InterviewSolution
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InterviewSolution
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If the equation whose roots are the squares of the roots of the cubic `x^3-a x^2+b x-1=0`is identical with the given cubic equation, then`a=0,b=3`b. `a=b=0`c. `a=b=3`d. `a ,b ,`are roots of`x^2+x+2=0`A. ` a = 0, b = 3`B. a = b = 0C. a = b = 3D. a, b are roots of ` x^(2) + x + 2 = 0 ` |
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Answer» Correct Answer - 2,3,4 Given equation is `x^(3) - ax^(3) + bx - 1 = 0 ` . If roots of the equationo be ` alpha, beta , gamma`, then `alpha^(2) + beta^(2) + gamma^(2) = (alpha + beta + gamma)^(2) - 2 (alpha beta + beta gamma + gamma alpha)` ` = alpha ^(2) - 2a ` `alpha^(2)beta^(2) + beta^(2) gamma^(2)+ gamma^(2)alpha ^(2) = (alpha beta + beta gamma+ gammaalpha )^(2) - 2 alpha beta gamma (alpha + beta + gamma )` ` b^(2) - 2a ` ` alpha ^(2) beta^(2) gamma^(2) = 1` So, the equations whose roots are ` alpha ^(2) , beta^(2) , gamma^(2)` is given by `x^(3) - ax^(2) + bx - 1= 0 ` `rArr a^(2) - 2b = a and b^(2) - 2a = b` Eliminatng b, we get `((a^(2) - a)^(2))/(4) - 2a = (a^(2) - a)/(2)` or ` a{a (a - 1)^(2) - 8 - 2a (a - 1)} = 0 ` or ` a (a^(3) - 2a ^(2) + a + 2) = 0` or ` a(a - 3) (a^(2) + a + 2) = 0` `rArr a = 0 or a = 3 or a^(2) + a + 2 = 0` Which gives b = 0 or b = 3 `b^(2) + b+ 2 = 0 `.So a = b - 0 or a = b =3 or a a, b are roots of ` x^(2) + x + 2 = 0` |
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