2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ At least I would try to multiply out the parentheses and divide by $abc$, to see what happens. $\endgroup$ – Arthur Jul 18 '15 at 12:45
3
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ $\quad$ thanks. $\endgroup$ – R K Jul 18 '15 at 13:02
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Pls explain why “solving this equation, we get $1=\frac{c(a-b)}{a(b-c)}$”. $\endgroup$ – Mick Jul 18 '15 at 13:10
  • $\begingroup$ i have solved the equation for $x$ and i get the two solutions $x_1=1$ and $x_2=\frac{c(a-b)}{a(b-c)}$ $\endgroup$ – Dr. Sonnhard Graubner Jul 18 '15 at 13:11
  • $\begingroup$ the solution above is too complicated from mathlove $\endgroup$ – Dr. Sonnhard Graubner Jul 18 '15 at 13:12
  • 1
    $\begingroup$ 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. $\endgroup$ – Mick Jul 18 '15 at 14:02
  • $\begingroup$ for $x=1$ we have $ab-ac+bc-ab+ac-bc=0$ $\endgroup$ – Dr. Sonnhard Graubner Jul 18 '15 at 14:22
1
$\begingroup$

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 $c$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.