1. Sep 28, 2013

### Coco12

1. The problem statement, all variables and given/known data
Sqrt 5x2 +11= x+5

State the restrictions for x and solve.

2. Relevant equations
5x2+11 ≥ 0
x2≥-11/5
(then how can u take the sqrt of a negative number??)

3. The attempt at a solution

The answer at the back of the book is 7/2, -1

I don't know what I'm doing wrong but I keep getting 7 and -2 for x??? That is when I isolated the sqrt to one side of the equation, squared it and factored it.
Also pls explain how to get the restrictions

2. Sep 28, 2013

### tiny-tim

HiCoco12!
The LHS is a square-root, so it has to be positive

so the RHS must also be positive.
Show us how.

3. Sep 28, 2013

### Coco12

Show us how. [/QUOTE]

by squaring the sqrt 5x2+11 and the x+5 on the other side of the equal sign to get rid of the sqrt which will give me 5x2+11= x2+10x+25
Then i brought over the x2+10x+25 to the other side which gives me:
4x2-10x-14
Then i factored out a 2:
2(2x2-5x-7)
and i factored that to give me (x-7) (x+2)
x= 7,-2...

4. Sep 28, 2013

### eumyang

$2x^2 - 5x - 7$ does not factor into $(x - 7)(x + 2)$!
$(x - 7)(x + 2) = x^2 - 5x + 14$!

5. Sep 29, 2013

### tiny-tim

the clue's in the 2x2 !

6. Sep 29, 2013

### Coco12

Ohh now I get it.. Oops missed that.. However how do I find the restrictions??

7. Sep 29, 2013

### tiny-tim

The LHS is a square-root, so it has to be positive

so the RHS must also be positive.