- Thread starter Grimmie
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This is clearly a problem involving algebraic equation.Please help me solve the exercise:

Find the error in the following argument,sqrt(x^2+1)=x+1and so let x=2 sqrt(5)=9, therefore 5=9

Thanks in advance!!!

Why are you posting this in pre-algebra section?

How do you know that the given equation [

Is this a Home Work problem? Which grade?

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As you've presented it, the exercise is to locate mistake(s) in somebody else's work.Please help me solve the exercise

Above, I have highlighted (in red) an error. Focus on that, to start. Once you tell us the correct value, we can discuss the exercise further. :cool:Find the error in the following argument,

sqrt(x^2+1) = x+1

and so let x = 2

sqrt(5) = 9,

therefore 5 = 9

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It looks as though you are assuming that sqrt[x^2 + 1] is everywhere (that is, for all values of x) equal to x + 1. Why? You know that (x + 1)^2 is not equal to x^1 + 1 (since powers don't "distribute"), so why are you assuming that the wrong statement "works" "going backwards"?Find the error in the following argument, sqrt(x^2+1)=x+1 and so let x=2 sqrt(5)=9, therefore 5=9

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1st of all sqrt(5) is not 5, so even if sqrt(5) = 9 it does NOT follow that 5=9.Please help me solve the exercise:

Find the error in the following argument, sqrt(x^2+1)=x+1 and so let x=2 sqrt(5)=9, therefore 5=9

Thanks in advance!!!

You mean to ask how does sqrt(5) = 9?

The equation you wrote is what is called a condition equation. It is only valid for 1 or more x values, if any.

Consider x+3 = 5. If we let x=7, then we get 7+3 = 5 or 10 = 5. This is not true, SO X IS NOT 7.

On the other hand, if we let x=2, then 2+3=5 or 5 = 5 which is true, so X = 2 IS THE CORRECT ANSWER