SOLUTION: Please help me solve this equation? {{{sqrt(5x-1)-1=sqrt(x+2)}}} I have tried: {{{(sqrt(5x-1)-1)^2=sqrt((x+2)^2)}}} {{{5x-1-2sqrt(5x-1)+1=x+2}}}

Algebra ->  Radicals -> SOLUTION: Please help me solve this equation? {{{sqrt(5x-1)-1=sqrt(x+2)}}} I have tried: {{{(sqrt(5x-1)-1)^2=sqrt((x+2)^2)}}} {{{5x-1-2sqrt(5x-1)+1=x+2}}}       Log On


   

Question 105233: Please help me solve this equation?
sqrt%285x-1%29-1=sqrt%28x%2B2%29
I have tried: %28sqrt%285x-1%29-1%29%5E2=sqrt%28%28x%2B2%29%5E2%29
5x-1-2sqrt%285x-1%29%2B1=x%2B2
-5x+1 -1 = -5x
________________________
-2sqrt%285x-1%29=-4x=2
But from there I am stuck, the only thing I know is that x=2, but the teacher asks for the procedure...hope you can help me...thank you!!!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%285x-1%29-1=sqrt%28x%2B2%29 Start with the given equation

%28sqrt%285x-1%29-1%29%5E2=sqrt%28%28x%2B2%29%5E2%29 Square both sides


5x-1-2sqrt%285x-1%29%2B1=x%2B2 Foil the left side


cross%285x-5x%29-1-2sqrt%285x-1%29%2B1=x%2B2-5x Subtract 5x from both sides


-2sqrt%285x-1%29=-4x%2B2 Combine like terms on both sides


cross%28-2%2F-2%29sqrt%285x-1%29=%28-4x%2B2%29%2F-2 Divide both sides by -2


sqrt%285x-1%29=2x-1 Divide



%28sqrt%285x-1%29%29%5E2=%282x-1%29%5E2 Square both sides again (to get rid of the square root)


5x-1=4x%5E2-4x%2B1 Foil the right side


0=4x%5E2-4x%2B1-5x%2B1 Get all terms to one side


0=4x%5E2-9x%2B2 Combine like terms


Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for x:


Starting with the general quadratic


ax%5E2%2Bbx%2Bc=0


the general solution using the quadratic equation is:


x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve 4%2Ax%5E2-9%2Ax%2B2=0 ( notice a=4, b=-9, and c=2)





x+=+%28--9+%2B-+sqrt%28+%28-9%29%5E2-4%2A4%2A2+%29%29%2F%282%2A4%29 Plug in a=4, b=-9, and c=2




x+=+%289+%2B-+sqrt%28+%28-9%29%5E2-4%2A4%2A2+%29%29%2F%282%2A4%29 Negate -9 to get 9




x+=+%289+%2B-+sqrt%28+81-4%2A4%2A2+%29%29%2F%282%2A4%29 Square -9 to get 81 (note: remember when you square -9, you must square the negative as well. This is because %28-9%29%5E2=-9%2A-9=81.)




x+=+%289+%2B-+sqrt%28+81%2B-32+%29%29%2F%282%2A4%29 Multiply -4%2A2%2A4 to get -32




x+=+%289+%2B-+sqrt%28+49+%29%29%2F%282%2A4%29 Combine like terms in the radicand (everything under the square root)




x+=+%289+%2B-+7%29%2F%282%2A4%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%289+%2B-+7%29%2F8 Multiply 2 and 4 to get 8


So now the expression breaks down into two parts


x+=+%289+%2B+7%29%2F8 or x+=+%289+-+7%29%2F8


Lets look at the first part:


x=%289+%2B+7%29%2F8


x=16%2F8 Add the terms in the numerator

x=2 Divide


So one answer is

x=2




Now lets look at the second part:


x=%289+-+7%29%2F8


x=2%2F8 Subtract the terms in the numerator

x=1%2F4 Divide


So another answer is

x=1%2F4


So our solutions are:

x=2 or x=1%2F4





Check:

However, we need to check if these solutions satisfy the original equation

sqrt%285x-1%29-1=sqrt%28x%2B2%29 Start with the given equation


sqrt%285%282%29-1%29-1=sqrt%282%2B2%29 Plug in x=2


sqrt%289%29-1=sqrt%284%29 Simplify


3-1=2 Take the square root


2=2 Simplify. Since both sides are equal, the solution x=2 is verified


-------------------

sqrt%285x-1%29-1=sqrt%28x%2B2%29 Start with the given equation


sqrt%285%281%2F4%29-1%29-1=sqrt%281%2F4%2B2%29 Plug in x=1%2F4


sqrt%281%2F4%29-1=sqrt%289%2F4%29 Simplify


1%2F2-1=3%2F2 Take the square root


-1%2F2=3%2F2 Simplify. Since both sides are not equal, the solution x=1%2F4 is not really a solution. It is an extraneous solution


So our only solution is x=2