Question

Find the values of ‘p’ and ‘q’ if root 7 + root 2/root 7 - root 2 = p + q root 14 Where p, q are
Rational Numbers​

Answer 1

√7+√2/√7-√2=p + q√14

√7+√2/√7-√2 × √7+√2/√7+√2= p+q√14 [Rationalising the denominator]

(√7+√2)²/(√7)²-(√2)²= p + q√14

[(a+b)(a-b)=a²- b²]

(√7)²+(√2)²+2(√7)(√2)/7-2= p+q√14

[(a+b)²= a²+b²+2ab]

7+2+2√14/5=p+q√14

9+2√14/5=p+q√14

It can be clearly seen that

p=9/5=1.8

q=2/5=0.4

Answer 2

Answer :

$p=\frac{9}{5} ,q=\frac{2}{5}$

Step-by-step explanation :

$\underline{\underline{\bf Rationalizing \ factor:}}$

=>   The factor of multiplication by which rationalization is done, is called as rationalizing factor.

=>   If the product of two surds is a rational number, then each surd is a rationalizing factor to other.

=>   For example, rationalizing factor of (3 + √2) is (3 - √2)

______________________________

$\bf =\frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}} =p+q\sqrt{14}$

Rationalizing factor = √7 + √2

$\bf =\frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}} \times \frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}+\sqrt{2}} \\\\\\ =\frac{(\sqrt{7}+\sqrt{2})(\sqrt{7}+\sqrt{2})}{(\sqrt{7}-\sqrt{2})(\sqrt{7}+\sqrt{2})} \\\\\\ =\frac{\sqrt{7}(\sqrt{7}+\sqrt{2})+\sqrt{2}(\sqrt{7}+\sqrt{2})}{\sqrt{7}(\sqrt{7}+\sqrt{2})-\sqrt{2}(\sqrt{7}+\sqrt{2})} \\\\\\ =\frac{\sqrt{7}^2+\sqrt{14}+\sqrt{14}+\sqrt{2}^2}{\sqrt{7}^2+\sqrt{14}-\sqrt{14}-\sqrt{2}^2}$

$\bf =\frac{7+2\sqrt{14}+2}{7-2} \\\\\\ =\frac{9+2\sqrt{14}}{5} \\\\\\ =\frac{9}{5} +\frac{2}{5} \sqrt{14} \\\\\\ =p+q\sqrt{14}$

∴ p = 9/5 , q = 2/5