Question

3. If the roots of the equation (a-b)x2 + (b-c)X + (c-a)=0 are equal,
Prove that b + c = 2a​

Answer 1

$\large{\underline{\boxed{\green{ANSWER}}}}$

GIVEN EQUATION:-

${(a - b)}x^{2} + (b - c)x + (c - a) = 0$

Also, roots of the equation are equal.

SOLUTION:-

We know that when roots of a quadratic equation are equal , then the discriminant of the equation is equal to 0

$\large{\underline{OR}}$

D = b² - 4ac = 0

here, a = (a - b) , b = (b - c) , c =(c - a)

NOW, putting the values in the above equation , we get

D = (b - c)² - 4(a - b)(c - a) = 0

=> b² + c² - 2bc - 4(ac - a² - bc + ba) = 0

=> b² + c² + 4a² + 2bc - 4ac - 4ba = 0

=> (b + c - 2a)² = 0

=> b + c - 2a = 0

=> b + c = 2a

Answer 2

plz refer to this attachment