Find the relation between $a,b$ and $c$ in quadratic equation.

If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0,\ \ \{a,b,c,x\}\in \mathbb{R}$ are equal, then $a,b,c$ are in

Options

$a.)\ AP\\ b.)\ GP\\ \color{green}{c.)\ HP}\\ d.)\ \text{cannot be determined}\\$

by using discriminant property

$[b(c-a)]^2-4ac(b-c)(a-b)=0$

I cannot reach any conclusion and is also cumbersome , though I don't know how wolfram reached this

$[b(c-a)]^2-4ac(b-c)(a-b)=0\ \Longleftrightarrow \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$

I look for a short and simple way .

I have studied maths up to $12$th grade.

• At least I would try to multiply out the parentheses and divide by $abc$, to see what happens. – Arthur Jul 18 '15 at 12:45

\begin{align}&b^2(c-a)^2=4ac(b-c)(a-b)\\&\Rightarrow b^2(c^2-2ac+a^2)=4ac(ab-b^2-ca+bc)\\&\Rightarrow b^2c^2-2ab^2c+a^2b^2=4a^2bc-4ab^2c-4a^2c^2+4abc^2\\&\Rightarrow b^2c^2+a^2b^2=4a^2bc-2ab^2c-4a^2c^2+4abc^2\end{align}Dividing the both sides by $a^2b^2c^2$, we have\begin{align}&\Rightarrow \frac{1}{a^2}+\frac{1}{c^2}=\frac{4}{bc}-\frac{2}{ac}-\frac{4}{b^2}+\frac{4}{ab}\\&\Rightarrow \frac{1}{a^2}-\frac{4}{ab}+\frac{4}{b^2}+\frac{1}{c^2}+\frac{2}{ac}-\frac{4}{bc}=0\\&\Rightarrow \left(\frac 1a-\frac 2b\right)^2+\frac 2c\left(\frac 1a-\frac 2b\right)+\frac{1}{c^2}=0\\&\Rightarrow \left(\frac 1a-\frac 2b+\frac 1c\right)^2=0\\&\Rightarrow \frac 1a-\frac 2b+\frac 1c=0\end{align}

• $\quad$ thanks. – R K Jul 18 '15 at 13:02

solving this equation we get $$1=\frac{c(a-b)}{a(b-c)}$$ and this is equivalent to $$2ac=b(a+c)$$

• Pls explain why “solving this equation, we get $1=\frac{c(a-b)}{a(b-c)}$”. – Mick Jul 18 '15 at 13:10
• i have solved the equation for $x$ and i get the two solutions $x_1=1$ and $x_2=\frac{c(a-b)}{a(b-c)}$ – Dr. Sonnhard Graubner Jul 18 '15 at 13:11
• the solution above is too complicated from mathlove – Dr. Sonnhard Graubner Jul 18 '15 at 13:12
• Instead of using a complicated process to find the solution, one can use (1) Apply factor theorem to find $x_1 = 1$ as a root; then (2) Use product of root to find $1. x_2 = \frac{c(a-b)}{a(b-c)}$”. Of course, this method depends on whether we can “recognize” 1 is root or not. – Mick Jul 18 '15 at 14:02
• for $x=1$ we have $ab-ac+bc-ab+ac-bc=0$ – Dr. Sonnhard Graubner Jul 18 '15 at 14:22

Hint

Consider the function $$f(x)=a(b-c)x^2+b(c-a)x+c(a-b)$$ It is clear that $x=1$ is a solution ($f(1)=0$).

Now, since you want the two roots of $f(x)=0$ to be equal, that is to say $x_1=x_2=1$, we then have $$x_1+x_2=2=-\frac{b(c-a)}{a(b-c)}$$ $$x_1 \times x_2=1=\frac{c(a-b)}{a(b-c)}$$ From the first equation you can extract $b$ as a function of $a$ and $c$ to get $$b=\frac{2 a c}{a+c}$$ Plugging into the second equation, this gives the stupid $1=1$ result.

From here you arrive to the result $$\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$$ that is to say $$b=\frac 2 {\frac 1a+\frac 1c}$$ so $b$ is the harmonic mean of $a$ and $c$.