The roots of the equatiion `(a+sqrt(b))^(x^(2)-15)+(a-sqrt(b))^(x^(2)-15)=2a` where `a^(2)-b=1` are

Options

A. `+-2,+-sqrt(3)`
B. `+-4,+-sqrt(14)`
C. `+-3,+-sqrt(5)`
D. `+-6, +- sqrt(20)`

Answer

B

Explanation

We have, `a-sqrt(b)=((a-sqrt(b))(a + sqrt(b)))/(a + sqrt(b))=(a^(2)-b)/(a+sqrt(b))=(1)/(a+sqrt(b))[because a^(2)-b = 1]`
So, by puttiong `(a + sqrt(b))^(x^(2)-15)=y`, the given equation becomes `y+(1)/(y) = 2a`
`rArr" "y^(2) - 2 ay + 1 = 0`
`rArr" "(y-a)^(2) = a^(2) - 1`
`rArr" "y - a = +-sqrt(a^(2)-1)`
`rArr" "y-a = +- sqrt(b)" "[because a^(2) - 1 = b]`
`rArr" " y = a +- sqrt(b)`
`rArr" "(a + sqrt(b))^(x^(2)-15) = a + sqrt(b) , a - sqrt(b)`
`rArr" "x^(2) - 15 = 1 or, x^(2) - 15 = - 1 rArr x = +- 4, x = +- sqrt(14)`
ALITER We have, `(a + sqrt(b))^(x^(2)-15)+(a-sqrt(b))^(x^(2)+15)=(a+sqrt(a))^(1)+(a-sqrt(b))^(1)`
`rArr" "x^(2) - 15 = +- 1 rArr x = +- 4, x = +- sqrt(14)`