Question

Question #5802c

Answer 1

See the answer below...

Explanation:

As I explained before , for the equation `color(red)(ax^2+bx+c=0` , if the roots of the eqn are `color(red)(R_1` and `color(red)(R_1`...
Then `color(darkred)(R_1+R_2=-b/a` and `color(darkred)(R_1cdotR_2=c/a`

For the given equation of your question `R_1=R_2=R` as the roots are same...

So,

`R+R=-(b-c)/(a-b)` `color(white)(mmmmm`; `color(white)(m)` `RcdotR=(c-a)/(a-b)`

`=>2R=(c-b)/(a-b)` `color(white)(mmmmmm)`; `R^2=(c-a)/(a-b)`

From the equations we can write,

`((c-b)/(2(a-b)))^2 =(c-a)/(a-b)`
`=>(c-b)^2/(4(a-b))=c-a`
`=>b^2-2bc+c^2=4(a-b)(c-a)`
`=>b^2-2bc+c^2=4(ac-a^2-bc+ab)`
`=>b^2-2bc+c^2=4ac-4a^2-4bc+4ab`
`=>4a^2+b^2+c^2-4ab+2bc-4ac=0`
`=>(2a-b-c)^2=0`
`=>2a-b-c=0`
`=>2a=b+c` (PROVED)

Hope it helps...
Thank you...

Answer 2

since the roots of the given quadratic equation ,`(a-b).x^2+(b-c)x+(c-a)=0` are same, its discriminant will be zero

So `(b-c)^2-4(a-b)(c-a)=0`

`=>b^2+c^2-2bc-4ac+4bc+4a^2-4ab=0`

`=>b^2+c^2-4ac+2bc+4a^2-4ab=0`

`=>b^2+c^2+2(-2a)c+2bc+(-2a)^2+2(-2a)b=0`

`=>(b+c-2a)^2=0`

`=>2a=b+c`