Question

028. If the roots of the equation (a - b)x' + (b-c)x + (c-a) = 0 are equal, show that c, a and b are in AP.​

Answer 1

Step-by-step explanation:

a−b)x

2

+(b−c)x+(c−a)=0

rootsareequal.sodeterminant=0

(b−c)

2

−4(a−b)(c−a)=0

⇒(b

2

+c

2

−2bc)−4(ac−a

2

−bc+ab)=0

⇒b

2

+c

2

−2bc−4ac+4a

2

+4bc−4ab=0

⇒b

2

+c

2

+4a

2

−4ac+2bc−4ab=0

\begin{gathered}\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 \end{gathered}

⇒(b)

2

+(c)

2

+(−2a)

2

+2×(−2a)×c+2bc+2×(−2a)×b=0

⇒(b+c−2a)

2

=0

⇒b+c−2a=0

⇒b+c=2a

⇒b−a=a−c

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

Thus, c, a and b are in AP