I will post the answer tomorrow. 😊 The correct answer is B) Nonpositive only

Did he really just give people over the internet homework?

√(a)² = |a| Definition of absolute value: If a>0 or a=0, |a| = a If a

I wish I had him in the fourth grade.

He's good, Wish I had him as a teacher when i was in school. Would have saved me from hating math back then.

B) Nonpositive only

This is great. Very smart idea. Make people participate and learn. This is really how you will get people to learn because your teaching makes peoppe want to know the answer. Only working it out yourself will make you remember.

It's nice to be reminded when these assertions are first posed. Thanks.

Answer B is “nonpositives” - this includes all negatives AND zero. This is the correct answer. Remember “-x” on the right would always become positive, if you plugged in any negative number, say “-5”, because -(-5) would always be a positive number. And the left side would always be a positive number with any negative x.

I think its gotta be non postives only (but 0 also cuz it works)... cuz like if you take the root to RHS, making it "x² = (-x)²" and then plugging in a negetive number like (-2), we get "(-2)² = (-(-2))²" which simplifies to "4 = 2² = 4" great question!

Your channel has quickly become one of my favorites ! 🎉 Keep it up, please!

B) non positive only, because the square root gives you non negative values by definition.

This heavily depends on your definition of square root. Many people in the comments section are saying non positive only because sqrt(x^2) = abs(x), however this does not hold true for all cases as it can be useful to define sqrt(x^2) as + or - x, in which case the answer would be all reals. Something like this is usually up to opinion and/or use case, but I am of the personal opinion (x^n)^1/n = abs(x) and root n of x^n = (e^2*pi*k*i/n) * x is a better delegation of definitions.

It helps to think of sqrt(x^2) as being equal to abs(x). Then, it is easy to visualize without graphing software, that y=-x and y=abs(x) are equal for all negative values.

X = -1 works right?!

i dont remember any of this. im gonna get a notebook and do two of these videos a day until I'm at a 6 grade math level.

People should memorize this property: √(x²) = |x| So |x| = - x if and only if x is a negative real number

X² ( with x real) is always positive, ✓of a positive number is always positive. So left side must be positive (or 0) -x is positive only if x is negative. So all non positive number is the solution.

Excellent video wonderful 😊😊😊😊😊❤❤❤❤❤

Netizens are stunned to see the negative zero.

me resisting the urge to type an imaginary number

Since sqrt(x^2) is equal to absval(x), you can simply solve by graphing and you get (-inf, 0]

A

It's just |x|=-x which is for negative numbers only, no need to substitute

The possible solutions shows x is less then or equal to zero. so I would say B because 0 should be included in nonpositive numbers.

Option E). because complex no. "-i" or "1/i" , 0 can satisfy it.

Left hand side is abs(x) , the equation abs(x)=-x is valid for all negative nunbers,its the definition of the abs(x) for x

B as, √x² = x if x is positive = -x if x is negetive now, if x is positive then x= -x ( it's impossible) and if x is negetive then x= x and for x= 0 0= 0 so, x would be 0 and negetive numbers or non positive .......... thank you 💕

sqrt(x^2)=|x| If you graph |x| and -x you see two graphes intercept at negative numbers including zero Answer choice is B

B

A bit tricky!! But if you make the correct decision of where the negative really lies!!

All nonpositives. I’m glad your page stumbled into my shorts feed. I’m 32 and starting a second degree. And I’ve forgotten most of calculus. And calculus is easy if your remember your algebra, so thank you for these. Bringing my memory back one equation at a time.

Negatives only. The negative part of the left will automatically equal the right.

All of them The √4 can be 2 and -2 And the √9 can be 3 and -3 The minus on √-3^2 is √9 = -3 or 3 And √2^2 is √4 = 2 and -2

B) sqrt(x^2) can be written as |x| so the equation becomes |x| = -x -> |x| + x = 0 Create two possible scenarios x + x = 0, x>= 0 -x + x = 0, x= 0 x€R, x

To solve this equation, you only need to find the domain of the equation which is x

Guessing before looking at the answer: x is equal to any real number

B, the square root of a squared number is basically the absolute value of it, problem is rhst if you take x > 0 -x is the opposite of it but if x < 0 -x will be the positive of it

A is the correct answer Because result of the square root should be positive or zero but if the number is imaginary number

sqrt(x^2) isn't something we're used to seeing. If you rewrite it as absolute value of x, |x|, then the equation becomes: |x| = -x In fact, we can get rid of the absolute value sign by looking at two different cases: 1. When x is positive, the equation is x = -x, which has only one solution at x = 0. 2. When x is negative, the equation is x = x, which is true for all negative numbers. So the solution is x

its B. the square root and power of 2 cancel eachother out and also turn the number positive and the negative on the right side only gets canceled if x is also negative.

My answer is B), but something I don't understand is that adding a minus sign to 0, is like adding a minus sign to a negative number, because 0 is neither positive nor negative, it's simply neutral, so I don't think (0²)^½ can equal -0

the squareroot and square cancel out so for the x to equal -x the x value has to always be negative. Or it could just be 0.

its b coz square root of x square would be mod x and hence mod x is equals to -x so x must be a non positive no so when we apply mod a - get multiplied to make it non negative

the left side of the equation is basically |x| which defined to be -x for non positives

I think the negative 3 could work as well as we are looking for a negative x, and a negative by the power of itself would also return to being a negative. That’s just my answer however!

the square root of any number is +/- x. For example, sqr(25) = 5 or -5. The answer is D. All numbers will make the statement true.. For example, if I put -3 or 3 as x, the equation will be sqr(9). The square root of 9 could equal -3. This is a trick question because there are always two answers. He just illustrated one of the possibilities to confuse ppl.

my ahh when I put "x is equal to plus or minus 0"

Non of the above the multiplication operations is a e log algorithm increment matlab responses from viewers

d because every number squared becomes posative then when you squareroot it it becomes itself again and its negative.

√x² = |x| • |x| = -x Therefore non-positive only ( B )

Professor, to me the answer is C) only positives. Any number with even exponents will always be positive!

That's above my pay grade. I'd go with negetives but can't explain it well.

Hint: make graphics and watch the points where these functions intersect P.D.: sqrt(x^2) = absolute value of x

Lot of people in the comments are posting examples of some solutions that work for this equation, however the real way to solve this problem is to first note is that sqrt(x^2) =|x|>or equal to 0, which means -x must also be greater than or equal to 0 as well for this equation to hold. Therefore-x>or equal to 0 means x

I may be wrong but my understanding that it is about balance. So if -0 will inherently distribute itself with no further action then (A will be my answer.

The answer is only non-positive numbers

hi can you try the proof equation 13 in the paper "new highly anti-interference regularization method for ill-posed problems"

There he is again doing his wizardry, LoL

All reals because a squareroot can be posative or negative. You can tell that they are both valid because seting i = -i forms a automorphism on the complex numbers

D), because the right side of the equation equals x and -x, but by the equation, it is chosen the negative, forcing one of the possibilities. But I think that, watching others of your presentation, you would say that the solution is A). But think of sin(x)=0, it has infinite solutions (0, pi, 2pi..., and the negative side). If an equation is in a physic problem that has as results multiple nums, it is required to dischard the absurd ones, that is all, but mathematicment the solution is correct for each result.

Multiply both sides by -1. Then x always equals a non-positive number.

Without seeing the video or reading the other comments: There is no valid solution to the equation √(x²) = -x. A square root that does not have a negative sign before the square root sign can only have a unique non-negative result (the result can only be positive or zero). This is because the result of a square root is an absolute value. Definition: √(x²) = |x| An absolute value is always positive (or zero). To write it very clearly: |+x| = +x |-x| = +x The positive signs only serve to clarify the relationship. Mathematically, they can also be omitted. EDIT OK. zero goes. However, the real question is to what extent -0 is a meaningful notation. I would always interpret -x as being a true negative value. That rules out zero from the start. And then what I wrote above applies. EDIT 2 I think I have to correct myself. I thought like this: On the left side of the equation is a square root. On the right side of the equation is a negative value, which is actually a simple negative number that can be specified. So something like: √(2x + 8) = -12 Here you can see immediately that there are no valid solutions for this. The reason is that a square root can only have a non-negative result. That’s because the result of a square root is an absolute number. Definition: √(x²) = |x| Things are different in the equation √(x²) = -x because the right-hand side is not a specifiable negative value, but basically the function y = -x This is a function that corresponds to a straight line with slope -1 and y-intercept 0. There is also a function on the left side, namely y = √(x²) or y = |x| This function gives an exact V with the vertex at the origin of coordinates. Now both functions are superimposed in the range x ≤ 0. Since x ∈ ℝ is supposed to apply, this probably means that there are uncountably infinitely many valid solutions. However, I may be wrong again. Now I checked that with Wolfram Alpha. If you enter: sqrt(x^(2))=-x, x is real Wolfram Alpha returns x ≤ 0 as the solution. So I’m right. Best regards Marcus 😎

E is true

√x²=-x make √x² into |x| since x² on reals only gives positive values and it gets square rooted |x|=-x Now lets look at the possibilities of this equation x=-x this is only true for 1 -x=-x this is true for all values So it's B

sqrt(x^2) = -x or x = -x or 2x=0 or x = 0 and this is a linear exp as when you remove the fractional powers so answer is 0 only

I'm not keen on doing it by elimination, it doesn't tell you anything, the point is sqrt(x^2) is the same as |x|, so ignoring 0 for a moment, X must be negative to make -x positive. But 0 works too, so X is non positive.

A and B.

Non positive because when you have x squared with square root it came out both positive and negative and in the other hand you have only negative answer so you have to choose non positive

People using calculus and stuff that I don't know to get the answer but I only know how to cancel and integers 😅😢

Bro literally😊

A and B

D

√(x^2)=-x √(2^2)=-2 we times both side by power of two then one two will be finished by square root √(2^2)^2=-2^2 4=4

On one it works

Nonpositive only

@husinamintulu well, at least as far as I can figure... if you attended school like everyone else, you either got held back about 12 times 😢, or have spent a lot of time in Juvenile Hall and then Prison and are getting rehabilitated and making up for lost time, or have been in special ED classes and are a potential candidate for the "special" Olympics, or perhaps dropped out of school to support yourself and your parents. In either of these cases, good luck to you and godspeed. It's a long way to that PhD in math, if that be your goal. The odds of making it aren't in your favor... statistically speaking (sorry about that... you probably haven't taken stats yet). *** SPOILER ALERT*** There isn't a general formula for X^5 polynomials. X^4 is the highest the formulas go. Look up the proof for a mind blowing afternoon! As a little homework assignme, try to prove the Collatz conjecture, and we'll see you in another 25 years.

Umm √( x² ) = | x | and modulus can only be positive or 0 | x | = -x means only negative or 0 It is to be noted that if It were to be √ ( x² ) = -a , where a is constant Then | x | = -a then , modulus will open accordingly and then range will extend to all no.s

After reading some comments i still do not get it. First it is mentioned at the top that x is a real number and except zero it doesn't matter which integer you are inserting in the equation, you will end up with a negative number on the one side. So, it has to be C.

A option

Null set, baby.

8th grade! This is 6th grade introductory stuff.

How is it b? All real numbers work. If its written like that i can see why people say its only nonpositives but if you square both sides you get x^2 = x^2 which is obviously true for all real numbers

It is i mustly say positive

C positive

I didn't know that negative zero was a thing.

Is it still considered review when I took the course 40 years ago? I will say Mr. Barnes would be happy I still remember most of these rules.

X on LHS has the power of x²°½ = x¹ so any number you will put whether be negtive or positive it will become negative If x = 1 Then 1 = -1 so we get -1 If x = -1 (X is already negative) Then -1 = -(-1) ==> -1 = 1.

I think people might struggle a bit with the - (-x) = postive x.

Option B

sqrt(x^2) is (one of) the definitions of the absolute value so: |x|=-x if x > 0: |x|=x=-x => x=0 which contradicts x>0 so x

B.. i think.😊

Rewrite that as x^(2/2) therefore x can only be nonpositive only to equal to =x

A principle root can't be negative. Therefore the answer is x = 0.

Clearly negative numbers work to make the expression valid, but since 0 also works it does not seem valid to say that B is correct because 0 is not considered a nonpositive value. Hence, I think the answer should be E, none of the above. If there was an answer that included 0 and nonpositive numbers then it would be correct but we don’t have that option.

It’s actually E because the correct answer would be real numbers except positive because zero isn’t considered a non-positive number

I think the option a and c are correct al negative numbers are valide to

Non positives only because: Square root of any number CANNOT be negative. So: Square root((-3)^2) = -(-3) = 3 Because: square root((-3)^2)= square root of 9 = 3, NOT (-3).

I sometimes confuse myself by seeing "-x" as a negative number 😓

That's a tough question