Question

If a=√3+1

Find the value of a^4+16/a^4

Answer 1

Answer:

this is ur solution

MARK AS BRAINLIEST

Answer 2

We need to recall the following Formula

1. $(a+b)^2=a^2+b^2+2ab$

2. $(a+b)(a-b)=a^2-b^2$

And the concept of conjugate

Let $a+b\sqrt{c}$ is expression then  $a-b\sqrt{c}$ is conjugate of this expression

This problem is about the linear equation.

Given  $a=\sqrt{3}+1$

We have to find the value of  $a^4+\frac{16}{a^4}$

Substitute a value in an expression $a^4+\frac{16}{a^4}$

then,

$({\sqrt{3}+1 })^4+\frac{16}{({\sqrt{3}+1 })^4}={({\sqrt{3}+1 })^2}^2+\frac{16}{{({\sqrt{3}+1 })^2}^2}$

                                $=(3+1+2\sqrt{3})^2+\frac{16}{(3+1+2\sqrt{3})^2}$

                                $=(4+2\sqrt{3})^2+\frac{16}{(4+2\sqrt{3})^2}$

                                $=16+12+4\sqrt{3} +\frac{16}{16+12+4\sqrt{3} }$

Now take the LCM

 $16+12+4\sqrt{3} +\frac{16}{16+12+4\sqrt{3} }=28+4\sqrt{3} +\frac{16}{28+4\sqrt{3}}$

                                              $=\frac{(28+4\sqrt{3} )^2+{16}}{28+4\sqrt{3}}$

                                              $=\frac{(784+48+8\sqrt{3})+16 }{28+4\sqrt{3} }$

                                              $=\frac{848+8\sqrt{3} }{28+4\sqrt{3} }$

Multiply the conjugate $28-4\sqrt{3}$ to simplify the expression

$\frac{848+8\sqrt{3} }{28+4\sqrt{3} }=\frac{848+8\sqrt{3} }{28+4\sqrt{3} }\times \frac{28-4\sqrt{3} }{28-4\sqrt{3} }$

              $=\frac{848+8\sqrt{3} \times28-4\sqrt{3} }{28^2-(4\sqrt{3})^2}$

              $=\frac{23648-3168\sqrt{3} }{736}$

Hence the value of $a^4+\frac{16}{a^4}$ is $\frac{23648-3168\sqrt{3} }{736}$.