Question

3+√7/3 -√7=a+b√7 find the volue of a and b.

Answer 1

GIVEN :–

$\\ \implies {\bold{ \dfrac{3 + \sqrt{7} }{3 - \sqrt{7} } = a + b \sqrt{7} }}$

TO FIND :–

• Value of 'a' and 'b' = ?

SOLUTION :–

$\\ \implies {\bold{ \dfrac{3 + \sqrt{7} }{3 - \sqrt{7} } = a + b \sqrt{7} }}$

• Rationalization –

$\\ \implies {\bold{ \dfrac{3 + \sqrt{7} }{3 - \sqrt{7} } \times \dfrac{3 + \sqrt{7} }{3 - \sqrt{7} } = a + b \sqrt{7} }}$

$\\ \implies {\bold{ \dfrac{(3 + \sqrt{7})^{2} }{ {(3)}^{2} - ( \sqrt{7})^{2} }= a + b \sqrt{7} }}$

$\\ \implies {\bold{ \dfrac{(3 + \sqrt{7})^{2} }{ {(3)}^{2} - ( \sqrt{7})^{2} }= a + b \sqrt{7} \: \: \: \: \left[ \: \because \: (a + b)(a - b) = {a}^{2} - {b}^{2} \: \right]}}$

$\\ \implies {\bold{ \dfrac{(3 + \sqrt{7})^{2} }{9-7}= a + b \sqrt{7} \: \: \: \: \left[ \: \because \: (a + b)^{2}= {a}^{2} + {b}^{2} + 2ab \: \right]}}$

$\\ \implies {\bold{ \dfrac{(3)^{2} + (\sqrt{7})^{2} + 2(3)( \sqrt{7})}{9-7}= a + b \sqrt{7}}}$

$\\ \implies {\bold{ \dfrac{9 + 7+ 6( \sqrt{7})}{2}= a + b \sqrt{7}}}$

$\\ \implies {\bold{ \dfrac{16+ 6( \sqrt{7})}{2}= a + b \sqrt{7}}}$

$\\ \implies {\bold{ \dfrac{2 \{8+ 3( \sqrt{7}) \}}{2}= a + b \sqrt{7}}}$

$\\ \implies {\bold{8 + 3 \sqrt{7} = a + b \sqrt{7}}}$

• Now compare –

$\\ \implies \large { \boxed {\bold{a = 8 \: \: , \: \: b = 3}}}$

Answer 2

Step-by-step explanation:

${\red{\underline{\underline{\bold{Given:-}}}}}$

$\frac{3 + \sqrt{7} }{3 - \sqrt{7} } = a + b \sqrt{7 }$

${\blue{\underline{\underline{\bold{Concept \: used\:here:-}}}}}$

• Rationalising of the numbers.

${\green{\underline{\underline{\bold{Solution:-}}}}}$

$\frac{3 + \sqrt{7} }{3 - \sqrt{7} } = a + b \sqrt{7}$

By rationalising:-

$\frac{3 + \sqrt{7} }{3 - \sqrt{7} } \times \frac{3 + \sqrt{7} }{3 + \sqrt{7} } = a + b \sqrt{7} \\ \\ \implies \frac{ {(3 + \sqrt{7} )}^{2} }{ {3}^{2} - {( \sqrt{7} )}^{2} } = a + b \sqrt{7} \: \: \: \: [(a+b) (a-b) = {a}^{2}-{b}^{2}]\\ \\ \implies \frac{ {3}^{2} + { (\sqrt{7}) }^{2} + 2(3)( \sqrt{7)} }{9- 7} = a + b \sqrt{7}\: \: \: \: [{(a+b)}^{2} = {a}^{2}+{b}^{2}+2ab] \\ \\ \implies \frac{9 + 7 + 6 \sqrt{7} }{ 2} = a + b \sqrt{7} \\ \\ \implies \frac{16 + 6 \sqrt{7} }{2} = a + b \sqrt{7}$

Take 2 as a common term in the numerator

$\implies \frac{2(8 + 3 \sqrt{7} )}{2} = a + b \sqrt{7} \\ \\ \implies8 + 3 \sqrt{7} = a + b \sqrt{7}$

Now comparing the values of a+ b√7 with 8+3√7

$\boxed{a = 8}$

$\boxed{b = 3 }$