The roots of `(a-b)^(2)x^(2)+2(a+b-2c)x+1=0` are real or imaginary according as c does not or does lie between a and b, `a lt b`. True False.

Answer

To determine whether the statement "The roots of \((a-b)^2x^2 + 2(a+b-2c)x + 1 = 0\) are real or imaginary according as \(c\) does not or does lie between \(a\) and \(b\), \(a < b\)" is true or false, we will analyze the quadratic equation and its discriminant.

### Step-by-Step Solution:

1. **Identify the coefficients** of the quadratic equation:
The given quadratic equation is:
\[ (a-b)^2x^2 + 2(a+b-2c)x + 1 = 0 \]
Here, we can identify:
- \(A = (a-b)^2\)
- \(B = 2(a+b-2c)\)
- \(C = 1\)

2. **Calculate the discriminant**:
The discriminant \(D\) of a quadratic equation \(Ax^2 + Bx + C = 0\) is given by:
\[ D = B^2 - 4AC \]
Substituting the values of \(A\), \(B\), and \(C\):
\[ D = [2(a+b-2c)]^2 - 4[(a-b)^2 \cdot 1] \]
Simplifying this:
\[ D = 4(a+b-2c)^2 - 4(a-b)^2 \]
Factoring out the 4:
\[ D = 4\left[(a+b-2c)^2 - (a-b)^2\right] \]

3. **Use the difference of squares**:
The expression \((a+b-2c)^2 - (a-b)^2\) can be factored using the difference of squares:
\[ D = 4\left[(a+b-2c - (a-b))(a+b-2c + (a-b))\right] \]
Simplifying further:
\[ D = 4\left[(b+c-2c)(b+c+2a-2b)\right] = 4\left[(b-c)(a+c-b)\right] \]

4. **Analyze the sign of the discriminant**:
The roots of the quadratic equation will be real if \(D \geq 0\) and imaginary if \(D < 0\).
- For \(D \geq 0\):
\[ (b-c)(a+c-b) \geq 0 \]
- For \(D < 0\):
\[ (b-c)(a+c-b) < 0 \]

5. **Determine the intervals for \(c\)**:
- If \(c < a\) or \(c > b\), then \(b-c > 0\) and \(a+c-b > 0\) (real roots).
- If \(c\) lies between \(a\) and \(b\) (i.e., \(a < c < b\)), then \(b-c < 0\) and \(a+c-b < 0\) (imaginary roots).

6. **Conclusion**:
The statement is **true**. The roots of the quadratic equation are real if \(c\) does not lie between \(a\) and \(b\) (i.e., \(c < a\) or \(c > b\)), and the roots are imaginary if \(c\) lies between \(a\) and \(b\).