Solution To Equation Involving Square Root: Extraneous Solution?

[rebo1984]

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:)

 

[Greg]

Re: Solution to equations

It's just seven. You're squaring the RHS which introduces an extraneous solution.

 

[HallsofIvy]

Re: Solution to equations

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.

 

[rebo1984]

Re: Solution to equations

Thank you for your very detailed response.