If a,b are real, then the roots of the quadratic equation `(a-b)x^(2)-5 (a+b) x-2(a-b) =0` are

Options

A. Real and equal
B. Non-real complex
C. Real and unequal
D. None of these

Answer

To determine the nature of the roots of the quadratic equation \((a-b)x^2 - 5(a+b)x - 2(a-b) = 0\), we will follow these steps:

### Step 1: Identify the coefficients
The standard form of a quadratic equation is \(Ax^2 + Bx + C = 0\). Here, we can identify:
- \(A = a - b\)
- \(B = -5(a + b)\)
- \(C = -2(a - b)\)

### Step 2: Calculate the discriminant
The discriminant \(D\) of a quadratic equation is given by the formula:
\[ D = B^2 - 4AC \]
Substituting the values of \(A\), \(B\), and \(C\):
\[ D = [-5(a + b)]^2 - 4(a - b)(-2(a - b)) \]

### Step 3: Simplify the discriminant
Calculating \(B^2\):
\[ B^2 = 25(a + b)^2 \]
Calculating \(4AC\):
\[ 4AC = 4(a - b)(-2(a - b)) = -8(a - b)^2 \]
Thus, the discriminant becomes:
\[ D = 25(a + b)^2 + 8(a - b)^2 \]

### Step 4: Analyze the discriminant
Now we need to analyze \(D = 25(a + b)^2 + 8(a - b)^2\). Both terms \(25(a + b)^2\) and \(8(a - b)^2\) are squares multiplied by positive constants, hence they are both non-negative.

Since \(a\) and \(b\) are real numbers, \(D\) will always be greater than or equal to zero. Specifically:
- If \(a + b \neq 0\) or \(a - b \neq 0\), then \(D > 0\).

### Step 5: Conclusion about the roots
Since \(D > 0\), the roots of the quadratic equation are real and unequal.

### Final Answer
The roots of the quadratic equation \((a-b)x^2 - 5(a+b)x - 2(a-b) = 0\) are real and unequal.

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