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