Question

If the roots of the equation (a-b)x^2+(b-c)x+(c-a)=0 are equal show that c, a and b are in AP

Answer 1

$(a-b)x^2+(b-c)x+(c-a)=0 \\ \\ roots \: are \: equal. \: so \: determinant \: = 0 \\ \\ {(b - c)}^{2} - 4(a - b)(c - a) = 0 \\ \\ \Rightarrow ( {b}^{2} + {c}^{2} - 2bc) - 4(ac - {a}^{2} - bc + ab) = 0\\ \\ \Rightarrow {b}^{2} + {c}^{2} - 2bc - 4ac + 4 {a}^{2} + 4 bc - 4 ab = 0\\ \\ \Rightarrow {b}^{2} + {c}^{2} + 4 {a}^{2} - 4ac + 2 bc - 4 ab = 0$

$\Rightarrow {(b)}^{2} + {(c)}^{2} + {( - 2a)}^{2} + 2 \times ( - 2a ) \times c+ 2 bc + 2 \times ( - 2 a) \times b = 0 \\ \\ \Rightarrow {(b + c - 2a)}^{2} = 0 \\ \\ \Rightarrow b + c - 2a = 0 \\ \\ \Rightarrow b + c = 2a \\ \\ \Rightarrow b - a = a - c$

$\boxed{ \textbf{Thus, \red{c, a and b} are in AP}}$

Answer 2

well,

c,a,b are in A.P if 2a = b + c.

Also,if roots of a quadratic polynomial is equal, then its discriminant must be zero.