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  • I. x^2 + (√7x)^2 + 12 = 0II. y^2 + 3y – (√4)^2 = 0
Question 1:
I. x^2 + (√7x)^2 + 12 = 0
II. y^2 + 3y – (√4)^2 = 0
A. if x > y
B. if x ≤ y
C. if x ≥ y
D. if x = y or relationship between x and y can't be established
Feedback According to the given equations :
I. x^2 +(√7x)^2 + 12 = 0
x^2 + 7x + 12 = 0
x^2 + 3x + 4x + 12 = 0
x (x + 3) + 4 (x + 3) = 0
(x + 4) (x + 3) = 0
x = – 3,– 4
II. y^2 + 3y –(√4)^2 = 0
y^2 + 3y – 4 = 0
y^2 + 4y – y – 4 = 0
y(y + 4) – 1(y + 4) = 0
(y – 1) (y + 4) = 0
y = 1, – 4
After comparison of both equations, the conclusion is, x = y or no relation is obtained
Hence, option D is correct.

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