In each of the following, two quadratic equations (I) and (II) are given. Solve both equations and determine the correct relationship between *x* and *y*.

I. $5x^2 + 11x + 2 = 0$

II. $4y^2 + 13y + 3 = 0$

Question

In each of the following, two quadratic equations (I) and (II) are given. Solve both equations and determine the correct relationship between *x* and *y*.

I. $5x^2 + 11x + 2 = 0$

II. $4y^2 + 13y + 3 = 0$

Accepted Answer

**Solution:**

1. Solve equation (I): $5x^2 + 11x + 2 = 0$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=5$, $b=11$, $c=2$:

$$x = \frac{-11 \pm \sqrt{11^2 - 4 	imes 5 	imes 2}}{2 	imes 5} = \frac{-11 \pm \sqrt{121 - 40}}{10} = \frac{-11 \pm \sqrt{81}}{10}$$

$$x = \frac{-11 \pm 9}{10}$$

So, the roots are:

- $x_1 = \frac{-11 + 9}{10} = \frac{-2}{10} = -0.2$
   - $x_2 = \frac{-11 - 9}{10} = \frac{-20}{10} = -2$

2. Solve equation (II): $4y^2 + 13y + 3 = 0$

Here, $a=4$, $b=13$, $c=3$:

$$y = \frac{-13 \pm \sqrt{13^2 - 4 	imes 4 	imes 3}}{2 	imes 4} = \frac{-13 \pm \sqrt{169 - 48}}{8} = \frac{-13 \pm \sqrt{121}}{8}$$

$$y = \frac{-13 \pm 11}{8}$$

Roots are:

- $y_1 = \frac{-13 + 11}{8} = \frac{-2}{8} = -0.25$
   - $y_2 = \frac{-13 - 11}{8} = \frac{-24}{8} = -3$

3. Now, compare roots of $x$ and $y$:

- $x$ roots: $-0.2$, $-2$
   - $y$ roots: $-0.25$, $-3$

4. Since both equations have two roots, and no specific root correspondence is given, we cannot definitively say $x > y$, $x < y$, or any other strict inequality.

5. Also, the roots overlap in ranges, so no unique relation can be established.

**Answer:** If $x = y$ or no relation can be established between $x$ and $y$

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