Quadratic

[Richard B]

Find the product of all real values of r for which

has exactly one real solution.

I tried getting rid of the fractions by multipying both sides by 14x, but it didn't really work.

 

[firemath]

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?

 

[Richard B]

no

 

[firemath]

Umm....where are you stuck? Have you been taught the Quadratic formula?

 

[Richard B]

7=2rx-2x^2
7-2rx+2x^2=0
2x^2-2rx+7=0

 

[MarkFL]

7=2rx-2x^2
7-2rx+2x^2=0
2x^2-2rx+7=0

What can we say about the disciminant when there is only 1 real (repeated) root?

 

[Steven G]

7=2rx-2x^2
7-2rx+2x^2=0
2x^2-2rx+7=0

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???

 

[Otis]

… I tried getting rid of the fractions by multipying both sides by 14x, but it didn't really work.

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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