Question

if a is equal to 2 + root 3 find a square + 1 by a square​

Answer 1

Answer:

hope it helps you...........

Answer 2

Answer:

14

Step-by-step explanation:

Given that ;

$\sf a = 2 + \sqrt{3}$

Now, find the value of 1/a.

$\sf \frac{1}{a} = \frac{1}{2 + \sqrt{3} } \\ \\ \sf \frac{1}{a} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \sf \frac{1}{a} = \frac{2 - \sqrt{3} }{(2) {}^{2} - ( \sqrt{3}) {}^{2} } \\ \\ \sf \frac{1}{a} = \frac{2 - \sqrt{3} }{4 - 3} \\ \\ \sf \frac{1}{a} = 2 - \sqrt{3}$

Find the value of a + (1/a);

$\implies \sf a + \frac{1}{a} = 2 + \sqrt{3} + 2 - \sqrt{3} \\ \\ \implies \sf a + \frac{1}{a} =2 + 2 \\ \\ \implies \sf a + \frac{1}{a} = 4$

On squaring both sides ;

$\sf (a + \frac{1}{a}) {}^{2} = (4)^{2} \\ \\ \implies \sf {a}^{2} + \frac{1}{a {}^{2} } + 2 \:. \:a \: . \: \frac{1}{a} = 16 \\ \\ \\ \implies \sf {a}^{2} + \frac{1}{a {}^{2} } + 2 = 16 \\ \\ \\ \implies \sf {a}^{2} + \frac{1}{a {}^{2} } = 16 - 2 \\ \\ \\ \boxed{ \bf \therefore \: {a}^{2} + \frac{1}{a {}^{2} } = 14}$