√(5x-4)=x-2 extraneous solution

[Fisiodes]

After solving you get: (x-8)(x-1)=0. Why 1 is an extraneous solution? I can verify it by squaring √1=-1 and obtain 1=1 . Why it is wrong?

 

[lookagain]

Why 1 is an extraneous solution? I can verify it by squaring √1=-1 and obtain 1=1 . Why it is wrong?

No, you cannot square both sides after that. You
are evaluating each side separately in the check to see if they are equal. \(\displaystyle \ \sqrt{1} = 1. \ \ \) That ends
the simplifying on the left-hand side. Now you
have (are looking at) 1 = -1, which is false. So,
you must discard x = 1 as a solution.

 

[Dr.Peterson]

After solving you get: (x-8)(x-1)=0. Why 1 is an extraneous solution? I can verify it by squaring √1=-1 and obtain 1=1 . Why it is wrong?

The reason 1 is not a solution is that √(5x-4)=x-2 is not true when x is replaced by 1: √(5*1-4)=1, while 1-2 = -1.

The fact that the squares of two numbers are equal does not imply that the numbers themselves are equal. And the radical symbol represents only one number, the positive square root.

 

[Fisiodes]

No, you cannot square both sides after that. You
are evaluating each side separately in the check to see if they are equal. \(\displaystyle \ \sqrt{1} = 1. \ \ \) That ends
the simplifying on the left-hand side. Now you
have (are looking at) 1 = -1, which is false. So,
you must discard x = 1 as a solution.

Thank you dr. Peterson and lookagain for your answers. So basically if there aren't hidden terms i can't square equations because that could lead to extraneous values. Is it an algebra rule? or common sense?

 

[Fisiodes]

*unknown terms

 

[Fisiodes]

Thank you i thought about it and understood the loss of information when squaring. Now everything is clear!

 

[Steven G]

Suppose x=-4.
Squaring both sides yields x^2 = 16. Then x= +/- 4. Surely x=4 is wrong. After all, we were told that x=-4.

 

[HallsofIvy]

While there are 2 values of x such that \(\displaystyle x^2= a^2\), x= a and x= -a, \(\displaystyle \sqrt{a^2}\) is specifically defined as the positive root, |a|.