Question

If a=8+3√7 and b=1/a, then find the value of (a+b)²​

Answer 1

ANSWER :-

• (a+b)² = 256

GIVEN :-

$\\ \sf \bullet \: a = 8 + 3 \sqrt{7} \: \\ \\ \sf \bullet \: b = \dfrac{1}{a}$

TO FIND :-

$\\ \sf \: \bullet \: (a + {b)}^{2}$

SOLUTION :-

We have ,

$\sf \: a = 8 + 3 \sqrt{7}$

Taking reciprocal , we get...

$\\ \sf \: \dfrac{1}{a} = \dfrac{1}{8 + 3 \sqrt{7} }$

Rationalising the denominator ,

Rationalising factor is (8 - 3√7)

$\\ \implies \sf \dfrac{1}{a} = \dfrac{1}{8 + 3\sqrt{7} } \times \dfrac{8 - 3 \sqrt{7} }{8 - 3 \sqrt{7} } \\ \\ \\ \implies\sf \dfrac{1}{a} = \dfrac{8 - 3 \sqrt{7} }{(8 + 3 \sqrt{7})(8 - 3 \sqrt{7} ) } \\ \\ \\ \implies \sf \dfrac{1}{a} = \dfrac{8 - 3 \sqrt{7} }{ {8}^{2} -( {3 \sqrt{7}) }^{2} } \\ \\ \\ \implies \sf \dfrac{1}{a} = \dfrac{8 - 3 \sqrt{7} }{64 - 63} \\ \\ \\ \implies\sf \boxed{ \sf\dfrac{1}{a} = 8 - 3 \sqrt{7} }$

$\\ \dashrightarrow\boxed{ \sf \: \frac{1}{a} = b = 8 - 3 \sqrt{7} }$

We have ,

• a = 8 + 3√7
• b = 8 - 3√7

$\\ \to \sf \: (a + {b)}^{2} = (8 + \cancel{ 3 \sqrt{7} }+ 8 - { \cancel{3 \sqrt{7}}) } \: ^{2} \\ \\ \to \sf (a + {b)}^{2} = {16}^{2} \\ \\ \sf \to \boxed{ \sf(a + {b)}^{2} = 256} \:$