Question

if the roots of theeqiation (b-c)x^2 (c-a)x+(a-b)=0 are equal then prove 2b=a+c​

Answer 1

Correction:

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

Answer:

Given that the quadratic equation (b-c)x²+(c-a)x+(a-b)=0 has real and equal roots.

If a quadratic equation has equal roots,it implies that the discriminant of that equation is zero. →D=0

To prove:

2b=c+a

On comparing with general form of a quadratic equation,

Ax²+Bx+C=0

Here,

A=(b-c),B=(c-a) and C=(a-b)

Now,

D=0

→B²-4AC=0

→(c-a)²-4(b-c)(a-b)=0

→(a²+c²-2ac)-4(ab-ac+b²+bc)=0

→a²-4b²+c²-2ac+4ac-4ab-4bc=0

→a²-4b²+c²+2ac-4ab-4bc=0

•Of the form (p+q+r)²=p²+q²+r²+2pq+2qr+2pr [p=a,b= -2b and r=c]

→(a)²+(-2b)²+(c)²+2(a)(c)+2(-2b)(a)+2(-2b)(c)=0

→(a-2b+c)²=0

→a-2b+c=0

→2b=a+c

Henceforth proved that 2b=a+c

Answer 2

QUESTION:

if the roots of theeqiation (b-c)x^2+ (c-a)x+(a-b)=0 are equal then prove 2b=a+c

Answer:

Given,

Quadtatic equation (b-c)x² +(c-a)x+(a-b) = 0 has real and equal roots.

Then,

Discriminant(D) = 0

As we know,

$\boxed{<strong>D</strong><strong> </strong>= {b}^{2}-4ac}$

here,

a = (b-c)

b = (c-a)

c = (a-b)

By sub the values,

(c-a)² -4(b-c)(a-b) = 0

c²+a²-2ac -4(ab-b²-ac+bc) = 0

c²+a²-2ac-4ab+4b²+4ac-4bc = 0

a²+4b²+c²-4ab-4bc+2ac = 0

Above equation is in the form-

x²+y²+z²+2xy+2yz+2zx = (x+y+z)²

Here,

x = a

y = -2b

z = c

By sub. the values,

[a+(-2b)+c]² = 0

a+(-2b)+c = √0

a-2b+c = 0

a+c = 2b

2b = a+c

Hence proved!