Question

b
If x = 7+4√3 then find the value of x
X​

Answer 1

$7 + 4 \sqrt{3} + \frac{1}{7 + 4 \sqrt{3} } \\ 7 + 4 \sqrt{3 } + \frac{1(7 - 4 \sqrt{3} )}{(7 + 4 \sqrt{3} )(7 - 4 \sqrt{3}) } </p><p></p><p>$

$7 + 4 \sqrt{3} + \frac{7 - 4 \sqrt{3} }{ {7}^{2} - (4 \sqrt{3}) {}^{2} } \\ 7 + 4 \sqrt{3} + \frac{7 - 4 \sqrt{3} }{49 - 48} \\ = 7 + 4 \sqrt{3} + 7 - 4 \sqrt{3}$

$cancel \: + 4 \sqrt{3 \:} \: and \: - 4 \sqrt{3} \\ 7 + 7 = 14$

$x + \frac{1}{x} = 14$

Answer 2

ANSWER:

Given:

• x = 7 + 4√3

To Find:

• Value of x + 1/x

Solution:

We are given that,

$\implies\sf x=7+4\sqrt3$

So,

$\implies\sf\dfrac{1}{x}=\dfrac{1}{7+4\sqrt3}$

On rationalising,

$\implies\sf\dfrac{1}{x}=\dfrac{1}{7+4\sqrt3}\times \dfrac{7-4\sqrt3}{7-4\sqrt3}$

$\implies\sf\dfrac{1}{x}=\dfrac{7-4\sqrt3}{(7+4\sqrt3)(7-4\sqrt3)}$

We know that,

$\hookrightarrow (a-b)(a+b)=a^2-b^2$

$\implies\sf\dfrac{1}{x}=\dfrac{7-4\sqrt3}{(7)^2-(4\sqrt3)^2}$

$\implies\sf\dfrac{1}{x}=\dfrac{7-4\sqrt3}{49-48}$

$\implies\sf\dfrac{1}{x}=\dfrac{7-4\sqrt3}{1}$

$\implies\sf\dfrac{1}{x}=7-4\sqrt3$

We need to find the value of,

$\implies x+\dfrac{1}{x}$

So,

$\implies x+\dfrac{1}{x}=(7+4\sqrt3)+(7-4\sqrt3)$

$\implies x+\dfrac{1}{x}=7+4\sqrt3+7-4\sqrt3$

Cancelling 4√3,

$\implies x+\dfrac{1}{x}=14$

Hence,

$\implies\bf x+\dfrac{1}{x}=14$