Question

if a:b:c=1:m:n, where a,b,c are real numbers, and the expressions $x^2+2(a+b+c)x+3(bc+ca+ab)$ is a perfect square, then the value of m+n is

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Answer 1

The given expression is a perfect square
$\Rightarrow$Discriminant=0
$\Rightarrow \left(2\right(a+b+c){)}^2-4(3\left)\right(abc+ca+ab)=0$
$\Rightarrow (a+b+c){)}^2-3bc-3ca-3ab=0$
$\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca-3bc-3ca-3ab$
$\Rightarrow a^2+b^2+c^2-ab-ca-bc=0$
$\Rightarrow$$\frac{1}{2}$$\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]=0$
This is possible only when a=b=c
$\therefore a:b:c=1:1:1$
$=1:m:n\left(given\right)$
or m=1, n=1 $\Rightarrow$ m+n=2