The number of negative real roots of the equation ` (x ^(2)+5x) ^(2) -24 =2 (x^(2) +5x)` is :

Options

A. 4
B. 3
C. 2
D. 1

Answer

To find the number of negative real roots of the equation \((x^2 + 5x)^2 - 24 = 2(x^2 + 5x)\), we will follow these steps:

### Step 1: Substitute \(t\)
Let \(t = x^2 + 5x\). Then, the equation becomes:
\[ t^2 - 24 = 2t \]

### Step 2: Rearrange the equation
Rearranging gives us:
\[ t^2 - 2t - 24 = 0 \]

### Step 3: Factor the quadratic equation
Next, we can factor this quadratic equation. We can add and subtract 1 to complete the square:
\[ t^2 - 2t + 1 - 25 = 0 \]
This simplifies to:
\[ (t - 1)^2 - 25 = 0 \]

### Step 4: Solve for \(t\)
Setting the equation to zero gives:
\[ (t - 1)^2 = 25 \]
Taking the square root of both sides results in:
\[ t - 1 = \pm 5 \]
Thus, we have two cases:
1. \(t - 1 = 5 \implies t = 6\)
2. \(t - 1 = -5 \implies t = -4\)

### Step 5: Substitute back for \(x\)
Now we substitute back for \(t\):
1. For \(t = 6\):
\[ x^2 + 5x = 6 \implies x^2 + 5x - 6 = 0 \]
Factoring gives:
\[ (x - 1)(x + 6) = 0 \]
So, the roots are \(x = 1\) and \(x = -6\).

2. For \(t = -4\):
\[ x^2 + 5x = -4 \implies x^2 + 5x + 4 = 0 \]
Factoring gives:
\[ (x + 1)(x + 4) = 0 \]
So, the roots are \(x = -1\) and \(x = -4\).

### Step 6: Count the negative roots
From the solutions:
- From \(t = 6\): one negative root, \(x = -6\).
- From \(t = -4\): two negative roots, \(x = -1\) and \(x = -4\).

### Conclusion
Thus, the total number of negative real roots is:
\[ 1 + 2 = 3 \]

The final answer is **3**.

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