1. ## equation

sqrt (x-5) + 14 = 3x + 2

4x - x/3 + (x + 1)/9 = 2x

2. Originally Posted by tara
sqrt (x-5) + 14 = 3x + 2
Isolate the square root:
$\sqrt{x - 5} = 3x - 12$

Now square both sides.
$x - 5 = (3x - 12)^2$
and solve for x normally from there.

Always always always plug your solutions back into the original equation. This method occasionally gives extra solutions that don't work. If none of the solutions work, then there is no solution.

-Dan

3. Originally Posted by tara

sqrt (x-5) + 14 = 3x + 2

4x - x/3 + (x + 1)/9 = 2x
$4x - \frac{x}{3} + \frac{x + 1}{9} = 2x$

I'm not sure what the problem is. Either add up the left hand side as fractions, or simply multiply both sides of the equation by 9:
$36x - 3x + (x + 1) = 18x$

Or did I misread the question?

-Dan

4. So, here is what I got on that problem Dan....

x-5 + 14 = (3x+2)^2
x-5 + 14 = (3x + 2) (3x + 2)
x -5 + 14 = 9x^2 + 12x + 4
set one side to zero
0= -x + 5 - 14 + 9x^2 + 12x + 4
0= 9x^2 + 11x -5

and that as far as I can get... ???

5. Originally Posted by tara
So, here is what I got on that problem Dan....

x-5 + 14 = (3x+2)^2
x-5 + 14 = (3x + 2) (3x + 2)
x -5 + 14 = 9x^2 + 12x + 4
set one side to zero
0= -x + 5 - 14 + 9x^2 + 12x + 4
0= 9x^2 + 11x -5

and that as far as I can get... ???
Remember you have a square root on that x - 5. It does not cover the 14! See my first response for what I think is the best approach. If you do it your way you do not get rid of the square root:
$\left ( \sqrt{x - 5} + 14 \right ) ^2 = (x - 5) + 2 \cdot 14 \cdot \sqrt{x - 5} + 14^2$

So continuing from where I left off in my first post:
$x - 5 = (3x - 12)^2$

$x - 5 = 9x^2 - 72x + 144$

$9x^2 - 73x + 149 = 0$

It might factor, it might not. For something with coefficients that look like this I'd just skip right to the quadratic formula to finish it.

(Hint: Check the discriminant. What does that tell you about the possible solutions for x?)

-Dan

6. Originally Posted by tara

sqrt (x-5) + 14 = 3x + 2
First, after rearranging, the equation equals $\sqrt{x-5}=3x-12.$ We require $x-5\ge0$ & $3x-12\ge0,$ from here, if the solution does exist, it must lie in the interval $[5,\infty[.$ Now note that the equation equals $\sqrt{x-5}-3(x-5)=3,$ and for $x\ge5\implies\sqrt{x-5}\le3(x-5),$ hence the LHS is negative, so the equation has no solution.

7. ## thanks

To both of you for posting. I am new on here and am just getting the "rules" down on how to do things. I appreciate your assistance...