In the following questions, two equations numbered I and II are given. Solve both equations and give the answer. I. $2x^2 + 7x + 3 = 0$ II. $4y^2 + 16y + 15 = 0$

x > y
x = y or no relation can be established between x and y
x < y
x < y

Correct answer: x = y or no relation can be established between x and y

Solution

Equation I factors as $(2x+1)(x+3)=0$, giving $x=-\tfrac12$ or $x=-3$. Equation II factors as $(2y+3)(2y+5)=0$, giving $y=-\tfrac32$ or $y=-\tfrac52$. Since some values of $x$ are greater than some values of $y$, while others are not, no definite relation can be established.

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