Quite a few questions tonight...

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spear33

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Jan 30, 2006
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Question number 1

sqrt (4r + 13) = 2r - 1

I raise each side to the 2nd power and get
4r + 13 = (2r)^2 + 1

Assuming this is right, I simplify it to
(2r)^2 - 4r - 12 = 0

Still assuming I didnt screw up, I'm not sure what to do next or if this is even solvable.

Thanks in advance for any help. I'll post some of the other questions after this one is answered and if I can't figure them out.
 
spear33 said:
Question number 1

sqrt (4r + 13) = 2r - 1

I raise each side to the 2nd power and get
4r + 13 = (2r)^2 + 1
Click to expand...
Raising both sides to the 2nd power is:
\(\displaystyle \mbox{ \left(\sqrt{4r + 13}\right)^2 = \left(2r - 1\right)^2}\)

You have the left-hand side fine, but notice the RHS is a quadratic - we can't simply square each term individually.

The next step is to expand the RHS, and simplify the equation to a quadratic.

Solve for r, and then check if each solution works in the original equation (as squaring a square root can result in extraneous solutions).
 
Ok, when I simplifiy the RHS of the equation I get
(4r)^2 - 4r + 1

I tried solving it like a quadratic, but I cant seem to get that to work. Am I supposed to set that equal to the 4r + 13 and solve or am I just not seeing the numbers I'm supposed to put in?
 
Ok. So we agree that we square both sides to have

\(\displaystyle \mbox{ 4r + 13 = (2r - 1)^2}\)

And the right-hand side expands to: \(\displaystyle \mbox{ 4r^2 - 4r + 1}\) (you basically had this, but be careful that \(\displaystyle \mbox{(2r)^2 = 4r^2}\), not \(\displaystyle \mbox{ (4r)^2}\).)

So our equation is now

\(\displaystyle \mbox{ 4r + 13 = 4r^2 - 4r + 1}\)

Now you can put this into the form \(\displaystyle \mbox{ar^2 + br + c = 0}\) which you can then factorise (pulling out a factor of 4 will help), and solve for r, not forgetting to check each solution.
 
I used the quadratic formula and got it reduced to
1 +- 2, I plugged in 3 and it worked.... So I'm hoping its 3.

Question 2

(-3k / 2) + (9k - 5 / 3) = 11k + 8 / 6k [All the "/"s are fractions, not sure if theres a better way to type them out]

I have no idea if any of my work is right or not, but I'll show you what I did anyways.

I multiplied each side by 6k and when its simplified its
-9k - 18k^2 + 54k^2 - 10k = 66k^2 + 48k

Even if it is somehow right, I'm still stuck.
 
You're spot on for the earlier question.

For this one, I suspect it's (-3k)/2 + (9k - 5)/3 = (11k + 8)/(6k) but could you clarify.
 
Cheers. Multiplying both sides by 6k is a good idea.

\(\displaystyle \mbox{ 6k\left(-\frac{3k}{2} + \frac{9k - 5}{3}\right) = 6k\left(\frac{11k + 8}{6k}\right)}\)

Now very carefully distribute and simplify the LHS; the RHS is 11k + 8.
 
I'm distributing wrong or something, because I only get what I showed you in the post above. I do see what I did wrong on the right side however.
 
Actually I think I see my problem.

-9k^2 + 18k^2 - 10k = 11k + 8
and then solve for that I'm guessing.
 
Yeah Im really confusing myself tonight, I wrote -10 instead of -8 in the Quad. Eq. I guess Ill try it again. But isnt it still a negative 9k^2?
 
Ah I see, Yeah I'm really going to fast tonight... I got 7 +- 3 and havent plugged any of it in yet to test, but does that seem right?
 
\(\displaystyle \mbox{ 9k^2 - 21k - 8 = 0}\)

\(\displaystyle \mbox{ k = \frac{-(-21) \pm \sqrt{(-21)^2 - (4)(9)(-8)}}{18} = \frac{21 \pm \sqrt{729}}{18} = \frac{21 \pm 27}{18} = \frac{8}{3}, -\frac{1}{3}}\)
 
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