Given the two quadratic equations:

I. $1.2x^2 - 17x + 36 = 0$

II. $2y^2 - 19y + 44 = 0$

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

Question

Given the two quadratic equations:

I. $1.2x^2 - 17x + 36 = 0$

II. $2y^2 - 19y + 44 = 0$

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

Accepted Answer

**Solution:**

1. **Rewrite the equations:**

- Equation I: $1.2x^2 - 17x + 36 = 0$

- Equation II: $2y^2 - 19y + 44 = 0$

2. **Convert coefficients to fractions for easier calculation:**

- Equation I: Multiply entire equation by 5 to clear decimal:

$6x^2 - 85x + 180 = 0$

- Equation II remains as is.

3. **Calculate roots of Equation I:**

Using quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=6$, $b=-85$, $c=180$:

- Discriminant:

$D = (-85)^2 - 4 \times 6 \times 180 = 7225 - 4320 = 2905$

- $\sqrt{2905} \approx 53.91$

- Roots:

$x = \frac{85 \pm 53.91}{12}$

- $x_1 = \frac{85 + 53.91}{12} = \frac{138.91}{12} \approx 11.58$

- $x_2 = \frac{85 - 53.91}{12} = \frac{31.09}{12} \approx 2.59$

4. **Calculate roots of Equation II:**

For $2y^2 - 19y + 44 = 0$, $a=2$, $b=-19$, $c=44$:

- Discriminant:

$D = (-19)^2 - 4 \times 2 \times 44 = 361 - 352 = 9$

- $\sqrt{9} = 3$

- Roots:

$y = \frac{19 \pm 3}{4}$

- $y_1 = \frac{19 + 3}{4} = \frac{22}{4} = 5.5$

- $y_2 = \frac{19 - 3}{4} = \frac{16}{4} = 4$

5. **Compare roots:**

- $x$ roots: approximately 11.58 and 2.59

- $y$ roots: 5.5 and 4

- Comparing pairs:

- $x_1 = 11.58 > y_1 = 5.5$

- $x_2 = 2.59 < y_2 = 4$

Since one $x$ root is greater than a $y$ root and the other is smaller, no definite relationship between $x$ and $y$ can be established.

**Answer:** (e) $x = y$ or relationship cannot be established

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