Solution To Equation Involving Square Root: Extraneous Solution?

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Discussion Overview

The discussion centers around solving the equation sqrt{x+2} = x-4, specifically addressing the solution set and the potential for extraneous solutions. Participants explore the validity of the solutions obtained and the implications of squaring both sides of the equation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant claims to have found the solutions 2 and 7 for the equation.
  • Another participant asserts that only 7 is a valid solution, suggesting that squaring the right-hand side introduces an extraneous solution.
  • A later reply provides a verification of both proposed solutions, showing that 7 satisfies the equation while 2 does not.
  • The same reply emphasizes that the square root function yields a single value, reinforcing the argument against 2 being a valid solution.

Areas of Agreement / Disagreement

Participants express disagreement regarding the validity of the solution set, with some asserting that only 7 is correct while others initially propose both 2 and 7. The discussion remains unresolved as to whether 2 can be considered a solution.

Contextual Notes

Participants note the importance of understanding the implications of squaring both sides of an equation and the definition of the square root function, which may lead to confusion regarding extraneous solutions.

rebo1984
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Hi everyone,

What is the solution set of the equation: sqrt{x+2}= x-4

I got 2 and 7.

Is it correct or is it just 7. If so why?

Thanks:)
 
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Re: Solution to equations

It's just seven. You're squaring the RHS which introduces an extraneous solution.
 
Re: Solution to equations

rebo1984 said:
Hi everyone,

What is the solution set of the equation: sqrt{x+2}= x-4

I got 2 and 7.

Is it correct or is it just 7. If so why?

Thanks:)
Check:
if x= 7 then sqrt(x+ 2)= sqrt(7+ 2)= sqrt(9)= 3 while x- 4= 7- 4= 3. Those are the same so x= 7 satisfies sqrt(x+ 2)= x- 4.

If x= 2, sqrt(x+ 2)= sqrt(2+ 2)= sqrt(4)= 2 while x- 4= 2- 4= -2. Those are not the same so x= 2 does not satisfy sqrt(x+ 2)= x- 4.

Note that the square root function, sqrt(4), cannot give both "2" and "-2" because a function, by definition, must give a single value. Further, suppose you were asked to solve the equation x^2= a. There are two numbers that satisfy that equation which you would write as x= +/- sqrt(a). The reason you need the "+/-" is because sqrt(a) is only the positive solution.
 
Re: Solution to equations

Thank you for your very detailed response.
 

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