If the roots of a(b-c)x^2 +b(c-a)x + c(a-b) are equal, th

Question

If the roots of a(b-c)x^2 +b(c-a)x + c(a-b) are equal, then a,b,c are in ?

A. A.P
B. H.P
C. A.G.P
D. G.P
E. None

Is A.G.P part of the gmat.. what difficulty level would you rate this questions ?

Vips0000 · Answer

Two points: 1) The question is incomplete, roots are of an equation, this has no RHS. ( = 0 part is missing) question should be modified as : If the roots of a(b-c)x^2 +b(c-a)x + c(a-b) =0 are equal, then a,b,c are in ? 2) I dont think this could be part of GMAT. Probably only at a difficulty level of 800+

However, having said that, lets look at solving such problem: First of all, important concept: if in equation Ax^2+Bx+C=0 roots are equal that would mean B^2 -4AC =0 or B^2 =4AC . Applying this in the given equation a(b-c)x^2 +b(c-a)x + c(a-b)=0 we get: ^2 = 4 * a(b-c) *c(a-b) So we need to work with this equation. Here you can observe first - if a b c were in gp then b^2 =ac or b/a = c/b = sqrt(c/a) and if a b c were in AP then 2b = a+c or b-a = c-b = (c-a)/2 Also, for AGP, terms are multiples of AP and GP. To start with I'm not considering HP as I dont see any term in the form of a+b or b+c or c+a etc. Lets, rewrite, our original equation in a form we can identify our sequences. b^2 * (c-a)^2 = ac * 2(b-c) * 2(a-b) or b^2 *(c-a)* (c-a) = ac * 2(b-c) * 2(a-b) If you notice, terms in LHS and RHS are basically multiplication of AP and GP. Thus a, b and c are in A.G.P. Ans C it is. Hope it helps

If the roots of a(b-c)x^2 +b(c-a)x + c(a-b) are equal, th

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