Find interval of values of `' a '` for which one root of equation `x^2+a^2(x+1)+(a-2)=0` is negative and the other root is positive

Options

A. `(-5,-4)`
B. `(-oo,-2)`
C. `(-oo,2)`
D. `(-2,1)`

Answer

To solve the problem, we need to find the interval of values for \( a \) such that one root of the quadratic equation

\[ x^2 + a^2(x + 1) + (a - 2) = 0 \]

is negative and the other root is positive.

### Step 1: Rewrite the quadratic equation

The equation can be rewritten as:

\[ x^2 + a^2x + a^2 + a - 2 = 0 \]

### Step 2: Identify the conditions for the roots

For one root to be negative and the other to be positive, the following conditions must be satisfied:
1. The product of the roots (given by \( \frac{c}{a} \) for a quadratic \( ax^2 + bx + c \)) must be negative.
2. The value of the quadratic at \( x = 0 \) must be negative.

### Step 3: Calculate the product of the roots

The product of the roots \( r_1 \) and \( r_2 \) of the quadratic equation is given by:

\[ r_1 \cdot r_2 = \frac{c}{a} = \frac{a^2 + a - 2}{1} = a^2 + a - 2 \]

For one root to be negative and the other positive, we require:

\[ a^2 + a - 2 < 0 \]

### Step 4: Solve the inequality

To solve the inequality \( a^2 + a - 2 < 0 \), we first find the roots of the equation \( a^2 + a - 2 = 0 \) using the quadratic formula:

\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2} \]

This gives us:

\[ a = 1 \quad \text{and} \quad a = -2 \]

### Step 5: Test intervals

Now we test the intervals determined by the roots \( -2 \) and \( 1 \):

1. For \( a < -2 \) (e.g., \( a = -3 \)):
\[ (-3)^2 + (-3) - 2 = 9 - 3 - 2 = 4 \quad (\text{positive}) \]

2. For \( -2 < a < 1 \) (e.g., \( a = 0 \)):
\[ 0^2 + 0 - 2 = -2 \quad (\text{negative}) \]

3. For \( a > 1 \) (e.g., \( a = 2 \)):
\[ 2^2 + 2 - 2 = 4 + 2 - 2 = 4 \quad (\text{positive}) \]

Thus, the inequality \( a^2 + a - 2 < 0 \) holds for:

\[ -2 < a < 1 \]

### Step 6: Check the value of the quadratic at \( x = 0 \)

Next, we need to ensure that the value of the quadratic at \( x = 0 \) is negative:

\[ f(0) = a^2 + a - 2 < 0 \]

This condition is already satisfied by the interval \( -2 < a < 1 \).

### Conclusion

The interval of values of \( a \) for which one root of the equation is negative and the other root is positive is:

\[ \boxed{(-2, 1)} \]