Given the two equations: I. $3x^2 - 22x + 7 = 0$ II. $y^2 - 15y + 56 = 0$ Determine the correct relationship between $x$ and $y$. Choose the correct option: (a) $x y$ (b) $x \geq y$ (c) $x

Question

Given the two equations:

I. $3x^2 - 22x + 7 = 0$

II. $y^2 - 15y + 56 = 0$

Determine the correct relationship between $x$ and $y$. Choose the correct option:

(a) $x > y$

(b) $x \geq y$

(c) $x < y$

(d) $x \leq y$

Accepted Answer

**Solution:**

1. Solve equation I: $3x^2 - 22x + 7 = 0$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=3$, $b=-22$, $c=7$:

$$
   x = \frac{22 \pm \sqrt{(-22)^2 - 4 \times 3 \times 7}}{2 \times 3} = \frac{22 \pm \sqrt{484 - 84}}{6} = \frac{22 \pm \sqrt{400}}{6}
   $$

$$
   x = \frac{22 \pm 20}{6}
   $$

So the roots are:

- $x_1 = \frac{22 + 20}{6} = \frac{42}{6} = 7$

- $x_2 = \frac{22 - 20}{6} = \frac{2}{6} = \frac{1}{3}$

2. Solve equation II: $y^2 - 15y + 56 = 0$

Using the quadratic formula with $a=1$, $b=-15$, $c=56$:

$$
   y = \frac{15 \pm \sqrt{(-15)^2 - 4 \times 1 \times 56}}{2} = \frac{15 \pm \sqrt{225 - 224}}{2} = \frac{15 \pm 1}{2}
   $$

So the roots are:

- $y_1 = \frac{15 + 1}{2} = 8$

- $y_2 = \frac{15 - 1}{2} = 7$

3. Now compare the roots:

- $x$ can be either 7 or $\frac{1}{3}$

- $y$ can be either 8 or 7

4. Check all possible comparisons:

- $x = 7$ and $y = 8$ : $7 < 8$

- $x = 7$ and $y = 7$ : $7 = 7$

- $x = \frac{1}{3}$ and $y = 8$ : $\frac{1}{3} < 8$

- $x = \frac{1}{3}$ and $y = 7$ : $\frac{1}{3} < 7$

5. In all cases, $x \leq y$.

**Answer:** (d) $x \leq y$

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