I. x2 -Px + 32 = 0 (Roots of the equation is 4 and 4A.)

2(cube of root of Y) = Square of root of Y.

Quantity I: Find the value of 2a.
Z2 – KZ + 990 = 0 (a and b are root of the equation and the value of K = 5P+3.)

Quantity II: Find the value of Y+A.

Quantity III: Find the value of (Y+2) – Highest root of the equation I.

Question

I. x2 -Px + 32 = 0 (Roots of the equation is 4 and 4A.)

2(cube of root of Y) = Square of root of Y.

Quantity I: Find the value of 2a.
Z2 – KZ + 990 = 0 (a and b are root of the equation and the value of K = 5P+3.)

Quantity II: Find the value of Y+A.

Quantity III: Find the value of (Y+2) – Highest root of the equation I.

Accepted Answer

**Solution:**

1. From Equation I: $x^2 - Px + 32 = 0$ with roots 4 and $4A$.

Substitute root 4 into the equation:
   $$4^2 - P 	imes 4 + 32 = 0 \Rightarrow 16 - 4P + 32 = 0$$
   $$48 = 4P \Rightarrow P = 12$$

2. The roots of Equation I are 4 and $4A$. Using the sum of roots:
   $$4 + 4A = P = 12 \Rightarrow 4A = 12 - 4 = 8 \Rightarrow A = 2$$

3. For $Y$, given:
   $$2 	imes (\sqrt[3]{Y}) = (\sqrt{Y})$$

Let $\sqrt[3]{Y} = t$.
   Then $2t = t^3$ (since $\sqrt{Y} = t^{3/2}$ is not consistent, we interpret as square root of $Y$ is $t^2$).

Alternatively, rewrite as:
   $$2 	imes (\sqrt[3]{Y}) = (\sqrt{Y})$$
   $$2t = t^{3/2}$$

To solve, raise both sides to power 2:
   $$(2t)^2 = (t^{3/2})^2 \Rightarrow 4t^2 = t^3 \Rightarrow 4 = t$$

So,
   $$\sqrt[3]{Y} = 4 \Rightarrow Y = 4^3 = 64$$

4. Equation II: $Z^2 - KZ + 990 = 0$, where $K = 5P + 3 = 5 	imes 12 + 3 = 63$.

So,
   $$Z^2 - 63Z + 990 = 0$$

5. Find roots of Equation II:

Calculate discriminant:
   $$D = 63^2 - 4 	imes 990 = 3969 - 3960 = 9$$

Roots:
   $$Z = \frac{63 \pm 3}{2}$$
   $$Z_1 = 33, \quad Z_2 = 30$$

6. Quantity I: $2a$ where $a$ is one root of Equation II.

Taking the larger root $a = 33$ for maximum value,
   $$2a = 2 	imes 33 = 66$$

7. Quantity II: $Y + A = 64 + 2 = 66$

8. Quantity III: $(Y + A) -$ highest root of Equation I

Highest root of Equation I is $8$ (since roots are 4 and 8),
   $$66 - 8 = 58$$

9. Comparing quantities:

$$	ext{Quantity I} = 66$$
   $$	ext{Quantity II} = 66$$
   $$	ext{Quantity III} = 58$$

Therefore,
   $$	ext{Quantity I} = 	ext{Quantity II} > 	ext{Quantity III}$$

**Answer:** Quantity I ≥ Quantity II > Quantity III (Option d)

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