Advertisements
Advertisements
Question
If a2 - 3a - 1 = 0 and a ≠ 0, find : `"a"^2 - (1)/"a"^2`
Sum
Advertisements
Solution
`"a"^2 - (1)/"a"^2`
= `("a" + 1/"a")("a" - 1/"a")`
= `(±sqrt(13)) (3)`
= ±3`sqrt(13)`.
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
Factorise:
`2x^2 + y^2 + 8z^2 - 2sqrt2xy + 4sqrt2yz - 8xz`
Factorise the following:
8a3 – b3 – 12a2b + 6ab2
If 2x+3y = 13 and xy = 6, find the value of 8x3 + 27y3
Find the following product:
(4x − 5y) (16x2 + 20xy + 25y2)
If a + b + c = 9 and ab + bc + ca = 23, then a2 + b2 + c2 =
Use identities to evaluate : (101)2
Evaluate the following without multiplying:
(95)2
Simplify:
(3a + 2b - c)(9a2 + 4b2 + c2 - 6ab + 2bc +3ca)
Without actually calculating the cubes, find the value of:
`(1/2)^3 + (1/3)^3 - (5/6)^3`
Simplify (2x – 5y)3 – (2x + 5y)3.
