Question

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

Answer 1

Given :

• If roots of equation (b-c)x²+(c-a)x+(a-b) =0 are equal, then show that 2b=a+c

Show that :

•2b=a+c

Formula :

• If the equations has real and equal roots then

$\implies\tt{D=b²-4ac=0}$

Solution :

•The given equation is

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

This equation has the general form ie,

$\implies \sf \: a {x}^{2} + bx + c = 0$

Here,

$\bullet \tt \: a = b - c$

$\bullet \tt \: b = c - a$

$\bullet \tt \: c = a - b$

Given that the equation has real and equal roots

Hence,

$\\ \implies\sf{D=b²-4ac=0}$

$\\ \implies \sf \: {(c - a)}^{2} - 4 \times (b - c)(a - b) = 0 \\ \\ \\ \implies \sf \: {c}^{2} + {a}^{2} - 2ac = 4(ab - {b}^{2} - ac + cb) = 0 \\ \\ \\ \implies \sf \: {c}^{2} + {a}^{2} - 2ac - 4ab + 4 {b}^{2} + 4ac - 4cb = 0 \\ \\ \\ \implies \sf \: {c}^{2} + {a}^{2} + 2ac - 4ab + 4 {b}^{2} - 4cb = 0 \\ \\ \\ \implies \sf \: {(a + c)}^{2} - 4ab + 4 {b}^{2} + 4cb = 0 \\ \\ \\ \implies \sf \: {(c + a - 2b)}^{2} = 0 \\ \\ \\ \implies \sf \: c + a - 2b = 0 \\ \\ \\ \implies \boxed{ \sf{c + a = 2b}}$

Hence,

It is proved that c+a=2b

Answer 2

Answer:

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