Advertisements
Advertisements
Question
If \[x - \frac{1}{x} = - 1\] find the value of \[x^2 + \frac{1}{x^2}\]
Advertisements
Solution
In the given problem, we have to find `x^2 + 1/x^2`
Given `(x-1/x)=-1`
On squaring both sides we get,
`(x-1/x)^2=(-1)^2`
We shall use the identity `(x-y )^2 = x^2 - 2xy + y`
`x^2 +1/x^2 - 2 xx x xx 1/x =- 1 xx -1`
`x^2 +1/x^2 -2 =1`
`x^2 +1/x^2 = 1+2`
`x^2+1/x^2 =3`
Hence the value of ` x^2 +1/x^2`is 3 .
APPEARS IN
RELATED QUESTIONS
Factorise the following using appropriate identity:
4y2 – 4y + 1
Expand the following, using suitable identity:
(–2x + 5y – 3z)2
Simplify the following: 175 x 175 x 2 x 175 x 25 x 25 x 25
If 3x - 7y = 10 and xy = -1, find the value of `9x^2 + 49y^2`
Prove that a2 + b2 + c2 − ab − bc − ca is always non-negative for all values of a, b and c
Find the cube of the following binomials expression :
\[\frac{3}{x} - \frac{2}{x^2}\]
Find the cube of the following binomials expression :
\[4 - \frac{1}{3x}\]
Evaluate of the following:
(598)3
If \[x + \frac{1}{x} = 3\], calculate \[x^2 + \frac{1}{x^2}, x^3 + \frac{1}{x^3}\] and \[x^4 + \frac{1}{x^4}\]
If a + b = 10 and ab = 16, find the value of a2 − ab + b2 and a2 + ab + b2
If \[a^2 + \frac{1}{a^2} = 102\] , find the value of \[a - \frac{1}{a}\].
If a + b + c = 9 and ab + bc + ca = 23, then a2 + b2 + c2 =
(x − y) (x + y) (x2 + y2) (x4 + y4) is equal to ______.
If a + `1/a`= 6 and a ≠ 0 find :
(i) `a - 1/a (ii) a^2 - 1/a^2`
Use the direct method to evaluate the following products:
(x + 8)(x + 3)
Expand the following:
(3x + 4) (2x - 1)
Expand the following:
`(2"a" + 1/(2"a"))^2`
Simplify by using formula :
(a + b - c) (a - b + c)
If a2 - 3a - 1 = 0 and a ≠ 0, find : `"a"^2 - (1)/"a"^2`
