CAT 2015 Quantitative Ability

Question : 29 of 34

Marks: +1, -0

x^2 + (x + 1 )(x + 2) (x + 3) (x + 6) = 0

, where x is a real number, then one value of x that satisfies this equation is

Solution:

(x+1)(x+2)(x+3)(x+6)=-x^{2}

(x^{8}+7x+6)(x^{2}+5x+6)=-x^{2}

x^{2}(x+7+\;\;{6}/{x})(x+5+\;\;{6}/{x})=-x^{2}

\;\text"Put"\;x+\;\;{6}/{x}=y

(y + 7)(y + 5) = -1

y^{2}+12y+36=0

⇒(y+6)^{2}=0

⇒ y =-6

∴ \; x+\;\;{6}/{x}=-6

x^{2}+6x+6=0

x=\;\;{-6±√{36-4(6)}}/{2}

=\;\;{-6±2√{3}}/{2}

=-3±√{3}

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