Question

In the following question, two equations are given. Solve both the equations and mark the corresponding answer, if:

a) x is greater than y

b) x is less than y

c) x is greater than or equal than y

d) x is less than or equal to y

e) x = y or a definite relationship between x & y cannot be established

I. x${}^2$ + 4x + 4 = 0

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

• A. (a)
• B. (b)
• C. (c)
• D. (d)

Answer 1

The correct option is B

(b)

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

(x + 2)${}^2$ = 0 =$>$x = - 2

y${}^2$ - 8y + 16 = 0

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

=$>$y = 4

Thus, we see that x$<$y.