SOLUTION: Hello. I know it might sound a bit crazy but can you help me solve this radical equation? I have been so focused on other formulas that I forgot this one. My question is: {{{sqr

Algebra ->  Radicals -> SOLUTION: Hello. I know it might sound a bit crazy but can you help me solve this radical equation? I have been so focused on other formulas that I forgot this one. My question is: {{{sqr      Log On


   

Question 996998: Hello. I know it might sound a bit crazy but can you help me solve this radical equation? I have been so focused on other formulas that I forgot this one. My question is:
sqrt%28x%2B15%29-x=3 I should start by adding x to both sides so the equationb now looks like this: sqrt%28x%2B15%29=3%2Bx at this time, I square both sides to look like this: sqrt%28x%2B15%29%5E2=%28x%2B3%29%5E2 Since the squares eliminate themselves on the left, I foil the right so the equation looks like this:
x+15=x^2+6x+9. Here is where I stop because the answer key has an answer of 1 but I'm not getting that no matter which way I go. Can you advise me?
Found 3 solutions by Alan3354, ikleyn, ankor@dixie-net.com:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x%2B15%29-x=3 I should start by adding x to both sides so the equationb now looks like this: sqrt%28x%2B15%29=3%2Bx at this time, I square both sides to look like this: sqrt%28x%2B15%29%5E2=%28x%2B3%29%5E2 Since the squares eliminate themselves on the left, I foil the right so the equation looks like this:
x+15=x^2+6x+9.
---------------
Subtract (x+15)
x%5E2+%2B+5x+-+6+=+0
(x + 6)*(x - 1) = 0
x = 1
-------------
x = -6 (works if you use sqrt(9) = -3)

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
I will continue from this point:

x%2B15 = x%5E2%2B6x%2B9.

x%5E2+%2B+5x+-+6 = 0,

(x-1)*(x+6) = 0,

The roots are x = 1 and x = -6.

What is the problem?


Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x%2B15%29-x=3 I should start by adding x to both sides so the equation now looks like this: sqrt%28x%2B15%29=3%2Bx at this time, I square both sides to look like this: sqrt%28x%2B15%29%5E2=%28x%2B3%29%5E2
Since the squares eliminate themselves on the left, I foil the right so the equation looks like this:
x+15=x^2+6x+9.
don't stop here, combine like terms on the right
0 = x^2 + 6x - x + 15 - 9
0 = x^2 + 5x - 6
factors to
(x+6)(x-1) = 0
positive solution
x = 1 is the solution
and
x = -6, will not be a solution in the original equation