Question

If x+1/x=6 find the value of x+1/x

Answer 1

Correct Question :

If $x\:+\:\dfrac{1}{x}\:=\:6$ then find ${x}^{2} \: + \: \dfrac{1}{ {x}^{2} }$

Solution :

→ $x\:+\:\dfrac{1}{x}\:=\:6$

Now,

Squaring on both sides ..

→ $\bigg(x \: + \: \dfrac{1}{x}\bigg)^{2} \: = \: ( {6)}^{2}$

We know that,

(a + b)² = a² + b² + 2ab

Use the above identity to find the value of x² + 1/x².

→ ${x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: + \: 2x \: \times \: \dfrac{1}{x} \: = \: 36$

→ ${x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: + \: 2 \: = \: 36$

→ ${x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: = \: 36 \: - \: 2$

→ ${x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: = \: 34$

Answer :

On solving we get, x² + 1/x² = 34.

_______________________________

Answer 2

✯$\huge\huge\boxed{Answer:-}$✯

Correct Question is

$if \: x + \frac{1}{x} = 6 \: so....find \: \: x {}^{2} + \frac{1}{x {}^{2} }$

Answer:-

$x + \frac{1}{x} = 6$

We have to square both the sides:-

$x + \frac{1}{x {}^{2} } = ((6)) {}^{2}$

$\star\boxed{Identity/Rule:-}\star$

$((a + b)) {}^{2} = a {}^{2} + b {}^{2} + 2ab = > \: identity$

We can use this identity as given above to do this sum that is [x² + 1/x²]

STEPS:-

$= x {}^{2} + \frac{1}{x {}^{2} } + 2x×\frac{1}{x} = > 36$

$= x {}^{2} + \frac{1}{x {}^{2} } + 2 = > 36$

$= x {}^{2} + \frac{1}{x {}^{2} } = 36 - 2$

$= x {}^{2} + \frac{1}{x {}^{2} } => 34$

Therefore,

✯$x { }^{2} + \frac{1}{x {}^{2} } = 34$✯