Advertisements
Advertisements
प्रश्न
If x = `sqrt(5) + 2`, then find the value of `x^2 + 1/x^2`
उत्तर
`sqrt(5) + 2` ⇒ x2 = `(sqrt(5) + 2)^2`
= `(sqrt(5))^2 + 2 xx 2 xx sqrt(5) + 2^2`
= `5 + 4sqrt(5) + 4`
= `9 + 4sqrt(5)`
`1/x = 1/(sqrt(5) + 2)`
= `(sqrt(5) - 2)/((sqrt(5) + 2)(sqrt(5) - 2))`
= `(sqrt(5) - 2)/((sqrt(5))^2 - 2^2)`
= `(sqrt(5) - 2)/(5 - 4)`
= `sqrt(5) - 2`
`1/x^2 = (sqrt(5) - 2)^2`
= `(sqrt(5))^2 - 2 xx sqrt(5) xx 2 + 2^2`
= `5 - 4sqrt(5) + 4`
= `9 - 4sqrt(5)`
∴ `x^2 + 1/x^2 = 9 + 4sqrt(5) + 9 - 4sqrt(5)` = 18
The value of `x^2 + 1/x^2` = 18
APPEARS IN
संबंधित प्रश्न
Rationalize the denominator.
`5/sqrt 7`
Rationalize the denominator.
`11 / sqrt 3`
Write the lowest rationalising factor of 5√2.
Write the lowest rationalising factor of : √13 + 3
Find the values of 'a' and 'b' in each of the following:
`( sqrt7 - 2 )/( sqrt7 + 2 ) = asqrt7 + b`
Find the values of 'a' and 'b' in each of the following:
`3/[ sqrt3 - sqrt2 ] = asqrt3 - bsqrt2`
If x =`[sqrt5 - 2 ]/[ sqrt5 + 2]` and y = `[ sqrt5 + 2]/[ sqrt5 - 2 ]`; find :
x2
If x =`[sqrt5 - 2 ]/[ sqrt5 + 2]` and y = `[ sqrt5 + 2]/[ sqrt5 - 2 ]`; find : xy
If m = `1/[ 3 - 2sqrt2 ] and n = 1/[ 3 + 2sqrt2 ],` find mn
Rationalise the denominator `5/(3sqrt(5))`
