NEED HELP WITH ALGEBRA!!! (please provide steps/explanations!)?

Question

Equation #1 =
-u^2 = 6 + 7u

Equation #2=
-((2x+1)/x) = 6+7( sqrt ((2x+1)/x))

Solve for x.
(notice that equation 2 is equation 1 when sqrt ((2x+1)/x) is substituted for u)
(i got that "u"= -6,-7 and then i set it equal to sqrt ((2x+1)/x) but it doesn't work out... please help!)

i got -6 and -1 for u (sry typo in the previous detail), but it still doesn't work! arg

when making sqrt ((2x+1)/x) equal to -1, i get -1 for x and then when i substitute -1 for x in equation2, i get -1=13 (not right!)
and when i do the same to -6, it also isn't right! arg. need help!

Answer 1

-u^2 = 6 + 7u
u^2+7u+6 = 0
(u+1)(u+6)=0
u = -1 or -6

-((2x+1)/x) = 6+7( sqrt ((2x+1)/x))
-(2x+1)/x -6= 7sqrt((2x+1)/x)
-(2x+1)/x-6x/x = 7sqrt((2x+1)/x)
(-8x -1)/x = 7sqrt((2x+1)/x)
64x^2+16x +1 = 49((2x+1)/x)
64x^2+16x+1 = (98x +49)/x
64x^3 +16x^2 +x =98x +49
64x^3 +16x^2 -97x -49 = 0 <-- Eq 1
One root of this equation is -1
So divide Eq1 by x+1 and solve the resulting quadratic equation for the other two roots.

Answer 2

u²+7u+6=0
x= -6, -1 so your first problem was your second root wasn't right

The second equation has no solution. The restrictions on x to make the sqrt possible will render the right side of the equation + and the left side negative.

Answer 3

your method is right but ur mistake is in the solution of "u".

it forms the eqn u^2+7u+6=0 . so u= -6,-1

now solve it ...

Answer 4

sorry i really dont understand algerbra myself it confuses me why dont you look up on the internet "how to do algerbra ?" you might get some really useful answers with the steps and explanations you were looking for !! anyway sorry but good luck !! x