Given two 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, and 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 the value of $(p + q)$.

Question

Given two 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, and 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 the value of $(p + q)$.

Accepted Answer

**Solution:**

1. From Equation II: $q y^2 - 8y + 4 = 0$, one root is given as $\frac{2}{3}$.

2. Substitute $y = \frac{2}{3}$ into Equation II:
   $$q \left(\frac{2}{3}ight)^2 - 8 	imes \frac{2}{3} + 4 = 0$$

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

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

5. Simplify:
   $$4q - 12 = 0 \implies 4q = 12 \implies q = 3$$

6. Now, Equation II becomes:
   $$3 y^2 - 8 y + 4 = 0$$

7. Find roots of Equation II using quadratic formula:
   $$y = \frac{8 \pm \sqrt{64 - 4 	imes 3 	imes 4}}{2 	imes 3} = \frac{8 \pm \sqrt{64 - 48}}{6} = \frac{8 \pm \sqrt{16}}{6}$$

8. Roots are:
   $$y = \frac{8 + 4}{6} = 2 \quad 	ext{and} \quad y = \frac{8 - 4}{6} = \frac{4}{6} = \frac{2}{3}$$

The highest root of Equation II is $2$.

9. Let the roots of Equation I be $x_1$ and $x_2$. The equation is:
   $$p x^2 - 9 x + 7 = 0$$

10. Using quadratic formula:
    $$x = \frac{9 \pm \sqrt{81 - 28 p}}{2 p}$$

11. The roots depend on $p$. The highest root is:
    $$x_{max} = \frac{9 + \sqrt{81 - 28 p}}{2 p}$$

12. Given the ratio of highest roots is $\frac{x_{max}}{2} = \frac{7}{4}$.

13. So:
    $$x_{max} = 2 	imes \frac{7}{4} = \frac{14}{4} = 3.5$$

14. Equate and solve for $p$:
    $$\frac{9 + \sqrt{81 - 28 p}}{2 p} = 3.5$$

15. Multiply both sides by $2p$:
    $$9 + \sqrt{81 - 28 p} = 7 p$$

16. Isolate the square root:
    $$\sqrt{81 - 28 p} = 7 p - 9$$

17. Square both sides:
    $$81 - 28 p = (7 p - 9)^2 = 49 p^2 - 126 p + 81$$

18. Rearrange:
    $$81 - 28 p = 49 p^2 - 126 p + 81$$

$$0 = 49 p^2 - 126 p + 81 + 28 p - 81$$

$$0 = 49 p^2 - 98 p$$

19. Factor:
    $$p (49 p - 98) = 0$$

20. Since $p$ is positive integer, $p 
eq 0$, so:
    $$49 p - 98 = 0 \implies 49 p = 98 \implies p = 2$$

21. Finally, calculate $p + q$:
    $$p + q = 2 + 3 = 5$$

**Answer:** $5$

**Note:** The correct answer is option (a) 5.

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