Question

3. Find p and q, if √3 −1√3 +1 = p + q√3.

Answer 1

ATQ;

$\Longrightarrow \sf \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1} = p + q\sqrt{3}$

Rationalizing the denominator we get;

$\Longrightarrow \sf \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1} = p + q\sqrt{3}$

$\Longrightarrow \sf \dfrac{\Big(\sqrt{3} - 1\Big)^2}{\big(\sqrt{3}\big)^2 - \big(1\big)^2} = p + q\sqrt{3}$

$\Longrightarrow \sf \dfrac{\big(\sqrt{3}\big)^2 + \big(1\big)^2 - 2\big(\sqrt{3}\big)\big(1\big)}{\big(\sqrt{3}\big)^2 - \big(1\big)^2} = p + q\sqrt{3}$

$\Longrightarrow \sf \dfrac{3 + 1 - 2\sqrt{3}}{3 - 1} = p + q\sqrt{3}$

$\Longrightarrow \sf \dfrac{4 - 2\sqrt{3}}{2} = p + q\sqrt{3}$

$\Longrightarrow \sf \dfrac{2 \big(2 - \sqrt{3}\big)}{2} = p + q\sqrt{3}$

$\Longrightarrow \sf 2 - \sqrt{3} = p + q\sqrt{3}$

$\Longrightarrow \sf 2 + \Big(-1 \sqrt{3}\Big) = p + q\sqrt{3}$

Final answers:

⇒ p = 2

⇒ q = -1