What is the domain for solving the equation √(5+x)-x=3 and writing down D(g)?

[Jaco Viljoen]

Homework Statement

g(x)=√(5+x)-x

a) Write down D(g)

b) Solve the equation
√(5+x)-x=3

Homework Equations

√(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

 

[Utilite]

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

 

[PWiz]

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

 

[Jaco Viljoen]

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

Oops, I've added my answer,

Thank you,

 

[PWiz]

Oops, I've added my answer,

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.

 

[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.

 

[Mark44]

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 $\sqrt{0} = 0$.

 

[Jaco Viljoen]

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.

 

[PWiz]

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 Statement

g(x)=√(5+x)-x

a) Write down D(g)

b) Solve the equation
√(5+x)-x=3

Homework Equations

√(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

 

[Utilite]

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]

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

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

 

[PWiz]

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

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