# The function g(x)=√(5+x)-x

## Homework Statement

g(x)=√(5+x)-x

a) Write down D(g)

b) Solve the equation
√(5+x)-x=3

√(5+x)-x=3

## The Attempt at a Solution

a)
x∈ℝ: x>-5
b)[/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:

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

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

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

Thank you,

Thank you,
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.

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

Jaco Viljoen
Mark44
Mentor
you can't take sqroot of 1 -1. sqrt(x^2)=|x|
I don't know what you meant, but your first sentence above is incorrect. 1 - 1 = 0, and ##\sqrt{0} = 0##.

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

I meant to say x>-5 so the root will not be of a negative number.
Yup, that's right.

Jaco Viljoen
Thank you,

ehild
Homework Helper

## Homework Statement

g(x)=√(5+x)-x

a) Write down D(g)

b) Solve the equation
√(5+x)-x=3

√(5+x)-x=3

## The Attempt at a Solution

a)
x∈ℝ: x>-5[/B]

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.
√(5-4)--4=3
√(1)+4=3 False

√(5-1)--1=3
2+1=3 True
x=-1
Correct.
Is x=-1 inside the domain?

Jaco Viljoen
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)?
I did fix this it was a typo

Is x=-1 inside the domain?
Yes it is

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

SammyS
Staff Emeritus