Multiple choice

In the given question two equations I & II are given, you have to solve them and answer the question. $( (I) x^2 + 14x + 49 = 0 )$ $( (II) 6y^2 + 61y + 153 = 0 )$

  1. if x > y

  2. if x < y

  3. if x = y or relation can’t be establish

  4. if x ≥ y

  5. if x ≤ y
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving equation (I): x^2 + 14x + 49 = 0 is a perfect square - (x+7)^2 = 0, giving x = -7. Solving equation (II): 6y^2 + 61y + 153 = 0 factors to (2y+9)(3y+17) = 0, giving y = -9/2 = -4.5 or y = -17/3 ≈ -5.67. Since x = -7 and y is either -4.5 or -5.67, we have x < y in both cases (on the number line, -7 is to the left of both y values).

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