$b=\sqrt{63}+\sqrt{d}$

where c and d are positive integers.

To find the ratio a : b given that c : d = 1 : 9, begin by expressing c and d in terms of a common factor.

Since c : d = 1 : 9, we can write c = k and d = 9k, where k is the common factor.

Now, substitute these values into the expressions for a and b:

$a = \sqrt{7}+\sqrt{k}$

$b=\sqrt{63}+\sqrt{9k}$

Express the ratio of a : b in its fractional form:

$a:b=\dfrac{a}{b}=\dfrac{\sqrt{7}+\sqrt{k}}{\sqrt{63}+\sqrt{9k}}$

Rewrite 63 as 3² · 7 and 9 as 3² (prime factorization):

$\dfrac{a}{b}=\dfrac{\sqrt{7}+\sqrt{k}}{\sqrt{3^2\cdot 7}+\sqrt{3^2\cdot k}}$

$\textsf{Apply the radical rule:} \quad \sqrt{ab}=\sqrt{\vphantom{b}a}\sqrt{b}$

$\dfrac{a}{b}=\dfrac{\sqrt{7}+\sqrt{k}}{\sqrt{3^2}\sqrt{7}+\sqrt{3^2}\sqrt{k}}$

$\textsf{Apply the radical rule:} \quad \sqrt{a^2}=a, \quad a \geq 0$

$\dfrac{a}{b}=\dfrac{\sqrt{7}+\sqrt{k}}{3\sqrt{7}+3\sqrt{k}}$

Factor out the common term 3 from the denominator:

$\dfrac{a}{b}=\dfrac{\sqrt{7}+\sqrt{k}}{3\left(\sqrt{7}+\sqrt{k}\right)}$

Cancel the common factor (√7 + √k):

$\dfrac{a}{b}=\dfrac{1}{3}$

Therefore, the ratio a : b in its simplest form is:

$\huge\boxed{\boxed{1:3}}$