Question

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The roots of the equation
$\sf\:a(b-c) x^{2} + b(c-a)x + c(a-b) = 0$ are ,

Answer:$\sf\: 1,\frac{c(a-b)}{a(b-c)}$

Answer 1

✌️✌️ hey mate,

(b-c)x²+(c-a)x+(a-b)=0

Comparing with quadratic equation

Ax²+Bx+C=0

A=(b-c),B=c-a,C=a-b

Discriminate when roots are equal

D=B²-4AC=0

D=(c-a)²−4(b-c)(a-b)=0

D=(c²+a²−2ac)-4(ba-ac-b²+bc)=0

D=c²+a²−2ac-4ab+4ac+4b²-4bc=0

c²+a²+2ac-4b(a+c)+4b²=0

(a+c)²-4b(a+c)+4b²=0

[(a+c)-2b]²=0

a+c=2b


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Answer 2

Given ;- Zeroes of Given Equation are equal.

By Discriminant formula;- b^2-4ac=0

{b(c-a)}^2-4{a(b-c)}{c(a-b)}=0

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

b^2c^2+a^2b^2-2ab^2c-4{a^2bc-ab^2c-a^2c^2+abc^2}=0

b^2c^2+a^2b^2-2ab^2c-4a^2bc+4ab^2c+4a^2c^2-4abc^2=0

b^2c^2+a^2b^2+2ab^2c-4a^2bc +4a^2c^2-4abc^2=0

b^2c^2+a^2b^2+4a^2c^2+2ab^2c -4a^2bc-4abc^2=0

Then,,


a^2+b^2+c^2+2ab+2bc+2ac={a+b+c}^2

From above ,,

we Get,,


{bc+ab-2ac}^2=0

bc+ab-2ac=0

b(a+c)=2ac

b=2ac/(a+c)


Hence proved

$thanks \: me \: later$