If the roots of the equation x^3 - ...
Question
If the roots of the equation x^3 - 11x^2 + 36x - 36 = 0 are in H.P. then the middle root is
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If the roots of the equation x3 - 11x2 + 36x - 36 = 0 are in H.P. then the middle root is (A) an even number (B) a perfect square of an integer (C) a prime number (D) a composite number

JEE/Engineering Exams
Maths
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If the roots of the equation x^3 - 11x^2 + 36x - 36 = 0 are in H.P. then the middle root is
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( x^{3}-11 x^{2}+36 x-36=0 )
let the roots be a,b and ( x^{3}-11 x^{2}+36 x-36=(x-a)(x-b)(x-c) )
( (x-a)(x-b)(x-c)=x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c )
comparing with ( x^{3}-11 x^{2}+36 x-36=0 )
we get
( a+b+c=11 ldots ldots )
( a b+a c+b c=36 ldots ldots 2 )
( a b c=36 ldots ldots .3 )
also we know
( frac{2}{b}=frac{1}{a}+frac{1}{c} )
( frac{2}{b}=frac{a+c}{a c} )
( frac{2 a c}{a+c}=b )
eqn2/eqn3 ( frac{1}{c}+frac{1}{a}+frac{1}{b}=1 )
( frac{2}{b}+frac{1}{b}=1 )
( b=3 )
( a+c=8 )
( a c=12 )
( a=2 ) and ( c=6 )
roots are 2,3,6

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