Given the two quadratic equations:

I. $$2x^2 + 16x + 32 = 0$$

II. $$2y^2 + 45y - 98 = 0$$

Determine the relationship between the roots *x* and *y*.

Question

Given the two quadratic equations:

I. $$2x^2 + 16x + 32 = 0$$

II. $$2y^2 + 45y - 98 = 0$$

Determine the relationship between the roots *x* and *y*.

Accepted Answer

**Solution:**

1. **Solve equation (I):**
   $$2x^2 + 16x + 32 = 0$$
   Divide throughout by 2:
   $$x^2 + 8x + 16 = 0$$

Calculate the discriminant:
   $$D = b^2 - 4ac = 8^2 - 4 	imes 1 	imes 16 = 64 - 64 = 0$$

Since $$D=0$$, the equation has one repeated root:
   $$x = \frac{-b}{2a} = \frac{-8}{2} = -4$$

2. **Solve equation (II):**
   $$2y^2 + 45y - 98 = 0$$

Calculate the discriminant:
   $$D = 45^2 - 4 	imes 2 	imes (-98) = 2025 + 784 = 2809$$

Since $$D > 0$$, two distinct roots exist:
   $$y = \frac{-45 \pm \sqrt{2809}}{2 	imes 2} = \frac{-45 \pm 53}{4}$$

Roots are:
   - $$y_1 = \frac{-45 + 53}{4} = \frac{8}{4} = 2$$
   - $$y_2 = \frac{-45 - 53}{4} = \frac{-98}{4} = -24.5$$

3. **Compare roots:**
   - $$x = -4$$ (single root)
   - $$y = 2$$ or $$-24.5$$

Since $$x$$ lies between the two roots of $$y$$, no definite relation such as $$x > y$$ or $$x < y$$ can be established for all roots.

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

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