Question

If the roots of the quadratic equation
(a - b)x^2 + (b - c)x + (c- a) = 0 are equal
prove that 2a = b + c​

Answer 1

Given :

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

To prove :

2a=b+c

Theory :

For a Quadratic equation of the form

ax²+bx+c= 0 , the expression b²-4ac is called the discriminant.

Nature of roots

The roots of a quadratic equation can be of three types.

If D>0, the equation has two distinct real roots.

If D=0, the equation has two equal real roots.

If D<0, the equation has no real roots.

Solution :

We have , (a-b)x²+(b-c)x+(c-a)=0

On comparing with the standard form of Quadratic equation ax²+bx+c= 0.

Here ,

• a= (a-b)
• b= (b-c)
• and c = (c-a)

When equation have equal ro ots then,

Discriminant = 0

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

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

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

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

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

$\sf{\implies b + c -2a = 0 }$

Hence proved.

Answer 2

Answer:

Given :

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

To prove :

2a=b+c

Theory :

For a Quadratic equation of the form

ax²+bx+c= 0 , the expression b²-4ac is called the discriminant.

Nature of roots

The roots of a quadratic equation can be of three types.

If D>0, the equation has two distinct real roots.

If D=0, the equation has two equal real roots.

If D<0, the equation has no real roots.

Solution :

We have , (a-b)x²+(b-c)x+(c-a)=0

On comparing with the standard form of Quadratic equation ax²+bx+c= 0.

Here ,

a= (a-b)

b= (b-c)

and c = (c-a)

When equation have equal ro ots then,

Discriminant = 0