Question

If the roots of the equation (b-c)x²+(a-c)x+(a-b)=0 are equal then prove that 2b=a+c.
please tell me the answer ​

Answer 1

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

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

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

Now the roots of the quadratic equation are equal

So

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

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

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

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

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

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

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

Hence proved

Answer 2

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