Advertisements
Advertisements
प्रश्न
If a - `1/a`= 8 and a ≠ 0 find :
(i) `a + 1/a (ii) a^2 - 1/a^2`
उत्तर
We know that,
( a + b )2 = a2 + 2ab + b2
Given that `a - 1/a` = 8 ; Substitute in equation (1), we have
`(8)^2 = a^2 + 1/a^2 - 2`
⇒ `a^2 + 1/a^2 = 64 + 2`
⇒ `a^2 + 1/a^2 = 66`
⇒ `(a + 1/a)^2 = a^2 + 1/a^2 + 2`
⇒ `(a + 1/a)^2 = 66 + 2`
⇒ `(a + 1/a)^2 = 68`
i) `a + 1/a = sqrt68 `
⇒ `sqrt(17xx4 )= _-^+2sqrt17`
ii) `a^2 - 1/a^2 = (a+1/a) (a - 1/a)`
⇒ `a^2 - 1/a^2 = _-^+2sqrt17 xx 8`
⇒ `a^2 - 1/a^2 = _-^+16sqrt17`
APPEARS IN
संबंधित प्रश्न
Factorise the following:
64m3 – 343n3
If `x + 1/x = sqrt5`, find the value of `x^2 + 1/x^2` and `x^4 + 1/x^4`
Write in the expand form: `(2x - y + z)^2`
If a + b + c = 9 and ab + bc + ca =23, then a3 + b3 + c3 − 3abc =
Use the direct method to evaluate the following products :
(8 – b) (3 + b)
Use the direct method to evaluate :
(2+a) (2−a)
Find the squares of the following:
`(7x)/(9y) - (9y)/(7x)`
If `x^2 + (1)/x^2 = 18`; find : `x - (1)/x`
Simplify:
(3a + 2b - c)(9a2 + 4b2 + c2 - 6ab + 2bc +3ca)
Find the following product:
`(x/2 + 2y)(x^2/4 - xy + 4y^2)`
