If `5^(k^(2))(25^(2k))(625)=25 sqrt(5)` and `k lt - 1`, what is the value of k ?

Options

A. `-3.581`
B. `-3.162`
C. `-2.613`
D. `-1.581`

Explanation

To solve the equation \( 5^{k^2} \cdot 25^{2k} \cdot 625 = 25 \sqrt{5} \) given that \( k < -1 \), we will follow these steps:

### Step 1: Rewrite the bases in terms of 5
We know that:
- \( 25 = 5^2 \)
- \( 625 = 25^2 = (5^2)^2 = 5^4 \)
- \( \sqrt{5} = 5^{1/2} \)

Thus, we can rewrite the equation as:
\[ 5^{k^2} \cdot (5^2)^{2k} \cdot 5^4 = 5^2 \cdot 5^{1/2} \]

### Step 2: Simplify the left-hand side
Now, simplify the left-hand side:
\[ 5^{k^2} \cdot 5^{4k} \cdot 5^4 = 5^{k^2 + 4k + 4} \]

### Step 3: Simplify the right-hand side
Now simplify the right-hand side:
\[ 5^2 \cdot 5^{1/2} = 5^{2 + 1/2} = 5^{2.5} = 5^{5/2} \]

### Step 4: Set the exponents equal to each other
Since the bases are the same, we can equate the exponents:
\[ k^2 + 4k + 4 = \frac{5}{2} \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ k^2 + 4k + 4 - \frac{5}{2} = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ 2k^2 + 8k + 8 - 5 = 0 \implies 2k^2 + 8k + 3 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2 \), \( b = 8 \), and \( c = 3 \).
\[ k = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2} \]
Calculating the discriminant:
\[ k = \frac{-8 \pm \sqrt{64 - 24}}{4} = \frac{-8 \pm \sqrt{40}}{4} = \frac{-8 \pm 2\sqrt{10}}{4} = \frac{-4 \pm \sqrt{10}}{2} \]

### Step 7: Evaluate the two possible values for k
This gives us two potential solutions:
\[ k = \frac{-4 + \sqrt{10}}{2} \quad \text{and} \quad k = \frac{-4 - \sqrt{10}}{2} \]

### Step 8: Determine which solution satisfies \( k < -1 \)
Calculating the approximate values:
- \( \sqrt{10} \approx 3.16 \)
- \( k = \frac{-4 + 3.16}{2} \approx \frac{-0.84}{2} \approx -0.42 \) (not valid since \( k \geq -1 \))
- \( k = \frac{-4 - 3.16}{2} \approx \frac{-7.16}{2} \approx -3.58 \) (valid since \( k < -1 \))

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{-3.58} \]