Question

$prove \: yourself \: brainlians$
If the roots of the equation (a-b)x²+(b-c)x+(c-a)=0 are equal, prove that 2a=b+c.​

Answer 1

HELLO MATE
HERE IS YOUR ANSWER

Given:
roots of the equation (a-b)x²+(b-c)x+(c-a)=0 are equal.

To prove: 2a = b + c

as it is given,
then, d=0 [here d= determinant]

also,
$d = {b}^{2} - 4ac$

$= > {( b - c)}^{2} - 4.(a - b).(c - a) = 0 \\ = > {b}^{2} + {c}^{2} - 2bc - 4ac + {4a}^{2} + 4bc - 4ab = 0$

SOLVE THIS EQUATION AND YOU WILL GET THAT VALUE.....

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BY SHIVAM SHAURYA

Answer 2

hello ✌✌

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let r & s be the roots, then: r + s = -(b - c)/(a - b) but r = s: 2r = -(b - c)/(a - b) r = -(b - c)/[2(a - b)] also: r * s = r^2 = (c - a)/(a - b) (b - c)^2/[4(a - b)^2] = (c - a)/(a - b) (b - c)^2/[4(a - b)] = (c - a) b^2 - 2bc + c^2 = 4ac - 4a^2 - 4bc + 4ab 4a^2 - 4ab + b^2 - 4ac + 2bc + c^2 = 0 (2a - b)^2 - 2c(2a - b) + c^2 = 0 [(2a - b) - c]^2 = 0 2a - b - c = 0 2a = b + c

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