Question

# 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

Solution

( 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