Question

Hi guys, can you please help me with this?
>> Given that,
$28 + p \sqrt{3} = (q + 2 \sqrt{3} ) {}^{2}$
where p and q are integers, find the values of p and q.
Answer should be: + or - 4 and + or - 16
Can I have the step by step explanation? Thanks :)​

Answer 1

Answer:

see explanation

Step-by-step explanation:

Expand the right side and then compare like terms with the left side.

(q + 2$\sqrt{3}$ )²

= q² + 4q$\sqrt{3}$ + 12

= q² + 12 + 4q$\sqrt{3}$

Compare like terms with 28 + p$\sqrt{3}$

Comparing the $\sqrt{3}$ terms gives

p = 4q

and comparing the constant terms gives

q² + 12 = 28 ( subtract 12 from both sides )

q² = 16 ( take the square root of both sides )

q = ± $\sqrt{16}$ = ± 4

Hence

p = 4q = 4 × ± 4 = ± 16