Given the two quadratic equations:

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

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

Determine the correct relationship between $x$ and $y$ from the options below.

Question

Given the two quadratic equations:

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

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

Determine the correct relationship between $x$ and $y$ from the options below.

Accepted Answer

**Solution:**

1. Solve the first equation for $x$:

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

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

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

$$
\sqrt{400} = 20
$$

So,

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

Two roots:

- $x_1 = \frac{22 + 20}{6} = \frac{42}{6} = 7$
- $x_2 = \frac{22 - 20}{6} = \frac{2}{6} = \frac{1}{3}$

2. Solve the second equation for $y$:

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

Here, $a=1$, $b=-15$, $c=56$.

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

$$
\sqrt{1} = 1
$$

Two roots:

- $y_1 = \frac{15 + 1}{2} = 8$
- $y_2 = \frac{15 - 1}{2} = 7$

3. Now, compare possible values of $x$ and $y$:

- $x$ can be $7$ or $\frac{1}{3}$
- $y$ can be $8$ or $7$

4. Check all combinations:

- $x=7$ 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 $y=7$ → $\frac{1}{3} < 7$

Since $x$ can be less than or equal to $y$, but also can be equal, and no strict inequality always holds, the exact relationship between $x$ and $y$ cannot be definitively established.

**Answer:** x = y or relationship cannot be established

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