If x-1x -2- Then the value of x2- 1x2 will be

Given, x 1x =2 On squaring both the sides we get, (x 1x)^2 =2^2 Appyling identity [(x y)^2=x^2 2xy+y^2] ⇒ x^2 2× x×1x+1x^2 =4 ⇒ x^2 2+1x^2 =4 ⇒ x^2+1x^2 =4+2 ⇒ ...

Byju's Answer

Standard VIII

Mathematics

Algebraic Identity: (x - y)^2

If x-1x =2, T...

Question

If $x - \frac{1}{x} = 2$ , Then the value of $x^{2} + \frac{1}{x^{2}}$ will be

Open in App

Solution

Given,
$x - \frac{1}{x} = 2$
On squaring both the sides we get,
$(x - \frac{1}{x})^{2} = 2^{2}$
Appyling identity $[(x - y)^{2} = x^{2} - 2 x y + y^{2}]$
$\Rightarrow x^{2} - 2 \times x \times \frac{1}{x} + \frac{1}{x^{2}} = 4$
$\Rightarrow x^{2} - 2 + \frac{1}{x^{2}} = 4$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = 4 + 2$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = 6$

Suggest Corrections

Similar questions

Q. If

α

and

β

are the zeros of

x^{2} + x - 2

then find value of

\frac{1}{α} - \frac{1}{β}

Q. If

α

and

β

are the zeros of the quadratic polynomial

f (x) = x^{2} + x - 2

, find the value of

\frac{1}{α} - \frac{1}{β}

Q. If

α

β

are the roots of the equation

x^{2} - p x + q = 0

and

α_{1}, β_{1}

be the roots of the equation

x^{2} - q x + p = 0

, then form the quadratic equation whose roots are

\frac{1}{α_{1} β} + \frac{1}{α β_{1}}

and

\frac{1}{α α_{1}} + \frac{1}{β β_{1}}

Q. If

y = {tan}^{- 1} \frac{1}{1 + x + x^{2}} + {tan}^{- 1} \frac{1}{x^{2} + 3 x + 3}

+ {tan}^{- 1} \frac{1}{x^{2} + 5 x + 7} + \dots +

upto

n

terms, then

y = x^{2} + \frac{1}{x^{2} + \frac{1}{\frac{1}{x^{2} + \dots}}}

, then

\frac{d y}{d x}

Explore more

Algebraic Identity: (x - y)^2

Standard VIII Mathematics

If x−1x=2, Then the value of x2+1x2 will be

Given, x−1x=2 On squaring both the sides we get, (x−1x)2=22 Appyling identity [(x−y)2=x2−2xy+y2] ⇒x2−2×x×1x+1x2=4 ⇒x2−2+1x2=4 ⇒x2+1x2=4+2 ⇒x2+1x2=6

If $x - \frac{1}{x} = 2$ , Then the value of $x^{2} + \frac{1}{x^{2}}$ will be