Radical Equation Extraneous Roots

In terms of radical equations, what are extraneous roots? Can you give an example by solving the following problem?

sqrt{x^2 + 4} = 20

 

An "extraneous root" may be defined as:
A solution to an equation that SEEMS to be right, but when we check it (by substituting it into the original equation) turns out NOT to be right.

Example: you work on an equation and come up with two roots (where it equals zero):
"a" and "b".
When you put "a" into the original equation it becomes zero, but when you put in "b" it doesn't.
So "b" is an extraneous root.

This often happens when we square both sides during our solution.

An extraneous solution is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation.
Example:

Solve for x , .





(x−2)+(x+2)=4

2x=4

x=2

But 2 is excluded from the domain of the original equation because it would make the denominator of one of the fractions zero--and division by zero is not allowed.
Therefore, it cannot be a root of the original equation. So, 2 is an extraneous solution. So, the equation has no solutions.

 

Very well-explained.

Let me see.

sqrt{x^2 + 4} = 20

[sqrt{x^2 + 4}]^2 = (20)^2

x^2 + 4 = 400

x^2 = 400 - 4

x^2 = 396

sqrt{x^2} = sqrt{336}

I should get two answers: x = positive, x = negative.

x = 18.330303

x = -18.330303

Yes?

If yes, I now need to check for extraneous roots.

Let x = 18.330303

sqrt{x^2 + 4} = 20

sqrt{(18.330303)^2 + 4} = 20

sqrt{336 + 4} = 20

sqrt{400} = 20

20 = 20

If I let x = -18.330303 in the original radical equation given, I also get 20 = 20.

I say x = 18.330303 and x = -18.330303 are not extraneous roots.

Is any of this here correct?

 

yes, that is correct

but, it is better to find exact solutions which are
x = 6sqrt(11)
x = -6sqrt(11)
since both solutions work, and none s excluded from domain
both turns out that both can be a root of the original equation, so none is an extraneous solution

 

In other words, it is best not to use the decimal representation of the answer for x.

 

it is best not to use the decimal representation because you cut lot of decimal places, so it will show you dfferent answer
-> turns out to be false
-> turns out to be true

 

Your method is correct but at one point you have
"x^2 = 396"
and the very next line is
"sqrt{x^2} = sqrt{336}".

The correct answer should be $\sqrt{396}= 19.8997..$​

Your check is wrong also. With c=18.330303... x^2+ 4 is 336+ 4= 340 and sqrt(x^2+ 4)= 18.493... but you wrote 400, probably without actually checking.​

 

Yes, I now see my typo. Huge difference between sqrt{396} and sqrt{336}.

 

Ok. I gotta start paying attention to detail.