Given the two quadratic equations:

(I) $$7x^2 + 5x - 18 = 0$$

(II) $$3y^2 + 4y - 20 = 0$$

Determine the correct relationship between $x$ and $y$ where $x$ and $y$ are the roots of their respective equations.

Question

Given the two quadratic equations:

(I) $$7x^2 + 5x - 18 = 0$$

(II) $$3y^2 + 4y - 20 = 0$$

Determine the correct relationship between $x$ and $y$ where $x$ and $y$ are the roots of their respective equations.

Accepted Answer

**Solution:** 1. Consider the first quadratic equation: $$$7x^2 + 5x - 18 = 0$$$ Calculate the discriminant $D_1$: $$$D_1 = 5^2 - 4 \times 7 \times (-18) = 25 + 504 = 529$$$ Since $D_1 > 0$, the roots are real and distinct. Find the roots: $$ $x = \frac{-5 \pm \sqrt{529}}{2 \times 7} = \frac{-5 \pm 23}{14}$ $$ So, $$$x_1 = \frac{-5 + 23}{14} = \frac{18}{14} = \frac{9}{7} \approx 1.2857$$$ $$$x_2 = \frac{-5 - 23}{14} = \frac{-28}{14} = -2$$$ 2. Consider the second quadratic equation: $$$3y^2 + 4y - 20 = 0$$$ Calculate the discriminant $D_2$: $$$D_2 = 4^2 - 4 \times 3 \times (-20) = 16 + 240 = 256$$$ Since $D_2 > 0$, the roots are real and distinct. Find the roots: $$ $y = \frac{-4 \pm \sqrt{256}}{2 \times 3} = \frac{-4 \pm 16}{6}$ $$ So, $$$y_1 = \frac{-4 + 16}{6} = \frac{12}{6} = 2$$$ $$$y_2 = \frac{-4 - 16}{6} = \frac{-20}{6} = -\frac{10}{3} \approx -3.3333$$$ 3. Now, compare the roots: - $x$ roots: $1.2857, -2$ - $y$ roots: $2, -3.3333$ 4. Since each equation has two roots, and the question does not specify which root of $x$ to compare with which root of $y$, we check all possibilities: - $$x_1 = 1.2857 < y_1 = 2$$ - $$x_1 = 1.2857 > y_2 = -3.3333$$ - $$x_2 = -2 > y_2 = -3.3333$$ - $$x_2 = -2 < y_1 = 2$$ 5. Because the roots overlap in values and no unique relation $x > y$ or $x < y$ can be established for all roots, the correct conclusion is: **Either $x = y$ or no definite relation can be determined between $x$ and $y$.** **Answer:** If $x = y$ or no relation can be established between $x$ and $y$.

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