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=b=0`B. `a=0,b=3`C. `a=b=3`D. `a,b` are roots of `x^(2)+x+2=0` |
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Answer» Correct Answer - A::C::D We have `x^(3)-ax^(2)+bx-1=0`……….i Then `alpha^(2)+beta^(2)+gamma^(2)=(alpha+beta=gamma)^(2)-2(alpha beta+beta gamma+gamma alpha)` `=a^(2)-2b` `alpha^(2) beta^(2)+beta^(2)gamma^(2)+gamma^(2) alpha^(2)=(alpha beta+beta gamma+gamma alpha)^(2)` `=-2 alpha beta gamma( alpha +beta+gamma)=b^(2)-2a` and `alpha^(2) beta^(2) gamma^(2)=1` Therefore, the equation whose roots are`alpha^(2),beta^(2)` and `gamma^(2)` is `x^(3)-(a^(2)-2b)x^(2)+(b^(2)-2a)x-1=0`...........ii `a^(2)-2b=a` and `b^(2)-2a=b` Eliminating `b` we have `((a^(2)-a)^(2))/4-2a=(a^(2)-a)/2` `impliesa{a(a-1)^(2)-8-2(a-1)}=0` `impliesa(a^(3)-2a^(2)-a-6)=0` `impliesa(a-3)(a^(2)+a+2)=0` `impliesa=0` or `a=3` or `a^(2)=a+2=0` `impliesb=0` or `b=3` or `b^(2)+b+2=0` `:.a=b=0` or `a=b=3` or a and b are roots of `x^(2)+x+2=0` |
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