Question

if the root of the equation,(a-b)xsquare+(b-c)x+(c-a)=0 are equal,prove that 2a=b+c

Answer 1

hope u get the answer

Answer 2

Step-by-step explanation:

$\huge{\boxed{\sf{SOLUTION-}}}$

$\huge{\boxed{\sf{GIVEN-}}}$

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

$\huge{\boxed{\sf{TO PROVE-}}}$

$\huge{\boxed{\sf{2b=a+c}}}$

$\huge{\boxed{\sf{PROOF-}}}$

$\huge{\boxed{\sf{METHOD(1)-}}}$

$\huge{\boxed{\sf{D=0}}}$

${b}^{2} - 4ac = 0 \\ = ) (b - c)^{2} - 4 \times( a - b)(c - a) = 0 \\ = ) {b}^{2} + {c}^{2} - 2bc - 4(ac - bc - {a}^{2} + b) = 0 \\ = ){b}^{2} + {c}^{2} + 4 {a}^{2} + 2bc - 4ac - 4ab = 0 \\ = )((b + c) - 2a)^{2} = 0 \\ = )b + c - 2a = 0 \\ = )b + c = 2a \\ = )2a = b + c$

$\huge{\boxed{\sf{PROVED}}}$

$\huge{\boxed{\sf{METHOD(2)}}}$

$sum \: of \: coefficient \\ = )a - b + b - c + c - a = 0 \\ = ) \alpha = 1 \\ product \: of \: roots( \alpha \beta ) = \frac{c}{a} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = )1 \times 1 = \frac{c - a}{a - b} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = )a - b = c - a \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = )a + a = b + c \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =)2a = b + c$

$\huge{\boxed{\sf{PROVED}}}$