Question

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

Answer 1

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

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

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

$\: \: \: \: \: \bull \: \: \sf \: \: 2a = b + c$

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

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

Let's solve the problem now!!

Given

• The quadratic equation (a - b)x² + (b - c)x + (c - a) = 0.

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

• A = a - b

• B = b - c

• C = c - a

Since,

It is given that

• given quadratic equation has equal roots.

So,

$\rm :\longmapsto\:Discriminant, \: D \: = \: 0$

$\rm :\longmapsto\: {B}^{2} - 4AC = 0$

On substituting the values of A, B and C, we get

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

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

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

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

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

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

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

$\boxed{ \sf \because \: {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca}$

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

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

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Answer 2

$\tt\huge\purple{hello!!!}$

HERE IS UR ANSWER

$_____________________________$

Given−

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

To prove−

2a = b + c

Solution−

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

Let's solve the problem now!!

Given

The quadratic equation (a - b)x² + (b - c)x + (c - a) = 0.

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

A = a - b

B = b - c

C = c - a

Since,

It is given that

given quadratic equation has equal roots.

So,

⟼Discriminant,D=0

⟼B^2−4AC=0

On substituting the values of A, B and C, we get

⟼(b−c)^2−4(a−b)(c−a)=0

⟼b^2+c^2−2bc−4(ac−a^2−bc+ab)=0

⟼b^2+c^2−2bc−4ac+4a^2+4bc−4ab=0

⟼b^2+c^2+2bc−4ac+4a^2−4ab=0

⟼b^2+c^2+4a^2+2bc−4ac−4ab=0

⟼b^2+c^2+(−2a)^2+2bc+2×(−2a)c+2×(−2a)b=0

⟼(b+c−2a)^2=0

∵(a+b+c)2=a2+b2+c2+2ab+2bc+2ca

⟼b+c−2a=0

⟹b+c=2a

${\boxed{\boxed{\bf{Hence, Proved}}}}$