An algebra problem by Wes Waruwu

$f=(b-c)x^{ 2 }+(c-a)x+(a-b)\\ f(1)=0\quad and\quad f\quad has\quad two\quad same\quad solution\quad =>\quad f={ (x-1) }^{ 2 }\\ So\quad (b-c)x^{ 2 }+(c-a)x+(a-b)={ x }^{ 2 }-2x+1\quad =>\quad b-c=a-b=1\quad <=>\quad b-a=c-b\\ Therefore\quad the\quad value\quad of\quad the\quad expression\quad is\quad above\quad is\quad 1.$

A quadratic equation has an equal root means it is a perfect square expression.

Like $x ^ { 2 } + 2x + 1 = ( x + 1 ) ^ { 2 }$ .

If $ax ^ { 2 } + bx + c$ is a perfect square expression, then $\displaystyle c = ( \frac { 2 } { b } ) ^ { 2 }$ .

Then above equation, $( b - c ) x ^ { 2 } + ( c - a ) x + ( -b + a )$ is a perfect square expression.

So, $\displaystyle -b + a = - ( b - a ) = ( \frac { c - a } { 2 } ) ^ { 2 }$ .

So, if $( b - c ) x ^ { 2 } + ( c - a ) x + ( a - b ) = 0$ is $x ^ { 2 } + 2x + 1 = 0$ , then we get $b - c = 1$ , $c - a = 2$ , and $a - b = 1$ .

We only get no solutions to the simultaneous equation.

But, $\displaystyle \frac { b - a } { c - b } = 1$ , because $b - a = c - b$ from $-b + a = - ( b - a )$ .

So, $\displaystyle \frac { b - a } { c - b } = \boxed { 1 }$ .