Quadratic

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Richard B

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Feb 7, 2020
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Find the product of all real values of r for which
$\frac{1}{2x}=\frac{r-x}{7}$
has exactly one real solution.

I tried getting rid of the fractions by multipying both sides by 14x, but it didn't really work.
 
Last edited:
You don't need to multiply both sides by 14x.
Here is what you need to do:

[MATH]\frac {1}{2x} = \frac {r-x}{7}[/MATH]
Then, multiplying each side by the other's denominator:

[MATH]7=2x(r-x)[/MATH]
[MATH]7=2rx-2x^2[/MATH]
Can you go on from here?
 
Umm....where are you stuck? Have you been taught the Quadratic formula?
 
Richard B said:
7=2rx-2x^2
7-2rx+2x^2=0
2x^2-2rx+7=0
Click to expand...
2x^2-2rx+7=0 is in the form of ax^2+bx+c=0. So a=2, b=-2r and c=7. These are the values you need to use in the quadratic equation. As MarkFL asked, when does the quadratic formula yield one solution???
 
Richard B said:
… I tried getting rid of the fractions by multipying both sides by 14x, but it didn't really work.
Click to expand...
Hi Richard. Multiplying each side by 14x is a valid way to start. Those multiplications will clear the fractions, so you must have done something wrong.

?\(\;\) Were you to have posted your steps, we could have (probably) found the mistake(s) for you.


\(\displaystyle \frac{14x}{1} \cdot \frac{1}{2x} = \frac{14x}{1} \cdot \frac{r - x}{7}\)

Cancel common factors, before multiplying.

\(\displaystyle \frac{\cancel{14}^{7}\cancel{x}}{1} \cdot \frac{1}{\cancel{2}_{1}\cancel{x}} = \frac{\cancel{14}^{2} x}{1} \cdot \frac{r - x}{\cancel{7}_{1}}\)

Now do the multiplications. On each side, we write numerator×numerator over denominator×denominator.

\(\displaystyle \dfrac{7 \cdot 1}{1 \cdot 1} = \dfrac{2x \cdot (r - x)}{1 \cdot 1}\\
\;\\
7 = 2x(r - x)\)
As you can see, that last equation is the same result we got by starting with cross-multiplication.

If you would like more help with this exercise, please show your latest work. Cheers!

?
 
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