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Radical equation

  • Thread starter Coco12
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  • #1
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Homework Statement


Sqrt 5x2 +11= x+5

State the restrictions for x and solve.






Homework Equations


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


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
 

Answers and Replies

  • #2
tiny-tim
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HiCoco12! :smile:
Also pls explain how to get the restrictions
The LHS is a square-root, so it has to be positive

so the RHS must also be positive. :wink:
… I keep getting 7 and -2 for x???
Show us how. :confused:
 
  • #3
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Show us how. :confused:[/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
eumyang
Homework Helper
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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...
[itex]2x^2 - 5x - 7[/itex] does not factor into [itex](x - 7)(x + 2)[/itex]!
[itex](x - 7)(x + 2) = x^2 - 5x + 14[/itex]!
 
  • #5
tiny-tim
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the clue's in the 2x2 ! :biggrin:
 
  • #6
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the clue's in the 2x2 ! :biggrin:
Ohh now I get it.. Oops missed that.. However how do I find the restrictions??
 
  • #7
tiny-tim
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Homework Helper
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… how do I find the restrictions??
The LHS is a square-root, so it has to be positive

so the RHS must also be positive. :wink:
 

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