Radical Equation Restrictions & Solution | Sqrt 5x2 +11= x+5 Homework

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Homework Help Overview

The problem involves solving the radical equation Sqrt(5x^2 + 11) = x + 5 and determining the restrictions on the variable x. The context includes exploring the implications of square roots and the conditions under which they are defined.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss isolating the square root and squaring both sides of the equation to eliminate the radical. There are questions about how to derive the restrictions for x, particularly regarding the positivity of both sides of the equation.

Discussion Status

Some participants have shared their attempts at solving the equation and expressed confusion about the results they obtained. There is an ongoing exploration of the correct factoring of the resulting polynomial, and multiple interpretations of the restrictions are being considered.

Contextual Notes

Participants note that the left-hand side of the equation involves a square root, which must be non-negative, leading to questions about how this affects the right-hand side. There is also mention of the answer provided in the textbook, which differs from the results some participants are obtaining.

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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
 
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HiCoco12! :smile:
Coco12 said:
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:
 
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...
 
Coco12 said:
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...
2x^2 - 5x - 7 does not factor into (x - 7)(x + 2)!
(x - 7)(x + 2) = x^2 - 5x + 14!
 
the clue's in the 2x2 ! :biggrin:
 
tiny-tim said:
the clue's in the 2x2 ! :biggrin:

Ohh now I get it.. Oops missed that.. However how do I find the restrictions??
 
Coco12 said:
… 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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