Multiple choice

In the following question, two equations numbered I and II are given. You have to solve both the equations and give answer if (I) $x^{2}+11x+28=0$ and (II) $ y^{2}+15y+56=0$

  1. $\displaystyle x>y$
  2. $\displaystyle x\geq y$
  3. $\displaystyle x< y$
  4. $\displaystyle x\leq y$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation I: x^2 + 11x + 28 = 0 => (x+4)(x+7) = 0, so x = -4, -7. Equation II: y^2 + 15y + 56 = 0 => (y+7)(y+8) = 0, so y = -7, -8. Comparing: -4 > -7, -4 > -8, -7 = -7, -7 > -8. Thus, x >= y.

AI explanation

For equation I, x^2 + 11x + 28 = 0, we factor the quadratic to get (x + 7)(x + 4) = 0, which gives the roots x = -7 and x = -4. For equation II, y^2 + 15y + 56 = 0, we factor it as (y + 7)(y + 8) = 0, yielding the roots y = -7 and y = -8. Comparing the real roots, we see that -4 is greater than both -7 and -8, and -7 is equal to -7 while being greater than -8. Therefore, we can conclude that x is greater than or equal to y, meaning x >= y.

Keep practicing — related questions