Question

If x^2-4x= 1, find x^2+1/x^2​

Answer 1

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

$\boxed{\sf{x^2 + \dfrac{1}{x^2} = 18}}$

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

Given :- $\sf{x^2 - 4x = 1}$

To find :- $\sf{x^2 + \dfrac{1}{x^2}}$

Solution :-

${x}^{2} - 4x = 1$

It can be written as :-

$x(x - 4) = 1$

$x - 4 = \dfrac{1}{x}$

$x - 4 - \dfrac{1}{x} = 0$

$x - \dfrac{1}{x} = 4$

Now squaring on both the sides

${(x - \dfrac{1}{x})}^{2} = {4}^{2}$

We know that, (x - y)² = x² + y² - 2xy

Here x = x, y = 1/x

By substituting the values in the identity we have :-

${x}^{2} + \dfrac{1}{ {x}^{2} } - 2(x)( \dfrac{1}{x}) = 16$

${x}^{2} + \dfrac{1}{ {x}^{2} } - 2 = 16$

${x}^{2} + \dfrac{1}{ {x}^{2} } = 16 + 2$

${x}^{2} + \dfrac{1}{ {x}^{2} } = 18$

$\Huge{\boxed{\sf{x^2 + \dfrac{1}{x^2} = 18}}}$

$\mathfrak{\large{\underline{\underline{Identity \: Used:-}}}}$

(x - y)² = x² + y² - 2xy

$\mathfrak{\large{\underline{\underline{Extra \: Information:-}}}}$

1. (x + y)² = x² + y² + 2xy

2. (x - y)² = x² + y² - 2xy

3. (x + y)(x - y) = x² - y²

4. (x + a)(x + b) = x² + (a + b)x + ab

5. (x + y)³ = x³ + y³ + 3xy(x + y)

6. (x - y)³ = x³ - y³ - 3xy(x - y)

7. x³ + y³ = (x + y)(x² - xy + y²)

8. x³ - y³ = (x - y)(x² + xy + y²)

9. (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2xz

10. (x + y + z)(x² + y² + z² - xy - yz - xz) = x³ + y³ + z³ - 3xyz