An algebra problem by Kshetrapal Dashottar
are real numbers such that and (where is real parameter). Determine the value of for which the difference between the maximum and minimum possible value of is equal to 8.
The answer is -9.
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1 solution
given x+y+z=3 xy+yz+zx=a
So we get y+z=3-x and yz=a-x(y+z) and (a+b)² ≥4ab so (y+z)²≥4(yz) i e (3-x)²≥4(a-x(3-x)) 9+x²-6x≥4a-12x+4x² => 9-4a≥3x²-6x dividing the whole eq by 3 we get 3-4a/3≥x²-2x adding 1 both the side 4-4a/3≥(x-1)² x≤1+(4-4a/3)^1/2 and similarly we get x≥1-(4-4a/3)^1/2 so we get the maximum and minimum values of x 1+(4-4a/3)^1/2 - 1+(4-4a/3)^1/2=8 2(4-4a/3)^1/2=8 4-4a/3=16 a=-9