Given the two equations:

I. $$2x^2 + 7x + k = 0$$

II. $$(ay + b)^2 = 0$$

It is known that the smaller root of equation I is $$-2$$ and the larger root of equation I is also a root of equation II.

Find the value of $$b$$.

Question

Given the two equations:

I. $$2x^2 + 7x + k = 0$$

II. $$(ay + b)^2 = 0$$

It is known that the smaller root of equation I is $$-2$$ and the larger root of equation I is also a root of equation II.

Find the value of $$b$$.

Accepted Answer

**Solution:**

1. From equation I: $$2x^2 + 7x + k = 0$$, the smaller root is given as $$-2$$.

2. Substitute $$x = -2$$ into equation I to find $$k$$:
   $$2(-2)^2 + 7(-2) + k = 0$$
   $$2 	imes 4 - 14 + k = 0$$
   $$8 - 14 + k = 0$$
   $$k = 6$$

3. So, equation I becomes:
   $$2x^2 + 7x + 6 = 0$$

4. Calculate the roots of equation I:
   Using quadratic formula,
   $$x = \frac{-7 \pm \sqrt{7^2 - 4 	imes 2 	imes 6}}{2 	imes 2} = \frac{-7 \pm \sqrt{49 - 48}}{4} = \frac{-7 \pm 1}{4}$$

5. Roots are:
   - Smaller root: $$\frac{-7 - 1}{4} = \frac{-8}{4} = -2$$ (given)
   - Larger root: $$\frac{-7 + 1}{4} = \frac{-6}{4} = -\frac{3}{2}$$

6. Equation II is $$(ay + b)^2 = 0$$, which means it has a repeated root at $$y = -\frac{b}{a}$$.

7. Since the larger root of equation I is also a root of equation II, we have:
   $$-\frac{b}{a} = -\frac{3}{2}$$

8. This implies:
   $$\frac{b}{a} = \frac{3}{2}$$

9. However, without knowing the values of $$a$$ or any additional information, the exact value of $$b$$ cannot be determined uniquely.

**Answer:** Cannot be determined

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