Question

If the roots of the equation (a - b)x^2 + (b - c) x + (c - a) = 0 are equal. Prove that 2a = b + c.​

Answer 1

$\huge\underline\mathbb{SOLUTION:-}$

$\mathsf \red{(a - b) x^2 + (b - c) x + (c - a) = 0}$

$\mathsf {T.P\:2a = b + c}$

$\mathsf {B^2 - 4AC = 0}$

$\mathsf {(b - c)^2 - [4(a - b)\:(c - a)] = 0}$

$\mathsf {b^2 - 2bc + c^2 - [4(ac - a^2 - bc + ab)] = 0}$

$\implies \mathsf {b^2 - 2bc + c^2 - 4ac + 4a^2 + 4bc - 4ab = 0}$

$\implies \mathsf {b^2 + 2bc + c^2 + 4a^2 - 4ac - 4ab = 0}$

$\implies \mathsf {(b + c - 2a)^2 = 0}$

$\implies \mathsf \blue {b + c = 2a}$

Answer 2

Answer:

(a−b)x

2

+(b−c)x+(c−a)=0

\mathsf {T.P\:2a = b + c}T.P2a=b+c

\mathsf {B^2 - 4AC = 0}B

2

−4AC=0

\mathsf {(b - c)^2 - [4(a - b)\:(c - a)] = 0}(b−c)

2

−[4(a−b)(c−a)]=0

\mathsf {b^2 - 2bc + c^2 - [4(ac - a^2 - bc + ab)] = 0}b

2

−2bc+c

2

−[4(ac−a

2

−bc+ab)]=0

\implies \mathsf {b^2 - 2bc + c^2 - 4ac + 4a^2 + 4bc - 4ab = 0}⟹b

2

−2bc+c

2

−4ac+4a

2

+4bc−4ab=0

\implies \mathsf {b^2 + 2bc + c^2 + 4a^2 - 4ac - 4ab = 0}⟹b

2

+2bc+c

2

+4a

2

−4ac−4ab=0

\implies \mathsf {(b + c - 2a)^2 = 0}⟹(b+c−2a)

2

=0

\implies \mathsf \blue {b + c = 2a}⟹b+c=2a

Step-by-step explanation:

hope it will be help you. ........