The value of (b - c) / (c - a)

[songoku]

Homework Statement

Please see below

Relevant Equations

Not sure

$$(b-a)^2-4(b-c)(c-a)=0$$
$$b^2-2ab+a^2=4(bc-ab-c^2+ac)$$
$$b^2-2ab+a^2+4ab=4bc-4c^2+4ac$$
$$(b+a)^2-4ac=4c(b-c)$$
$$b-c=\frac{(b+a)^2-4ac}{4c}$$

I don't know how to continue and not even sure what I did is useful.

Thanks

 

[fresh_42]

I would start from behind. But I have no idea whether this is any better. We want to find a number $x$ such that
\begin{align*}
\dfrac{b-c}{c-a}=x &\Longrightarrow (b-c)=x\cdot (c-a) \\
&\Longrightarrow 4(b-c)^2=4x(b-c)(c-a)=x(b-a)^2
&\Longrightarrow \ldots
\end{align*}

 

[Hill]

Hint.
$b-a=(b-c)+(c-a)$

 

[WWGD]

Or maybe dividing through by $(c-a)^2$, to get you a bit closer?

 

[PeroK]

This is the problem with multiple choice. Note that the equality holds if $b - a$ is twice $b - c$ and $c - a$. I.e. $c$ is halfway between $a$ and $b$. A simple example is $a = 0, b = 2, c = 1$.

You don't actually have to show the general case to get a multiple choice question right.

 

[Steve4Physics]

You could arbitrarily set $a=0, b=1$ and solve (the resulting quadratic) for $c$.

Inelegant - but practical in an exam' situation.

 

[jack action]

By definition, you already have:
$$\frac{b-c}{c-a}=x$$
Where $x$ is one of the answers.

You then need to transform the other equation into a function of $x$ as well. (@WWGD has given the best hint).

In each of these equations, you will be able to isolate one variable and then get 2 other equations containing $x$, both - of course - also being equal. Combining these 2 equations should give you a nice equation that should depend exclusively on $x$.

 

[SammyS]

Hint.
$b-a=(b-c)+(c-a)$

This will give a solution in a few easy steps.

 

[pasmith]

$(b-a)^2 - 4(b-c)(c-a) = 0$ is equivalent to the statement that the quadratic $$x^2 \pm (b - a)x + (b-c)(c-a) = 0$$ has a repeated root. So we have $$\begin{split}<br /> 2x &= (b - c) + (c - a) \\<br /> x^2 &= (b - c)(c - a) \end{split}$$ which together imply $$((b- c) - (c - a))^2 = 0$$ so that $(b - c) = (c - a)$.

 

[Hill]

$(b-a)^2 - 4(b-c)(c-a) = 0$ is equivalent to the statement that the quadratic $$x^2 \pm (b - a)x + (b-c)(c-a) = 0$$ has a repeated root. So we have $$\begin{split}<br /> 2x &= (b - c) + (c - a) \\<br /> x^2 &= (b - c)(c - a) \end{split}$$ which together imply $$((b- c) - (c - a))^2 = 0$$ so that $(b - c) = (c - a)$.

Even shorter. Substituting $(b-c)+(c-a)$ for $(b-a)$ leads straight to $((b-c)-(c-a))^2=0$.

 

[PeroK]

Even shorter. Substituting $(b-c)+(c-a)$ for $(b-a)$ leads straight to $((b-c)-(c-a))^2=0$.

That said, noting that (by inspection!), $c - a = b - c$ solves the equation is simplest of all!

 

[songoku]

I understand.

Thank you very much for all the help and explanation fresh_42, Hill, WWGD, PeroK, Steve4Physics, jack action, SammyS, pasmith