SOLUTION: I am to solve this problem finding all real or imaginary solutions. I believe I solved it, but I question whether it is correct. The question is: 6x=19x+25/x+1 My work: 6x(x

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Question 37613: I am to solve this problem finding all real or imaginary solutions. I believe I solved it, but I question whether it is correct.
The question is: 6x=19x+25/x+1
My work:
6x(x+1)=6x^2+6x
Then I put it into the equation form of -6x^2-25x+25=0
I used the b^2-4(a)(c) formula which =625-4(-6)(25)
The whole thing turned out to be 625-600=25
Because the outcome is positive, does it not have to have two solutions? I am unsure what the second one is?
Thank you
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
I am to solve this problem finding all real or imaginary solutions. I believe I solved it, but I question whether it is correct.
The question is: 6x=19x+25/x+1
My work:
6x(x+1)=6x^2+6x...OK
Then I put it into the equation form of -6x^2-25x+25=0
NO..YOU MISSED 19X
6X^2+6X=19X+25
6X^2-13X-25=0
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%2813+%2B-+sqrt%28+13%5E2-4%2A6%2A%28-25%29+%29%29%2F%282%2A6%29+
x+=+%2813+%2B-+sqrt%28+169%2B600%29+%29%2F%2812%29+
x+=+%2813+%2B-+sqrt%28+769+%29%29%2F%2812%29+
ARE THE 2 SOLUTIONS.
I used the b^2-4(a)(c) formula which =625-4(-6)(25)
The whole thing turned out to be 625-600=25
Because the outcome is positive, does it not have to have two solutions? I am unsure what the second one is?
Thank you