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Leela Bava
Asked: 2022-11-06T11:16:33+05:30 In:

the roots of the equation `(a+sqrt(b))^(x^2-15)+(a-sqrt(b))^(x^2-15)=2a` where `a^2-b=1` are
A. `+-2,+-sqrt(3)`
B. `+-4,+-sqrt(14)`
C. `+-3,+-sqrt(5)`
D. `+-6, +- sqrt(20)`

the roots of the equation `(a+sqrt(b))^(x^2-15)+(a-sqrt(b))^(x^2-15)=2a` where `a^2-b=1` are
A. `+-2,+-sqrt(3)`
B. `+-4,+-sqrt(14)`
C. `+-3,+-sqrt(5)`
D. `+-6, +- sqrt(20)`
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  1. Correct Answer – B
    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)`

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