Given the quadratic equations:

I. $$2x^2 - 11x + p = 0$$

II. $$2y^2 - 15y + q = 0$$

Note:

(i) One root of equation (I) is 2.

(ii) One root of equation (II) is three times the smaller root of equation (I).

Find the product of $$p$$ and $$q$$.

Question

Given the quadratic equations:

I. $$2x^2 - 11x + p = 0$$

II. $$2y^2 - 15y + q = 0$$

Note:

(i) One root of equation (I) is 2.

(ii) One root of equation (II) is three times the smaller root of equation (I).

Find the product of $$p$$ and $$q$$.

Accepted Answer

**Solution:**

1. From equation (I): $$2x^2 - 11x + p = 0$$, one root is given as $$x = 2$$.

2. Substitute $$x = 2$$ into equation (I) to find $$p$$:

$$2(2)^2 - 11(2) + p = 0$$

$$2 	imes 4 - 22 + p = 0$$

$$8 - 22 + p = 0$$

$$p = 14$$

3. Let the roots of equation (I) be $$2$$ and $$r$$ (the smaller root). Since one root is 2, the other root is $$r$$.

4. Sum of roots of equation (I) is:

$$rac{11}{2} = 2 + r$$

$$r = rac{11}{2} - 2 = rac{11}{2} - rac{4}{2} = rac{7}{2} = 3.5$$

5. The smaller root of equation (I) is $$2$$ or $$3.5$$? Since 2 < 3.5, the smaller root is $$2$$.

6. According to the problem, one root of equation (II) is three times the smaller root of equation (I):

$$3 	imes 2 = 6$$

7. Equation (II) is:

$$2y^2 - 15y + q = 0$$

Let the roots be $$6$$ and $$s$$.

8. Sum of roots of equation (II):

$$rac{15}{2} = 6 + s$$

$$s = rac{15}{2} - 6 = rac{15}{2} - rac{12}{2} = rac{3}{2} = 1.5$$

9. Product of roots of equation (II):

$$rac{q}{2} = 6 	imes 1.5 = 9$$

$$q = 18$$

10. Finally, find the product $$p 	imes q$$:

$$14 	imes 18 = 252$$

**Answer:** $252$

**Note:** The options provided have $252$ as option (c), but the correct product is $252$. However, the correct answer according to the calculations is $252$. Please verify the options carefully.

**Correction:** Since the product is $252$, the correct option is (c) $252$.

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