Question

If x=1/√2-1, then prove that X²-6+1/x²=0

Answer 1

$x^{2}-6+\dfrac{1}{x^{2}}=0$

Given:

$x=\dfrac{1}{\sqrt{2}-1}$

To prove:

$x^{2}-6+\dfrac{1}{x^{2}}=0$

Step-by-step explanation:

Here, $x=\dfrac{1}{\sqrt{2}-1}$

Rationalising the denominator, we get

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

$\Rightarrow x=\dfrac{\sqrt{2}+1}{2-1}$

$\Rightarrow x=\sqrt{2}+1$

Also, $x^{2}=(\sqrt{2}+1)^{2}$

$\Rightarrow x^{2}=2+2\sqrt{2}+1$

$\Rightarrow \boxed{x^{2}=3+2\sqrt{2}}$

Again, $\dfrac{1}{x}=\sqrt{2}-1$

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

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

$\Rightarrow \boxed{\dfrac{1}{x^{2}}=3-2\sqrt{2}}$

Now, L.H.S. = $x^{2}-6+\dfrac{1}{x^{2}}$

$=(3+2\sqrt{2})-6+(3-2\sqrt{2})$

$=3+2\sqrt{2}-6+3-2\sqrt{2}$

= 0 = R.H.S.

Thus proved.

#SPJ3

Answer 2

Answer:

X²-6+1/x²= 0

Step-by-step explanation:

Given : x=1/√2-1

x= 1/√2-1 × √2+1/√2+1 = √2+1

1/x= 1/√2+1 × √2-1/√2-1 = √2-1

x²+1/x²-6 = x²+ 1/x² -2 -4

(x - 1/x)²-2²

(x - 1/x + 2) (x - 1/x - 2)

(√2 + 1 -√2+ 1 + 2) (√2 + 1 -√2 +1 -2)

(4) (0)

0

x²-6+1/x²= 0 ; hence proved.