if x^2+2x(a+b+c)+3(ab+bc+ca)is a perfect square,then prove that,a=b=c

Question

we have to prove that   b^2-4ac=0

Answer 1

x²+2x(a+b+c)+(a+b+c)²=(x+a+b+c)², therefore, if (a+b+c)²=3(ab+bc+ca), then we have a perfect square.

(a+b+c)²=((a+b)²+2c(a+b)+c²=a²+2ab+b²+2ac+2bc+c², that is:

a²+b²+c²+2(ab+bc+ca), which has to equal 3(ab+bc+ca).

Therefore a²+b²+c²=ab+bc+ca. There are three terms on each side so let's equate them:

a²=ab, b²=bc, c²=ac⇒a=b, b=c, c=a, that is, a=b=c (a,b,c≠0).

Are there any other solutions? Suppose a²=bc, b²=ca, c²=ab, then:

a²-b²=bc-ca=c(b-a), (a-b)(a+b)=c(b-a)⇒a=b or c=-(a+b), or a+b+c=0;

b²-c²=(b-c)(b+c)=ca-ab=a(c-b)⇒b=c or a=-(b+c), or a+b+c=0;

c²-a²=(c-a)(c+a)=ab-bc=b(a-c)⇒a=c or b=-(a+c), or a+b+c=0.

There may be another solution: a+b+c=0 or c=-(a+b). If we substitute for c in x²+2x(a+b+c)+(a+b+c)², we get x², which is a perfect square.

Therefore there are two solutions: a=b=c or a+b+c=0.