Given the following quadratic equations:

Equation I: $p x^2 - 9x + 7 = 0$

Equation II: $q y^2 - 8y + 4 = 0$

where:
- $p$ and $q$ are positive integers,
- One root of Equation II is $\frac{2}{3}$,
- The ratio of the highest root of Equation I to the highest root of Equation II is $7 : 4$.

Find both roots of the equation:

$$(p+q)^2 a^2 + (3 p q + 1) a - 6 = 0$$

Question

Given the following quadratic equations:

Equation I: $p x^2 - 9x + 7 = 0$

Equation II: $q y^2 - 8y + 4 = 0$

where:
- $p$ and $q$ are positive integers,
- One root of Equation II is $\frac{2}{3}$,
- The ratio of the highest root of Equation I to the highest root of Equation II is $7 : 4$.

Find both roots of the equation:

$$(p+q)^2 a^2 + (3 p q + 1) a - 6 = 0$$

Accepted Answer

**Solution:**

1. **Analyze Equation II:**
   Given one root of Equation II is $\frac{2}{3}$.
   Equation II: $q y^2 - 8 y + 4 = 0$

Using the root $y = \frac{2}{3}$:
   Substitute into the equation:
   $$q \left(\frac{2}{3}ight)^2 - 8 	imes \frac{2}{3} + 4 = 0$$

Calculate:
   $$q 	imes \frac{4}{9} - \frac{16}{3} + 4 = 0$$

Multiply entire equation by 9 to clear denominators:
   $$4 q - 48 + 36 = 0 \implies 4 q - 12 = 0$$

Solve for $q$:
   $$4 q = 12 \implies q = 3$$

2. **Find roots of Equation II:**
   Equation II becomes:
   $$3 y^2 - 8 y + 4 = 0$$

Calculate discriminant:
   $$\Delta = (-8)^2 - 4 	imes 3 	imes 4 = 64 - 48 = 16$$

Roots:
   $$y = \frac{8 \pm \sqrt{16}}{2 	imes 3} = \frac{8 \pm 4}{6}$$

So roots are:
   $$y_1 = \frac{8 + 4}{6} = 2, \quad y_2 = \frac{8 - 4}{6} = \frac{4}{6} = \frac{2}{3}$$

3. **Analyze Equation I:**
   Equation I: $p x^2 - 9 x + 7 = 0$

Let roots be $x_1$ and $x_2$ with $x_1 > x_2$.

Sum of roots:
   $$x_1 + x_2 = \frac{9}{p}$$

Product of roots:
   $$x_1 x_2 = \frac{7}{p}$$

4. **Use the ratio of highest roots:**
   Highest root of Equation I to highest root of Equation II is $7 : 4$.

Highest root of Equation II is $2$.

So highest root of Equation I:
   $$x_1 = \frac{7}{4} 	imes 2 = \frac{14}{4} = \frac{7}{2} = 3.5$$

5. **Find $p$ using $x_1$:**
   Using sum of roots:
   $$x_1 + x_2 = \frac{9}{p}$$

Using product of roots:
   $$x_1 x_2 = \frac{7}{p}$$

Let $x_2 = r$, then:
   $$3.5 + r = \frac{9}{p}$$
   $$3.5 r = \frac{7}{p}$$

From the second equation:
   $$r = \frac{7}{3.5 p} = \frac{2}{p}$$

Substitute $r$ in the first equation:
   $$3.5 + \frac{2}{p} = \frac{9}{p}$$

Multiply both sides by $p$:
   $$3.5 p + 2 = 9$$

$$3.5 p = 7 \implies p = 2$$

6. **Verify roots of Equation I:**
   Sum of roots:
   $$\frac{9}{2} = 4.5$$

We have $x_1 = 3.5$, so
   $$x_2 = 4.5 - 3.5 = 1$$

Product of roots:
   $$3.5 	imes 1 = 3.5$$

Check product formula:
   $$\frac{7}{2} = 3.5$$

Correct.

7. **Find roots of the required equation:**
   Given:
   $$(p + q)^2 a^2 + (3 p q + 1) a - 6 = 0$$

Substitute $p = 2$, $q = 3$:
   $$(2 + 3)^2 a^2 + (3 	imes 2 	imes 3 + 1) a - 6 = 0$$

Calculate:
   $$5^2 a^2 + (18 + 1) a - 6 = 0$$

$$25 a^2 + 19 a - 6 = 0$$

8. **Solve quadratic for $a$:**
   Discriminant:
   $$\Delta = 19^2 - 4 	imes 25 	imes (-6) = 361 + 600 = 961$$

$$\sqrt{961} = 31$$

Roots:
   $$a = \frac{-19 \pm 31}{2 	imes 25} = \frac{-19 \pm 31}{50}$$

So,
   $$a_1 = \frac{-19 + 31}{50} = \frac{12}{50} = \frac{6}{25}$$

$$a_2 = \frac{-19 - 31}{50} = \frac{-50}{50} = -1$$

9. **Check options:**
   The roots are $\frac{6}{25}$ and $-1$.

However, the options list $\frac{4}{25}, -1$ as correct.

Re-examine step 7:

Wait, the calculation of $3 p q + 1$:
   $$3 	imes 2 	imes 3 + 1 = 18 + 1 = 19$$

So the quadratic is correct.

Roots are $\frac{6}{25}$ and $-1$.

The option with $\frac{6}{25}, -1$ is option (e).

**Answer:** Option (e) $\frac{6}{25}, -1$

**Note:** The initial marking of option (a) as correct was an error; the correct roots are $\frac{6}{25}$ and $-1$ matching option (e).

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