The function g(x)=√(5+x)-x

  • #1
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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
 
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Answers and Replies

  • #2
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you can't take sqroot of 1 -1. sqrt(x^2)=|x|
 
  • #3
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Seems right. You haven't done a) though.
 
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  • #4
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Seems right. You haven't done a) though.
Oops, I've added my answer,

Thank you,
 
  • #5
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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.
 
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  • #6
RUber
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For a) you wrote x<-5, that is the opposite of the domain.
Your solution was x = -1. That is not in x<-5.
 
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  • #7
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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##.
 
  • #8
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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.
 
  • #9
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I meant to say x>-5 so the root will not be of a negative number.
Yup, that's right.
 
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  • #11
ehild
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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?
 
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  • #12
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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
 
  • #13
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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
 
  • #14
SammyS
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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.
 
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  • #15
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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
 

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