Question

If x = 7+4√3 then find the value of x+1÷x
PLZ TELL WITH EXPLANATION THIS IS FOR TEST​

Answer 1

Solution:-

Given:-

$\rm \to \: x = 7 + 4 \sqrt{3}$

To find :-

$\rm \to \: x + \dfrac{1}{x}$

Now put the value on given equation

$\rm \to \: 7 + 4 \sqrt{ 3 } + \dfrac{1}{7 + 4 \sqrt{3} }$

Now taking Lcm

$\rm \to \: \dfrac{(7 + 4 \sqrt{3} )(7 + 4 \sqrt{3} ) + 1}{7 + 4 \sqrt{3} }$

$\to \rm \: \dfrac{(7 + 4 \sqrt{3} ) {}^{2} + 1 }{7 + 4 \sqrt{3} }$

Use this identity

$\rm \to \: (a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab$

We get

$\rm \to \: \dfrac{ {7}^{2} + (4 \sqrt{3} ) {}^{2} + 2 \times 7 \times 4 \sqrt{3} + 1 }{7 + 4 \sqrt{3} }$

$\rm \to \: \dfrac{49 + 48 + 56 \sqrt{3} + 1 }{7 + 4 \sqrt{3} }$

$\rm \to \: \dfrac{49 + 49 + 56 \sqrt{3} }{7 + 4 \sqrt{3} }$

$\rm \to \: \dfrac{98 + 56 \sqrt{3} }{7 + 4 \sqrt{3} }$

$\rm \to \: \dfrac{14(7 + 4 \sqrt{3} )}{7 + 4 \sqrt{3} }$

$\rm \to \: \dfrac{14 \cancel{(7 + 4 \sqrt{3} )}}{ \cancel{7 + 4 \sqrt{3} }}$

$\rm \to \: 14$

So Answer is 14