Really simple math question

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Any number times itself is a positive number (or zero), so you can't ever get to a negative number by squaring. Since square roots undo squaring, negative numbers can't have square roots. The square root of a number is defined to be the value such that .
 
remember that sqrt(x) = |x|
if you plug -1 into the original equation you get : 2 = -2 which is not true. the solution is x = 4
anymore questions feel free to ask.
 
(x+5)^(1/2) = x-1

Set the condition x + 5 > 0 since you can't square root a negative number, so x > -5

Square both side: x + 5 = (x-1)^2
x + 5 = x^2 -2x +1
x^2 -3x - 4 = 0
x = -1 or x = 4
Both roots satisfy the condition x > -5, but only x = 4 is correct.

*Fixed my mistake*
 
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x=-1 is what's called an erroneous root. you should always plug your answers back into the original equation to check.
when you deal with Logs and square roots it's a good idea to check your answers.
I learned the hard way 😉
 
remember that sqrt(x) = |x|
if you plug -1 into the original equation you get : 2 = -2 which is not true. the solution is x = 4
anymore questions feel free to ask.
Oh woops totally forgot. Plugging in the answer can get you -2 as well (if you assumed that sq root of 4 could be 2 or -2 which I learned is wrong now). But I think the rule of remembering roots as |x|=(x^2)^1/2 is best.

(x+5)^(1/2) = x-1

Set the condition x + 5 > 0 since you can't square root a negative number, so x > -5

Square both side: x + 5 = (x-1)^2
x + 5 = x^2 -2x +1
x^2 -3x - 4 = 0
x = -1 or x = 4
Since both roots satisfy the condition x > -5, you take both roots and x = -1 is one of them.


Ofcourse you can't square root a negative number, you will get an imaginary number if u do such, but square root of 4 is -2 and 2 or square root of 9 is -3 and 3 etc..
Destroyer said that the only answer was 4 thats why I was so tripped up
 
yes the only answer is 4 and not -1. You should be very careful.
when you do x+5>= 0 you get x>= -5. You cannot jump to the conclusion and say that since -1 and 4 are both greater than -5 then they are both solutions why?
because you have another equation on the opposite side of the equal sign which x-1.
I'm assuming you are talking about the domain of sqrt(x+5).
 
Ah, I got mixed up with something like x^2 = 9 so x = +/- 3. I apologize. Square root of x^2 = |x| so x = 4 is the only root for the question.
 
yes the only answer is 4 and not -1. You should be very careful.
when you do x+5>= 0 you get x>= -5. You cannot jump to the conclusion and say that since -1 and 4 are both greater than -5 then they are both solutions why?
because you have another equation on the opposite side of the equal sign which x-1.
I'm assuming you are talking about the domain of sqrt(x+5).
Yea I was mixed up on why it couldn't be -1 but that's because sqrt(x+5) would be -2 which is wrong since it can only be |x| if sqroot. Got it lol I will never forget this. Thanks !!!
Ah, I got mixed up with something like x^2 = 9 so x = +/- 3. I apologize. Square root of x^2 = |x| so x = 4 is the only root for the question.
same. Thanks for helping though!
 
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