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View Full Version : Help with algebra 2 question please


IKoRuPT
Nov 7, 2006, 07:51 AM
On one of my recent test this was one of the problem that I was unable to do. Can anyone help me with it, and show me the steps please? I'm familiar with fractional exponents, but when it came to this problem with a radical exponent I didn't know what to do:

http://img465.imageshack.us/img465/9418/problemei3.png


Any and all help is greatly appreciated. Thanks in advance.

What I tried to do was square both sides, and ended up with 81 = (x + 9)^2. From there I factored it to x^2 + 18x - 6480 = 0. It took some time to find the factors but I got (x + 90)(x - 72), so the answers came out to x= -90 and x = 72. When I check it, I don't get the right answer. What am I doing wrong?

Capuchin
Nov 8, 2006, 03:09 AM
You can't square both sides to get rid of the square root in an exponent.

as a rule (n^x)^y = n^(x*y), so if you square both sides, you get n^2root2, which doesn't help you.

root4 is 2 of course, so square the LHS first

you need to raise both sides to the power of root2, to get 81^root2 = (x+9)^2

81^root2 = 500 almost exactly (I myself am not sure how I would do this without a calculator - I don't know if you had one)

Then just solve as you did.

Hope this helps - let me know how you get along.

IKoRuPT
Nov 8, 2006, 05:55 PM
Thanks for the help.

When I solve 500 = (x+9)^2 I get 500 = x^2 + 18x + 81, and when I set it equal to zero I get x^2 + 18x - 419 = 0. I cannot factor this, so I plugged it into the quadratic equation and got -18 +/- root2000 over 2. When I go on to simplify that, I get (-9 +/- 10root5) - I simplified the radical of 2000 into 20root5, and then factored out the two which I divided. I'm not sure how to check this answer, but I tried solving it using a different method and my check worked.

What I did was instead of immediately reducing radical 4 into 2 and squaring it to get 81, I split it into 9^root2 * root2. Then I multiplied the other side by its reciprocal to get 1, and did the same for this side. I assume that the one of the root2 would then cancel out and I got 9^root2 over (x + 9). I then set this into a proportion, where the other side was 1 over 1, so then I got 9^root2 = x + 9. I subtracted 9 from both sides, and my answer is 9^root2 - 9. Now I didn't check this with a calculator, because I felt it wasn't necessary. I plugged in the value for x and got 9^root4 = (9^root2 - 9 + 9)^root2. Then +9/-9 cancels out and I get 9^root4 = 9^root2^root2, which by your rule would equal 9^root4, and if you calculate that you would get 81 = 81. I know it checks, but can you tell me if I made a mistake, or broke any law in my solving? Thanks.

dmatos
Nov 8, 2006, 10:13 PM
Looks okay to me, but I think you took the long route to get there. When you have an equation like that, you are allowed to raise both sides to the same power. That power can be rational, fractional, a complex number, just about anything. You could have saved yourself some time just by raising both sides of the equation to the power of (1/sqrt(2))

9^sqrt(4) = (x+9)^sqrt(2)
(9^sqrt(4))^(1/sqrt(2)) = ((x+9)^sqrt(2))^(1/sqrt(2))
9^sqrt(2) = x+9
x = 9^sqrt(2) - 9

Capuchin
Nov 9, 2006, 05:32 AM
Both ways are fun ;)

IKoRuPT
Nov 9, 2006, 05:31 PM
Thanks for the help, both of you. :)