Question

if x - 1/x = 2 then find the value of x ^4+1/x^4​

Answer 1

Answer:

x-1 = 2x

x-2x = 1

x= -1

putting valueof x in equation x^4+1/x^4

-1^4+1/-1^4

1+1/1

=1

Ans= 1

Answer 2

Question :

If $\bf{x - \dfrac{1}{x} = 2}$ , then find the value of

$\bf{x^{4} + \dfrac{1}{x^{4}}}$.

To Find :

The value of the $\bf{x^{4} + \dfrac{1}{x^{4}}}$.

Given :

• $\boxed{\bf{x - \dfrac{1}{x} = 2}}$

We Know :

• $\bf{(a + b)^{2} = a^{2} + b^{2} + 2ab}$

• $\bf{(a - b)^{2} = a^{2} + b^{2} - 2ab}$

Concept :

By using the identities , we can find the value of $\bf{x^{2} + \dfrac{1}{x^{2}}}$ , and by squaring it , we can find the required value.

Solution :

By Squaring the Equation , we get :

Identity used :- $\bf{(a - b)^{2} = a^{2} + b^{2} - 2ab}$]

$\implies \bf{x - \dfrac{1}{x}} \\ \\ \\ \\ \implies \bf{\bigg(x - \dfrac{1}{x}\bigg)^{2} = x^{2} + \bigg(\dfrac{1}{x}\bigg)^{2} - 2 \times x \times \dfrac{1}{x}} \\ \\ \\ \\ \implies \bf{\bigg(x - \dfrac{1}{x}\bigg)^{2} = x^{2} + \dfrac{1}{x^{2}} - 2 \times \not{x} \times \dfrac{1}{\not{x}}} \\ \\ \\ \\ \implies \bf{\bigg(x - \dfrac{1}{x}\bigg)^{2} = x^{2} + \dfrac{1}{x^{2}} - 2} \\ \\ \\ \\ \implies \bf{x^{2} + \dfrac{1}{x^{2}} = \bigg(x - \dfrac{1}{x}\bigg)^{2} + 2}$

Putting the value of $\bf{x - \dfrac{1}{x}}$ in the

$\bigg(x - \dfrac{1}{x}\bigg)^{2}$ , we get :

$\implies \bf{x^{2} + \dfrac{1}{x^{2}} = 2^{2} + 2} \\ \\ \\ \\ \implies \bf{x^{2} + \dfrac{1}{x^{2}} = 4 + 2} \\ \\ \\ \\ \implies \bf{x^{2} + \dfrac{1}{x^{2}} = 6} \\ \\ \\ \\ \therefore \purple{\bf{x^{2} + \dfrac{1}{x^{2}} = 6}}$

Hence the value of $\bf{x^{2} + \dfrac{1}{x^{2}}}$ is 6.

Now , on Squaring the $\bf{x^{2} + \dfrac{1}{x^{2}}}$ ,we get :

$\implies \bf{x^{2} + \dfrac{1}{x^{2}}} \\ \\ \\ \\ \implies \bf{\bigg(x^{2} + \dfrac{1}{x^{2}}\bigg)^{2}} = (x^{2})^{2} + \bigg(\dfrac{1}{x^{2}}\bigg)^{2} + 2 \times x^{2} \times \dfrac{1}{x^{2}} \\ \\ \\ \\ \implies \bf{\bigg(x^{2} + \dfrac{1}{x^{2}}\bigg)^{2} = x^{4} + \dfrac{1}{x^{4}} + 2} \\ \\ \\ \implies \bf{x^{4} + \dfrac{1}{x^{4}} = \bigg(x^{2} + \dfrac{1}{x^{2}}\bigg)^{2} - 2}$

Putting the value of $\bf{x^{2} + \dfrac{1}{x^{2}}}$ in the Equation we get :

$\implies \bf{x^{4} + \dfrac{1}{x^{4}} = 6^{2} - 2} \\ \\ \\ \\ \bf{x^{4} + \dfrac{1}{x^{4}} = 36 - 2} \\ \\ \\ \\ \bf{x^{4} + \dfrac{1}{x^{4}} = 34} \\ \\ \\ \therefore \purple{\bf{x^{4} + \dfrac{1}{x^{4}} = 34}}$

Hence the value of $\bf{x^{4} + \dfrac{1}{x^{4}}}$ is 34.