Question

if the roots of equation (b-c)x²+(c-a)x+(a-b)=0 are equal then 2b=a+c i.e., a, b, c are in AP . ​

Answer 1

$\large\underline{\sf{Given- }}$

↝ The roots of equation (b-c)x²+(c-a)x+(a-b)=0 are equal.

$\large\underline{\sf{To\:prove - }}$

↝ 2b = a + c

Concept Used :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac

$\red{\large\underline{\sf{Solution-}}}$

Given quadratic equation

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

On comparing with, Ax² + Bx + C = 0, we get

• A = b - c

• B = c - a

• C = a - b

Now, it is given that,

The roots of equation (b-c)x²+(c-a)x+(a-b)=0 are equal,

So, Discriminant, D = B² - 4AC = 0

So, on substituting the values, we get

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

$\rm :\longmapsto\: {c}^{2} + {a}^{2} - 2ac - 4(ab - ac - {b}^{2} + bc)= 0$

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

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

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

can be re-arranged as

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

can be rewritten as

$\rm :\longmapsto\: {a}^{2} + {( - 2b)}^{2} + {c}^{2} + 2(a)( - 2b) + 2( - 2b)(c) + 2(a)(c) = 0$

$\rm :\longmapsto\: {(a - 2b + c)}^{2} = 0$

$\rm :\longmapsto\: {a - 2b + c} = 0$

$\bf :\longmapsto\: {a + c} = 2b$

$\rm :\longmapsto\:a + c = b + b$

$\rm :\longmapsto\:c - b = b - a$

$\bf\implies \:a,b,c \: are \: in \: AP$

Answer 2

Answer:

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