# if $a(b-c)x^2+b(c-a)x+c(a-b)=0$ has repeated roots prove…

if the equation $$a(b-c)x^2+b(c-a)x+c(a-b)=0$$ has repeated roots prove that $${1\over a}, {1\over b},{1\over c}$$ are in Arithmetic Progression

• What is A.P. ?? – Brevan Ellefsen Oct 31 '15 at 5:58
• If a quadratic has a repeated root, what is its discriminant? – Empy2 Oct 31 '15 at 6:00
• Sorry A.P is Arithmetic Progression – user283172 Oct 31 '15 at 6:01
• I tried the discriminant method.. i got stuck.. there are 3 variables? – user283172 Oct 31 '15 at 6:02
• First thing is I'd try to figure out what the repeated root is and see if that gives me any information. – fleablood Oct 31 '15 at 6:02

With $a,b,c\ne 0$, observe that $1/a,1/b,1/c$ is an Arithmetic Progression iff $$K=0\text {, where}$$ $$K=a b+b c-2a c.$$ The condition that the quadratic has a repeated root is that the discriminant is zero. We have $$0=b^2(c-a)^2-4a c(a-b)(b-c)\iff$$ $$0=b^2c^2+b^2a^2-2b^2a c-4a c(-b^2+b c+b a-ac)\iff$$ $$0=b^2c^2+b^2a^2+2a c b^2-4a c(b c+b a-a c)\iff$$ $$0= b^2c^2+b^2a^2+2a c b^2-4 a c(b c+b a-2a c)-4a^2c^2\iff$$ $$0=b^2c^2+b^2a^2+2a c b^2-4a c K-4a^2c^2\iff$$ $$0=(b c+b a)^2-4a c K-4a^2c^2\iff$$ $$0=(b c+b a)^2 -(2a c)^2-4a c K\iff$$ $$0=(b c+b a-2a c)(b c+b a+2a c)-4a c K$$ $$0=K(b c+b a+2a c)-4a c K\iff$$ $$0=K(b c+b a-2a c)\iff$$ $$0=K^2.$$

clearly 1 is a root of the given equation. since it is of second degree and has repeated roots the the other root is also 1. we know that sum of roots is b(a-c)/(a(b-c))=2. manipulating in proper form we will get a, b, c are in H.P

• I don't understand why it is "1"? – user283172 Oct 31 '15 at 6:10
• the number 1 becuase if x=1 the equation result is 0. Hence a root – mea43 Oct 31 '15 at 6:11
• Awesome dude! got it thanks ! – user283172 Oct 31 '15 at 6:13
• Nice. Maybe a little easier, product of roots is $1$, so $a(b-c)=c(a-b)$, now divide by $abc$. – André Nicolas Oct 31 '15 at 6:15

1 is a root so, comparing with $x^2-2*x+1=0$ will directly prove that a,b.c are in HP.