Quantitative Aptitude Ques 1964

Question: Directions: In each of the following questions two equations are given. Solve these equations and give answer. [IBPS (PO) 2013]

I. $ x^{2}+4x+4=0 $
II. $ y^{2}-8y+16=0 $

Options:

A) If $ x\ge y $

B) If $ x>y $

C) If $ x\le y $

D) If $ x<y $

E) If $ x=y $ or no relation can be established between x and y

Show Answer

Answer:

Correct Answer: D

Solution:

  • I. $ x^{2}+4x+4=0 $

$ \Rightarrow $ $ x^{2}+2x+2x+4=0 $

$ \Rightarrow $ $ x(x+2)+2(x+2)=0 $

$ \Rightarrow $ $ (x+2)(x+2)=0 $

$ \Rightarrow $ $ {{(x+2)}^{2}}=0 $

$ \Rightarrow $ $ x=-2 $ II. $ y^{2}-8y+16=0 $

$ \Rightarrow $ $ y^{2}-4y-4y+16=0 $

$ \Rightarrow $ $ y(y-4)-4(y-4)=0 $

$ \Rightarrow $ $ (y-4)(y-4)=0 $

$ \Rightarrow $ $ {{(y-4)}^{2}}=0 $

$ \therefore $ $ y=4 $ Hence, $ x<y $