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

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The discussion revolves around the equation (b - c) / (c - a) and its implications in solving a quadratic equation. Participants explore various algebraic manipulations to express relationships between the variables a, b, and c, particularly focusing on the condition that c is the midpoint between a and b. They suggest substituting expressions to simplify the problem and emphasize that the quadratic has a repeated root, leading to the conclusion that (b - c) equals (c - a). Overall, the conversation highlights different strategies for approaching the problem while confirming that c - a = b - c is a straightforward solution.
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Homework Statement
Please see below
Relevant Equations
Not sure
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$$(b-a)^2-4(b-c)(c-a)=0$$
$$b^2-2ab+a^2=4(bc-ab-c^2+ac)$$
$$b^2-2ab+a^2+4ab=4bc-4c^2+4ac$$
$$(b+a)^2-4ac=4c(b-c)$$
$$b-c=\frac{(b+a)^2-4ac}{4c}$$

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

Thanks
 
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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*}
 
Hint.
##b-a=(b-c)+(c-a)##
 
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Likes songoku, SammyS, Frabjous and 1 other person
Or maybe dividing through by ##(c-a)^2##, to get you a bit closer?
 
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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.
 
You could arbitrarily set ##a=0, b=1## and solve (the resulting quadratic) for ##c##.

Inelegant - but practical in an exam' situation.
 
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##.
 
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Hill said:
Hint.
##b-a=(b-c)+(c-a)##
This will give a solution in a few easy steps.
 
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(b-a)^2 - 4(b-c)(c-a) = 0 is equivalent to the statement that the quadratic <br /> x^2 \pm (b - a)x + (b-c)(c-a) = 0 has a repeated root. So we have \begin{split}<br /> 2x &amp;= (b - c) + (c - a) \\<br /> x^2 &amp;= (b - c)(c - a) \end{split} which together imply <br /> ((b- c) - (c - a))^2 = 0 so that (b - c) = (c - a).
 
  • #10
pasmith said:
(b-a)^2 - 4(b-c)(c-a) = 0 is equivalent to the statement that the quadratic <br /> x^2 \pm (b - a)x + (b-c)(c-a) = 0 has a repeated root. So we have \begin{split}<br /> 2x &amp;= (b - c) + (c - a) \\<br /> x^2 &amp;= (b - c)(c - a) \end{split} which together imply <br /> ((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##.
 
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  • #11
Hill said:
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!
 
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  • #12
I understand.

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