Bunuel
If R is a root of the equation X^2 +3X – 54 = 0, than which of the following equations have also the root R ?
(A) X^2 – 12X +27 = 0
(B) X^2 – 6X – 16 = 0
(C) X^2 – 10X – 31.25 = 0
(D) X^2 – 15X + 54 = 0
(E) X^2 + 10X + 16 = 0
\(x^2+3x-54=0........x^2+9x-6x+54=0.......(x+9)(x-6)=0\)
So R can be -9 or 6.
You could do it three ways.
(I) Substitute x as -9 and 6 in each option and see if any of the options has one of these as roots.
(II) If in hurry, just check the product of roots in each option given by x^2+bx+(product of roots). If more than one fits in as a multiple of 9 or 6, substitute the exact values and check.
(III) Factor each option
I’ll use a mix of all three to get to the answer
(A) X^2 – 12X +27 = 0
Product of roots 27 does contain 9 as a factor. But the moment x is -9, all the terms in left hand side will be positive and will never be 0. \(9^2-12*(-9)+27=71+108+27\) is not 0.
OR
\(x^2-9x-3x+27=0.......(x-9)(x-3)=0\) None of the roots match
(B) X^2 – 6X – 16 = 0
\(x^2-8x+2x-16=0.....(x-8)(x+2)=0\) None match
(C) X^2 – 10X – 31.25 = 0
The product of roots is 31.25, which is neither multiple of 9 nor 6. So the roots are not 9 or 6.
(D) X^2 – 15X + 54 = 0
\(x^2-9x-6x+54=0......(x-9)(x-6)=0\)
So the root 6 matches. Possible answer.
(E) X^2 + 10X + 16 = 0
Again the product of roots, 16, is not a multiple of 9 or 6.
D