(Square root of 3x+1) =(3+sqaure root of x-4) ???? please help.

Question

im not sure how to find the asymptotes

Answer 1

√(3x+1) = 3 + √(x-4)

 

Square both sides to get:  3x + 1 = 9 + 6√(x-4) + x - 4

 

                                     2x - 4 = 6√(x-4)

 

Square both sides again:  4x² - 16x + 16 = 36(x-4)

 

                                    4x² - 16x + 16 = 36x - 144

 

                                    4x² - 52x + 160 = 0

 

                                      x² - 13x + 40 = 0

 

                                    (x-8)(x-5) = 0

 

                                     x = 8 or x = 5

 

Substituting each of these values into the ORIGINAL equation, we see that they both work.  

 

So, the solutions are 8 and 5.  

 

Answer 2

solving for x...

√(3x+1)=3+√(x-4)

square both sides

3x+1=9+6√(x-4)+x-4

3x+1=5+x+6√(x-4)

3x+1-5-x=6√(x-4)

2x-4=6√(x-4)

square both sides again

4x²-16x+16=36(x-4)

4x²-16x+16=36x-144

4x²-16x+16-36x+144

4x²-52x+160=0

divide both sides by 4

x²-13x+40=0

factor

40=8*5 and 8+5=13

(x-8)(x-5)=0

x-8=0

x=8 is one solution

x-5=0

x=5 is another solution