What went wrong with (-x)^2=x^2?

I have a very basic confusion that supports some basic elements of algebra. Being a high school student my teacher couldn't answer this, hope someone could help here.

We know this equation is true: (-x)^2=x^2
but once we square root both sides it becomes this: -x=x
we can see this equation was solved using the correct principles of algebra but we know this is false.
What's wrong here?

jeff einstein said:

we can see this equation was solved using the correct principles of algebra

I don't think that it was. Which principles are these?

##\sqrt{a^2} = |a|## not ##a##

jeff einstein said:

We know this equation is true: (-x)^2=x^2
but once we square root both sides it becomes this: -x=x

The square root of (-x)^2 is not unique; it's -x and x. We choose x because -x is a nonsensical answer (in basic algebra).

so both sides become absolute values of X?

That two numbers have the same square means NOT that they are equal... I think that your problem (and that of your moronic teacher!) is that you don't see the difference between "imply" and "is equivalent to" (bi-directional implication). Let me illustrate the conceptual mistake that you are making with another example: Lenin had a goatee, goats have one too, therefore Lenin was a goat (right?)

Sir please give respect to my teacher and plus i don't really understand your argument here please explain this with math

jeff einstein said:

We know this equation is true: (-x)^2=x^2

This equation means, ##(-x) \times (-x)=x \times x##.
There is no way to conclude from
##(-x) \times (-x)=x \times x##
that
##-x=x##,
"using the correct principles of algebra."

jeff einstein said:

Being a high school student my teacher couldn't answer this, hope someone could help here.

Your teacher is a high school student? Not surprised that he or she couldn't answer this.

jeff einstein said:

We know this equation is true: (-x)^2=x^2
but once we square root both sides it becomes this: -x=x

No, this isn't true. See below.

DaveC426913 said:

The square root of (-x)^2 is not unique; it's -x and x.

No, this is a very frequent misconception. The square root of a nonnegative real number is unique.
##(-x)^2 = x^2##
##\sqrt{(-x)^2} = \sqrt{x^2} = |x|##, which is always nonnegative.
While it is true that, for example, 4 has two square roots, 2 and -2, the expression ##\sqrt 4## is agreed on by convention to be the positive, or principal, square root.

jeff einstein said:

Being a high school student my teacher couldn't answer this.

Your teacher is a high school student?

DaveC426913 said:

We choose x because -x is a nonsensical answer (in basic algebra).

That's not what's really going on here. (Note that if ##x=-2##, then ##\sqrt{x^2} = -x##.) The mistake the OP is making is thinking ##\sqrt{(\text{something})^2} = \text{something}##. By definition, ##\sqrt x## is the principal square root of ##x##, which by convention we take as the non-negative square root. So the correct rule is ##\sqrt{x^2}=\lvert x \rvert##.

A function is given if no two values of ##f(x)## come from the same value ##x,## i.e. ##a=b## requires ##f(a)=f(b).## This is trivial you might think, but not all relations are functions. "Has a daughter" wouldn't be a function since it is possible for the same mom to have two different daughters. A function doesn't allow the assignment of different images to the same point of its domain.

If the opposite is also true, i.e. if ##f(a)=f(b)## implies ##a=b## then we call a function injective. In such a case, ##f## establishes a one-to-one correspondence between the set of ##\{a\}## and the set of ##\{f(a)\}.## Such functions have therefore a well-defined inverse: ##f^{-1}(f(a))=a.##

The function ##f\, : \,x\longmapsto x^2## is not injective since ##f(x)=f(-x)## whereas ##x\neq -x## for any ##x\neq 0.## This means, that there is no inverse function. We can write ##f^{-1}(\{y^2\})=\{\pm y\}## and mean sets and call ##f^{-1}(\{y^2\})## the pre-image of ##y^2## but that wouldn't establish a function anymore. In order to deal with roots, people often write ##f^{-1}(y^2)=\pm y## which is at least sloppy, more often confusing, and strictly spoken wrong. To use the square root anyway, we restrict ourselves to the positive value of ##y## so that the root is defined as ##\sqrt{y^2}=+|y|.## We ignore the negative branch. This makes it (a bit artificially) an injective function ##f\, : \,\mathbb{R}^+ \longrightarrow \mathbb{R}^+## that can therefore be inverted: ##f^{-1}(x)=+\sqrt{x}.##

If we want to use both values, then we use the set-theoretical language and speak of the pre-image of ##y^2## which is the set ##\{-|y|\, , \,+|y|\}.## This is not a function anymore, even less the inverse of squaring, it is only a so-called relation - like the mother with her daughters.

If you plot the numbers that can be squared to equal a result, you would have at X=1, two Y values, 1 and -1. At X=4 you would have two Y values, 2 and -2. At X=9 you would have two Y values, 3 and -3.

At none of those is the positive number equal to the negative number. The "square root" of 4 is defined as being the positive number that squares to give 4.

If you want to express what squared gives you a number, you have to give two answers on both sides of your equation.

For example
$$(-3)^2=3^2 \to 9=9$$
after applying the square root you have two options. Either
$$\sqrt{9}=\sqrt{9} \to 3=3$$
or
$$-\sqrt{9}=-\sqrt{9} \to -3=-3$$
Both are true.

jeff einstein said:

TL;DR Summary: I have a very basic confusion that supports some basic elements of algebra. Being a high school student my teacher couldn't answer this, hope someone could help here.

I have a very basic confusion that supports some basic elements of algebra. Being a high school student my teacher couldn't answer this, hope someone could help here.

We know this equation is true: (-x)^2=x^2
but once we square root both sides it becomes this: -x=x
we can see this equation was solved using the correct principles of algebra but we know this is false.
What's wrong here?

No. Not right.

Look carefully here:
(-x)^2------------------the left side
(-1)(x)(-1)(x)
(-1)(-1)xx
(+1)x^2
x^2------------------the right side

Just to clear up (i think this is where my confusion arises from), the square root of a number (negative or positive) is always the absolute value of the number?

jeff einstein said:

Just to clear up (i think this is where my confusion arises from), the square root of a number (negative or positive) is always the absolute value of the number?

Assuming you mean the square root of the square of a number (negative or positive) is always the absolute value, yes, by definition. The square root of a negative number introduces more complexity.

That way, we can say that if ##x^2=a## then ##x=\pm\sqrt{a}##, or if we know the answer is negative for some reason we can say ##x=-\sqrt{a}## and there's no ambiguity over what that means.

Mark44 said:

Your teacher is a high school student? Not surprised that he or she couldn't answer this.


No, this isn't true. See below.


No, this is a very frequent misconception. The square root of a nonnegative real number is unique.
##(-x)^2 = x^2##
##\sqrt{(-x)^2} = \sqrt{x^2} = |x|##, which is always nonnegative.
While it is true that, for example, 4 has two square roots, 2 and -2, the expression ##\sqrt 4## is agreed on by convention to be the positive, or principal, square root.

This makes the most sense to me. The way i perceive this root (-x)^2 is the same as (-1*x)^2 and so when squaring the negative 1 it becomes positive and that just makes the equation root x^2 and so the root of a (-x) squared number is always. Is this right?

The square root of any number x has two solutions: -sqrt(x) and +sqrt(x). This holds even for negative
numbers, where the solutions are in the complex number plane: -i sqrt(x) and +i sqrt(x). Therein,
i equals the square root of -1.
As a general law, the n-th root of a number x has n solutions, which are equidistant on a circle of
radius sqrt(|x|) in the complex number plane, spanned by the orthogonal vectors (1, 0) and (0, i).
To solve your problem, consider both solutions of the square roots to the left and to the right,
and combine them appropriately. Of the four resulting pairs, only two make sense; the remaining
two are artifacts of the calculation.

falsifying​

Have you been reading Popper, or watching Faux Noise?

?

jeff einstein said:

TL;DR Summary: I have a very basic confusion that supports some basic elements of algebra. Being a high school student my teacher couldn't answer this, hope someone could help here.

I have a very basic confusion that supports some basic elements of algebra. Being a high school student my teacher couldn't answer this, hope someone could help here.

We know this equation is true: (-x)^2=x^2
but once we square root both sides it becomes this: -x=x
we can see this equation was solved using the correct principles of algebra but we know this is false.
What's wrong here?

Thanks for this thought provoking question.

The problem is with the square root operator. If negative answers are allowed then it is a function that returns a set of two solutions. If negative answers are not allowed then it is a function that returns a set of one solution. As you will see, this makes a difference.

In the given example negative solutions are allowed. Then

A = sqrt(4) = { 2, -2 }
and
B = sqrt(4) = { 2,-2 }

A = B

So far, so good.

However as you can see it does not follow that every solution in A is equal to every solution in B. Indeed we are guaranteed that there will be a pair of solutions, one from A and one from B, that are not equal. This is because a set is a collection in which every member is unique. If your two sets of solutions are {a1, a2} and {b1, b2} then a1 = b1 implies a1<>b2.

This is always true if one of the solution sets is bigger than one. Only if both solution sets are of size one is it possible that every solution in A is equal to every solution in B.

Edgar53 said:

The square root of any number x has two solutions: -sqrt(x) and +sqrt(x).

No. For a nonnegative real number a, the square root of a is the number b such that ##b^2 = a##. So ##\sqrt a = b, the principal square root of a.
Where you seem to be confused is that every positive real number has two square roots, one positive and one negative, but the symbol ##\sqrt b## represents a single positive real number.

Edgar53 said:

This holds even for negative
numbers, where the solutions are in the complex number plane: -i sqrt(x) and +i sqrt(x). Therein,
i equals the square root of -1.

Again, no. Like positive numbers, negative numbers have two square roots, but the symbol ##\sqrt a## represents the principal square root. For example, ##\sqrt{-4} =i\sqrt 4 = 2i##.
The equation ##x^2 = -4## has two solutions: x = 2i and x = -2i, but only the first is the principal square root of -4.

Hornbein said:

The problem is with the square root operator. If negative answers are allowed then it is a function that returns a set of two solutions. If negative answers are not allowed then it is a function that returns a set of one solution. As you will see, this makes a difference.

In the given example negative solutions are allowed. Then

A = sqrt(4) = { 2, -2 }
and
B = sqrt(4) = { 2,-2 }

No for the reasons given above. ##\sqrt 4 = 2##. Period.

Since the thread has become filled with less than helpful information (I'm being charitable), I am closing it.