1. ## Solve for x

(3 + x)/(sqrt{3 + x}) = 1

I found x to be -2.

The book tells me that the answer is x = -1.

How can this be?

2. Your answer is correct. Just substitue the value of x in LHS of equation and see it yourself

3. ## math books

Originally Posted by arpitagarwal82
Your answer is correct. Just substitue the value of x in LHS of equation and see it yourself
I knew that I was right. I find that math books have a lot of wrongs answers in the answer section.

4. Originally Posted by magentarita
(3 + x)/(sqrt{3 + x}) = 1

]I found x to be -2.

The book tells me that the answer is x = -1.

How can this be?
$(3+x) = \sqrt{3+x}$

square both sides:

$x^2 + 6x + 9 = 3+x$

$x^2 + 5x +6 = 0$
$
(x+3)(x+2) = 0$

therefore both -3 and -2 are solutions

5. Originally Posted by e^(i*pi)
$(3+x) = \sqrt{3+x}$

square both sides:

$x^2 + 6x + 9 = 3+x$

$x^2 + 5x +6 = 0$
$
(x+3)(x+2) = 0$

therefore both -3 and -2 are solutions

If you plug -3 back into the original equation $\frac{3+x}{\sqrt{3+x}}=1$, you have $\frac{0}{0}=1$ which is impossible, since $\frac{0}{0}$ is an indeterminate value... XD

So magentarita's answer x=-2 is the only solution.

6. ## Yes...

The only answer that makes sense is x = -2. If we replace x with -3, we create division by zero, which is undefined.