Question

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

Answer 1

Answer:

Step-by-step explanation:

The equation is

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

In order to have equal roots, the discriminant i.e, $b^{2} -4ac$ should be equal to 0

Hence in the above equation

$(c-a)^{2} -4(b-c)(a-b)=0\\else,c^2+a^2-2ac-(4b-4c)(a-b)=0\\else,c^2+a^2-2ac-[4ab-4b^2-4ac+4bc]=0\\else,c^2+a^2-2ac-4ab+4b^2+4ac-4bc=0\\else,b^2+c^2+(-2a)^2+2bc+2c(-2a)+2(-2a)b=0\\else,(b+c-2a)^2=0\\else,b+c-2a=0\\<em><strong><u>else,2a=b+c</u></strong></em>\\<strong>Hence, proved</strong>$

Answer 2

plz refer to this attachment