SOLUTION: I have a problem that are driving me crazy. The problem has a quadratic solution, but for the life of me I can't figure it out. Please help! 5(3x-4)^6 + 7(3x-4)^3 -23 =0 I se

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I have a problem that are driving me crazy. The problem has a quadratic solution, but for the life of me I can't figure it out. Please help! 5(3x-4)^6 + 7(3x-4)^3 -23 =0 I se      Log On


   

Question 353887: I have a problem that are driving me crazy. The problem has a quadratic solution, but for the life of me I can't figure it out. Please help!
5(3x-4)^6 + 7(3x-4)^3 -23 =0
I set it up by substitution / y= (3x-4)^3
This gives me the form 5y^2 + 7y -13 =0
My problem is not knowing how to "unfold" the problem after I've done the alternate equation.
Thanks for your help
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assuming you meant to say 5y%5E2%2B7y-23=0


First, let's solve 5y%5E2%2B7y-23=0


5y%5E2%2B7y-23=0 Start with the given equation.


Notice that the quadratic 5y%5E2%2B7y-23 is in the form of Ay%5E2%2BBy%2BC where A=5, B=7, and C=-23


Let's use the quadratic formula to solve for "y":


y+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


y+=+%28-%287%29+%2B-+sqrt%28+%287%29%5E2-4%285%29%28-23%29+%29%29%2F%282%285%29%29 Plug in A=5, B=7, and C=-23


y+=+%28-7+%2B-+sqrt%28+49-4%285%29%28-23%29+%29%29%2F%282%285%29%29 Square 7 to get 49.


y+=+%28-7+%2B-+sqrt%28+49--460+%29%29%2F%282%285%29%29 Multiply 4%285%29%28-23%29 to get -460


y+=+%28-7+%2B-+sqrt%28+49%2B460+%29%29%2F%282%285%29%29 Rewrite sqrt%2849--460%29 as sqrt%2849%2B460%29


y+=+%28-7+%2B-+sqrt%28+509+%29%29%2F%282%285%29%29 Add 49 to 460 to get 509


y+=+%28-7+%2B-+sqrt%28+509+%29%29%2F%2810%29 Multiply 2 and 5 to get 10.


y+=+%28-7%2Bsqrt%28509%29%29%2F%2810%29 or y+=+%28-7-sqrt%28509%29%29%2F%2810%29 Break up the expression.


So the solutions in terms of 'y' are y+=+%28-7%2Bsqrt%28509%29%29%2F%2810%29 or y+=+%28-7-sqrt%28509%29%29%2F%2810%29



Now recall that you let y=%283x-4%29%5E3. So this means that


%283x-4%29%5E3+=+%28-7%2Bsqrt%28509%29%29%2F%2810%29 or %283x-4%29%5E3+=+%28-7-sqrt%28509%29%29%2F%2810%29



Start with the given pair of equations

%283x-4%29%5E3+=+%28-7%2Bsqrt%28509%29%29%2F%2810%29 or %283x-4%29%5E3+=+%28-7-sqrt%28509%29%29%2F%2810%29


Take the cube root of both sides to get:

3x-4+=root%283%2C+%28-7%2Bsqrt%28509%29%29%2F%2810%29%29 or 3x-4+=root%283%2C+%28-7-sqrt%28509%29%29%2F%2810%29%29

Note: I'm only taking the principal cube root


Add 4 to both sides to get:
3x+=root%283%2C+%28-7%2Bsqrt%28509%29%29%2F%2810%29%29%2B4 or 3x=root%283%2C+%28-7-sqrt%28509%29%29%2F%2810%29%29%2B4


Divide both sides by 3 to get:
x+=%28root%283%2C+%28-7%2Bsqrt%28509%29%29%2F%2810%29%29%2B4%29%2F3 or x=%28root%283%2C+%28-7-sqrt%28509%29%29%2F%2810%29%29%2B4%29%2F3


So the two real solutions are


x+=%28root%283%2C+%28-7%2Bsqrt%28509%29%29%2F%2810%29%29%2B4%29%2F3 or x=%28root%283%2C+%28-7-sqrt%28509%29%29%2F%2810%29%29%2B4%29%2F3