Question

roots of equation (a-b) x2 + (b-c) x + (c-a) = 0 are equal, prove that 2a= b+c​

Answer 1

Answer:

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Step-by-step explanation:

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answer given above

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

The roots of the quadratic equation

$p {x}^{2} + qx + r = 0$are said to be equal if

Discriminant $D = { {q}^{2} - 4pr} = 0$

CALCULATION

Comparing the given equation with

$p {x}^{2} + qx + r = 0$

We get

$p \: = a \: - b \: , \: q \: = b - c \:, r \: = c - a$

Now the roots of the quadratic equation are equal

So

$D = { {q}^{2} - 4pr} = 0$

$\implies \: \: {(b - c)}^{2} - 4 \times (a - b)(c - a) = 0$

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

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

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

$\implies \: {(2a - b - c)} = 0$

$\implies \: 2a = b + c$

Hence proved