Question

if the Roots of the quadratic equation(a-b)x2+(b-c)x+(c-a)=0are equal,prove that 2a=b+c.​

Answer 1

Hi,

We know that,

If the quadratic equation=ax²+bx+c=0 whose roots are equal then it's deteminant is equal to zero.

=>(a-b)x²+(b-c)x+(c-a)=0

Deteminant =0

=>(b-c)² -4(a-b)(c-a)=0

=>b²+c²-2bc-4ac+4a²+4bc-4ab=0

=>b²+c²+4a²+4bc-4ac-4ab=0

=>b²+c²(-2a)²+2bc+2c(-2a)+2(-2a)b=0

=>(b+c-2a)²=0

=>b+c-2a=0

Therefore,

b+c=2a

Hence proved.

I hope this helps you. :):):)

If yes then plz mark it as brainliest...

Answer 2

Answer:

Step-by-step explanation:

(b-c)x² + x(c-a) + (a-b) =0

To have real and equal roots, Discriminant, D =0

(c-a)² - 4(a-b)(b-c) =0

⇒c² +a² - 2ac - 4(ab -ac -b²+ bc) =0

⇒c² + a² -2ac - 4ab +4ac +4b² -4bc =0

⇒a² +4b² +c² - 4ab - 4bc +2ac =0

⇒(a -2b +c)² =0

∴(a -2b +c) =0

⇒(a+c) =2b (proved )

a,b and c are in A.P , where b is A.M.