If k=\frac{c}{2} (c\neq0), then the roots of the equation are:
A quadratic equation is given by (a+b)x^{2}-(a+b+c)x+k=0, where a, b, c are real.
- A. Real and equal
- B. Real and unequal ✓
- C. Real iff a>c
- D. Complex but <strong>NOT</strong> real
Correct Answer: B. Real and unequal
Explanation
For k = \frac{c}{2}, the discriminant is \Delta = (a+b+c)^2 - 4(a+b)(\frac{c}{2}) = (a+b+c)^2 - 2c(a+b). Expanding this gives (a+b)^2 + c^2 + 2c(a+b) - 2c(a+b) = (a+b)^2 + c^2. Since a,b,c are real and c \neq 0, (a+b)^2 \geq 0 and c^2 \gt 0, ensuring \Delta \gt 0. Therefore, the roots are real and unequal.
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