If ??^?+ ?? + ? = ? and ??^?+ ?? + ? = ? have minimum one common root and a, b & c are natural numbers then how many values of “a” is possible for ?∈[?,???]
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1
40 values of a is possible.
a=5,10,15,20,25,………200
equations are
5x^2 +2x+3=0, 10x^2+4x+6=0, 15x^2+6x+9=0,………,200x^2+80x+120=0
for 5x^2+2x+3=0 , discriminant < 0, so it has no real roots. It will have two roots which are complex conjugate. So to have at least one common root with this equation, ax^2+bx+c=0 has to have two common roots (becasue two roots will always be conjugates to each other).
SO, the condition for ax^2+bx+c=0 to have common roots with the given equation should be a/5 = b/2 = c/3 =k
As, ab,c are integers, and maximum possible value of a is 200 so, k = 1,2,3,….40
total 40 values of "a" is possible within the given range
40
200
20
40. Because first eqn discriminant is negative. So roots will be complex no.. We know that complex roots occur in pair. So both roots will be common.
40
200/5=40
40