Question

If the roots of the equation (a-b)x² + (b-c)x+
"(c-a) = 0 arě equal', then prove that 2a=b+c
​

Answer 1

Answer:

$d = b { }^{2} - 4ac \\ (b - c) {}^{2} - 4( a-b )( c-a ) = 0\\ b {}^{2} + c { }^{2} - 2bc - 4ac+ 4a {}^{2} - 4ab + 4bc = 0 \\ we \: can \: also \: write \: it \: as \\ 4a {}^{2} + b {}^{2} + c {}^{2} - 4ab + 2bc - 4ac = 0 \\ (2a {}^{2} ) + ( - b {}^{2} ) + ( - c {}^{2} ) + 2(2a)( - b) + 2( - b)( - c) + 2(2a)( - c) = 0\\ (2a - b - c) {}^{2} = 0 \\ 2 a- b - c = 0 \\ it \: means \\ 2a = b + c$

Answer 2

plz refer to this attachment