The function g(x)=√(5+x)-x

1. Oct 20, 2015

Jaco Viljoen

1. The problem statement, all variables and given/known data
g(x)=√(5+x)-x

a) Write down D(g)

b) Solve the equation
√(5+x)-x=3
2. Relevant equations
√(5+x)-x=3

3. The attempt at a solution
a)
x∈ℝ: x>-5
b)

√(5+x)-x=3
√(5+x)=x+3
(5+x)=(x+3)^2
(5+x)=(x+3)(x+3)
(5+x)=x^2+6x+9
x^2+5x+4=0
(x+4)(x+1)
x=-4 or x=-1

√(5-4)--4=3
√(1)+4=3 False

√(5-1)--1=3
2+1=3 True
x=-1

Last edited: Oct 20, 2015
2. Oct 20, 2015

Utilite

you can't take sqroot of 1 -1. sqrt(x^2)=|x|

3. Oct 20, 2015

PWiz

Seems right. You haven't done a) though.

4. Oct 20, 2015

Jaco Viljoen

Thank you,

5. Oct 20, 2015

PWiz

That's incorrect.(Assuming that you have the condition that the output must be real.) If x were to be less than -5, then you would have to take square roots of negative numbers, so your output would no longer be real.

6. Oct 20, 2015

RUber

For a) you wrote x<-5, that is the opposite of the domain.
Your solution was x = -1. That is not in x<-5.

7. Oct 20, 2015

Staff: Mentor

I don't know what you meant, but your first sentence above is incorrect. 1 - 1 = 0, and $\sqrt{0} = 0$.

8. Oct 20, 2015

Jaco Viljoen

I meant to say x>-5 so the root will not be of a negative number.

9. Oct 20, 2015

PWiz

Yup, that's right.

10. Oct 20, 2015

Jaco Viljoen

Thank you,

11. Oct 20, 2015

ehild

You wrote that x should be less than -5. Can you take the square root of x-5 then? So what is the correct domain of g(x)?
b)
√(5+x)-x=3
√(5+x)=x+3
(5+x)=(x+3)^2
(5+x)=(x+3)(x+3)
(5+x)=x^2+6x+9
x^2+5x+4=0
(x+4)(x+1)
x=-4 or x=-1[/QUOTE]
Correct.
Correct.
Is x=-1 inside the domain?

12. Oct 20, 2015

Jaco Viljoen

I did fix this it was a typo

Yes it is

13. Oct 20, 2015

Utilite

I
meant sqrt of 1 doesn't equal -1

14. Oct 20, 2015

SammyS

Staff Emeritus
The domain is not quite right. $\ \sqrt{ 0 } \$ is a real number, so -5 is in the domain.

15. Oct 20, 2015

PWiz

You're right. I overlooked that fact. The inequality should be weak @Jaco Viljoen