Question

if a is equals to 2 + under root 3 then find a square + 1 upon a square ​

Answer 1

Answer:

$\boxed{\red{\bf a^2 + \dfrac{1}{a^2} = 14}}$

Step-by-step explanation:

Given that -

• a = 2 + √3

To find :

• a² + $\sf \dfrac{1}{a^2}$

Solution :

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

So,

$\sf \implies \dfrac{1}{a} = \dfrac{1}{2 + \sqrt{3} }$

Rationlising the denominator ;

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

Adding both :

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

On squaring both sides :

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

Hence, the answer is 14.