Given the two quadratic equations:

(I) $$4x^2 - 21x + 27 = 0$$

(II) $$3y^2 - 19y + 28 = 0$$

Find the correct relation between $x$ and $y$ where $x$ and $y$ are the roots of the respective equations.

Question

Given the two quadratic equations:

(I) $$4x^2 - 21x + 27 = 0$$

(II) $$3y^2 - 19y + 28 = 0$$

Find the correct relation between $x$ and $y$ where $x$ and $y$ are the roots of the respective equations.

Accepted Answer

**Solution:**

1. Consider the first quadratic equation:
   $$$4x^2 - 21x + 27 = 0$$$
   Using the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$, where $a=4$, $b=-21$, and $c=27$:

$$
  $ x = \frac{21 \pm \sqrt{(-21)^2 - 4 \times 4 \times 27}}{2 \times 4} = \frac{21 \pm \sqrt{441 - 432}}{8} = \frac{21 \pm \sqrt{9}}{8}$
   $$

So,
   $$
  $ x_1 = \frac{21 + 3}{8} = 3, \quad x_2 = \frac{21 - 3}{8} = \frac{18}{8} = 2.25$
   $$

2. Now, consider the second quadratic equation:
   $$$3y^2 - 19y + 28 = 0$$$
   Here, $a=3$, $b=-19$, and $c=28$.

$$
 $  y = \frac{19 \pm \sqrt{(-19)^2 - 4 \times 3 \times 28}}{2 \times 3} = \frac{19 \pm \sqrt{361 - 336}}{6} = \frac{19 \pm \sqrt{25}}{6}$
   $$

Thus,
   $$
 $  y_1 = \frac{19 + 5}{6} = 4, \quad y_2 = \frac{19 - 5}{6} = \frac{14}{6} \approx 2.33$
   $$

3. Comparing the roots:
   - $x_1 = 3$, $x_2 = 2.25$
   - $y_1 = 4$, $y_2 \approx 2.33$

We see that $x_1 < y_1$ but $x_2 < y_2$ is false since $2.25 < 2.33$ is true.

However, since the roots are distinct and do not maintain a consistent inequality (some $x$ roots are less than $y$ roots, some are not), no definite relation $x < y$ or $x > y$ holds for all roots.

**Answer:** $x = y$ or no relation between $x$ and $y$ (option **d**).

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