Let `Delta=|(1,-4,20),(1,-2,5),(1,2,5x^(2))|` Solution set of `Delta=0` is

Options

A. `{-2,-3}`
B. `{-3,4}`
C. `{4,-6}`
D. `{-2,-1}`

Answer

To solve the determinant \( \Delta = \begin{vmatrix} 1 & -4 & 20 \\ 1 & -2 & 5 \\ 1 & 2 & 5x^2 \end{vmatrix} \) and find the solution set for \( \Delta = 0 \), we will follow these steps:

### Step 1: Calculate the Determinant
We can calculate the determinant using the formula for a 3x3 matrix:

\[ \Delta = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our determinant:

\[ \Delta = 1 \begin{vmatrix} -2 & 5 \\ 2 & 5x^2 \end{vmatrix} - (-4) \begin{vmatrix} 1 & 5 \\ 1 & 5x^2 \end{vmatrix} + 20 \begin{vmatrix} 1 & -2 \\ 1 & 2 \end{vmatrix} \]

### Step 2: Calculate Each Minor
1. Calculate \( \begin{vmatrix} -2 & 5 \\ 2 & 5x^2 \end{vmatrix} \):
\[ = (-2)(5x^2) - (5)(2) = -10x^2 - 10 \]

2. Calculate \( \begin{vmatrix} 1 & 5 \\ 1 & 5x^2 \end{vmatrix} \):
\[ = (1)(5x^2) - (5)(1) = 5x^2 - 5 \]

3. Calculate \( \begin{vmatrix} 1 & -2 \\ 1 & 2 \end{vmatrix} \):
\[ = (1)(2) - (-2)(1) = 2 + 2 = 4 \]

### Step 3: Substitute Back into the Determinant Formula
Now substitute these values back into the determinant expression:

\[ \Delta = 1(-10x^2 - 10) + 4(5x^2 - 5) + 20(4) \]

### Step 4: Simplify the Expression
\[ \Delta = -10x^2 - 10 + 20x^2 - 20 + 80 \]
\[ = (20x^2 - 10x^2) + (-10 - 20 + 80) \]
\[ = 10x^2 + 50 \]

### Step 5: Set the Determinant Equal to Zero
Now, we set \( \Delta = 0 \):

\[ 10x^2 + 50 = 0 \]

### Step 6: Solve for \( x^2 \)
\[ 10x^2 = -50 \]
\[ x^2 = -5 \]

### Step 7: Analyze the Solution
Since \( x^2 = -5 \) has no real solutions (as the square of a real number cannot be negative), we conclude that there are no real values of \( x \) that satisfy the equation.

### Final Answer
The solution set of \( \Delta = 0 \) is that there are no real values of \( x \).

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