Did he really just give people over the internet homework?
@adrielotic
10 ай бұрын
Some front benchers already worked out the sum in their comment notebook and submitted it online 😅
@gamtax
10 ай бұрын
I'm cool with this. His video shorts are like a gym for my brain.
@farahchishty2673
10 ай бұрын
Haha!😂
@michaelparella9407
10 ай бұрын
Yes
@NationalistFirst
10 ай бұрын
Yaa
@mrhtutoring
10 ай бұрын
I will post the answer tomorrow. 😊 The correct answer is B) Nonpositive only
@random19911004
10 ай бұрын
Sqrt(x^2) = abs(x) The left branch of abs(x) is y = -x So answer is nonpositive
@Mr23143sir
10 ай бұрын
I think it May be only 0 because sqrtx^2 = - x and x is every where The same so if we plug fe. - 2 we get that sqrt-2^2 = -(-2) we get that - 2 = 2 which is not correct
@_humanbeing_Homosapien.
10 ай бұрын
What but I studied √+ve number will always be positive and can't be negative 💀 But here👁️
@ahmedbenmbarek9938
10 ай бұрын
a) sqrt((-3)^2)= srqrt (9)= 3 Isn't sqrt defined for positive reel numbers only? Keep in mind this is a grade 8 algebra so no imaginary numbers here
@wernergamper6200
10 ай бұрын
But I already did 🙂
@f.r.y5857
9 ай бұрын
√(a)² = |a| Definition of absolute value: If a>0 or a=0, |a| = a If a
@schioncalzanzi2019
8 ай бұрын
absolute zero
@hetanshthakore5886
7 ай бұрын
but why is square root of (a)² = |a| ?
@turtlemaster-gz6dc
7 ай бұрын
I believe that because the square root does say plus/minus, it’s assumed that it’s the positive square root. In which case the absolute value is used to ensure it’s positive
@SonGoku-ok5lk
6 ай бұрын
@@hetanshthakore5886imagine x equal -2, -2^2 would be 4? So square root of 4 is 2. |-2| would be 2 too. So square root of (x^2) is |x|.
@aeesqu
6 ай бұрын
@@hetanshthakore5886square root result is always positive, by definition
@MiaMonique
10 ай бұрын
I wish I had him in the fourth grade.
@jimmy6535
9 ай бұрын
He's good, Wish I had him as a teacher when i was in school. Would have saved me from hating math back then.
@murphybed7919
10 ай бұрын
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.
@oddlyspecificmath
9 ай бұрын
It's nice to be reminded when these assertions are first posed. Thanks.
@DavidMauas
6 ай бұрын
Your channel has quickly become one of my favorites ! 🎉 Keep it up, please!
@mrhtutoring
6 ай бұрын
Thanks! Will do!
@guidoeduardo
10 ай бұрын
B) Nonpositive only
@radhamroun5304
6 ай бұрын
People should memorize this property: √(x²) = |x| So |x| = - x if and only if x is a negative real number
@wernergamper6200
10 ай бұрын
B) non positive only, because the square root gives you non negative values by definition.
@GiovanniSanguedolce
10 ай бұрын
True
@solutionencryption968
10 ай бұрын
this comment on a t-shirt.
@ronndan2004
10 ай бұрын
Why is aquare root always positive by definition?
@ultrajaywalker
10 ай бұрын
What are you smoking? Sqrts give negative and positive answers. √4= {2,-2}
@GiovanniSanguedolce
10 ай бұрын
@@ultrajaywalker I suggest you return back to school. The sqrt function returns a non-negative value by definition. Without the need of having to smoke or drink anything.
@bartholomew9999
4 ай бұрын
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.
@abdo450gaming3
7 ай бұрын
X = -1 works right?!
@ThatUnknownDude_
6 ай бұрын
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!
@mrhtutoring
6 ай бұрын
👍
@pawebandulet5119
3 ай бұрын
Now also plug in a positive number.
@alqahharmunjilul196
2 ай бұрын
@@pawebandulet5119 plugging in a positive number will not give you -x
@bottlecapbrony366
9 ай бұрын
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.
@marcusgloder8755
9 ай бұрын
For y = |x| there are no negative solutions. |x| is either positive or zero. Additionally, -0 is not a meaningful notation. A negative sign corresponds to multiplication a positive value by -1. The following applies: -1 × 0 = 0. Zero is the only unsigned value on the number line. All other numbers are either positive or negative. The term -x should be reserved for true negative values. This excludes zero a priori. Best regards Marcus 😎
@ravinaashetty
5 ай бұрын
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.
@Valentinathevamp
5 ай бұрын
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.
@UnderscoreYTG
6 ай бұрын
Guessing before looking at the answer: x is equal to any real number
@Alex_Cara77
6 ай бұрын
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.
@mrhtutoring
6 ай бұрын
👍
@nono-bq9be
6 ай бұрын
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.
@Mesa_Mike
3 ай бұрын
The square root symbol is defined as giving only the positive square root. I guess you can have your own private definition of the square root symbol if you want, but I'd avoid that myself.
@darthtardis5465
8 ай бұрын
Since sqrt(x^2) is equal to absval(x), you can simply solve by graphing and you get (-inf, 0]
@3dsaulgoodman
2 ай бұрын
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
@ARS-fi5dp
10 ай бұрын
sqrt(x^2)=|x| If you graph |x| and -x you see two graphes intercept at negative numbers including zero Answer choice is B
@kikilolo6771
10 ай бұрын
if it wasnt a mcq you would lose points
@bsfighter4721
9 ай бұрын
Simplify to x=-x. Solving this is 0 only
@heinrich.hitzinger
6 ай бұрын
@@bsfighter4721That's not the only solution.
@ItsPREPP98
6 ай бұрын
@@heinrich.hitzingerwhat else?
@kirikouvarken9389
6 ай бұрын
@@heinrich.hitzingerbro you’re supposed to say what else 😭
@jayasuryaassassin
10 ай бұрын
Netizens are stunned to see the negative zero.
@unknown-eq8yj
10 ай бұрын
yaaaa .....y zero negative kbse aane laga
@AlfonsoNeilJimenezCasallas
8 ай бұрын
Hint: make graphics and watch the points where these functions intersect P.D.: sqrt(x^2) = absolute value of x
@cegexen8191
5 ай бұрын
me resisting the urge to type an imaginary number
@UnNamedMonster
9 ай бұрын
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.
@robtaylor6806
10 ай бұрын
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.
@mrhtutoring
10 ай бұрын
Wonderful!
@yairkaz
7 ай бұрын
It's just |x|=-x which is for negative numbers only, no need to substitute
@SteveMathematician-th3co
10 ай бұрын
To solve this equation, you only need to find the domain of the equation which is x
@bhaskarsaha6251
10 ай бұрын
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 💕
@user-kj4pn5vy9l
10 ай бұрын
I said the same thing 😊❤👏🏻
@bhaskarsaha6251
10 ай бұрын
@@user-kj4pn5vy9l where are u from bro ?
@user-kj4pn5vy9l
10 ай бұрын
@@bhaskarsaha6251 Kurdistan 🥰
@user-kj4pn5vy9l
10 ай бұрын
@@bhaskarsaha6251 do you know anything about my region?
@bhaskarsaha6251
10 ай бұрын
@@user-kj4pn5vy9l no
@christianmosquera9044
5 ай бұрын
Excellent video wonderful 😊😊😊😊😊❤❤❤❤❤
@sonambhalsolanki5873
9 ай бұрын
Option E). because complex no. "-i" or "1/i" , 0 can satisfy it.
@itachi5271
9 ай бұрын
Bro read the 1st line , x is a real number .......😅
@abdoezzat8427
6 ай бұрын
A is the correct answer Because result of the square root should be positive or zero but if the number is imaginary number
@sonic_force
5 ай бұрын
ur statement is true for this example √-2 but the above equation is √(x²) = -x notice the x². if x = -2 LHS: √(x²) = √(-2)² = √4 = 2 ---> (1) RHS: -x = -(-2) = 2 ----> (2) from (1) and (2) the correct option is (b)
@isjosh8064
10 ай бұрын
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
@finite1731
10 ай бұрын
Does non positive include 0?
@isjosh8064
10 ай бұрын
@@finite17310 is neither positive nor negative so nonpositive does include 0 since it isn’t positive
@realmuru6365
10 ай бұрын
The answer is All Reals
@myster4775
10 ай бұрын
@@realmuru6365How is that?
@random19911004
10 ай бұрын
Just think about it graphically. The left part of abs(x) is -x Left part is where x is nonpositive.
@tadiosbelay2307
13 күн бұрын
hi can you try the proof equation 13 in the paper "new highly anti-interference regularization method for ill-posed problems"
@user-rk4nm9yf7d
6 ай бұрын
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
@tondgvr
6 ай бұрын
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
@FloraLemonYT
5 ай бұрын
Square root of 4 cannot be -2. The definition of “square root” implies only the positive number.
@foxsins314
6 ай бұрын
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!
@thomasdewierdo9325
6 ай бұрын
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.
@trucid2
4 ай бұрын
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
@philrobson7976
4 ай бұрын
Multiply both sides by -1. Then x always equals a non-positive number.
@nizogos
8 ай бұрын
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
@hanslepoeter5167
8 ай бұрын
left hand side is x or -x. The real domain is the correct answer imho.
@nizogos
8 ай бұрын
@@hanslepoeter5167 Left hand side is always positive ( hence abs(x) ),since you can't get a negative output from the square root in the real numbers.
@hanslepoeter5167
8 ай бұрын
@@nizogos Not what I learned. Hey, where do we find the answers to all those questions this teacher poses ?
@Tommybotham
10 ай бұрын
Negatives only. The negative part of the left will automatically equal the right.
@user-iu8uk5tc9s
10 ай бұрын
What about 0? He said 0 works. So negatives and 0 which is nonpositive only
@Tommybotham
10 ай бұрын
@@user-iu8uk5tc9s Yeah you're right I forgot the 0 :)
@weebsplayz
6 ай бұрын
√x² = |x| • |x| = -x Therefore non-positive only ( B )
@mrhtutoring
6 ай бұрын
👍
@luisclementeortegasegovia8603
10 ай бұрын
Professor, to me the answer is C) only positives. Any number with even exponents will always be positive!
@Juwar1974
9 ай бұрын
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.
@xipher5896
8 ай бұрын
agreed
@joe-ib1wn
8 ай бұрын
√25 = √5² = |5| = 5 The result of a square root (while working with real values) is always contained within [0 ; +∞) x² = 25 has two solutions, however. Both -5 and 5 are solutions to x² = 25, but not to x = √25
@clmnyng
7 ай бұрын
thats where im at, is this poorly written or do i need to involve square -1 @@joe-ib1wn
@davidbroadfoot1864
6 ай бұрын
@@clmnyng It's not poorly written, and @joe-ib1wn describe the situation very well.
@maxnolife_
6 ай бұрын
my ahh when I put "x is equal to plus or minus 0"
@5Stars49
10 ай бұрын
A
@gdmathguy
6 ай бұрын
√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
@bradleymartinez4876
7 ай бұрын
A bit tricky!! But if you make the correct decision of where the negative really lies!!
@GameSharkBlue
Ай бұрын
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.
@sarfarazsarkar7551
10 ай бұрын
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
@Aditya_196
9 ай бұрын
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
@fightbleedrepeat
4 ай бұрын
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
@bablema7808
8 ай бұрын
d because every number squared becomes posative then when you squareroot it it becomes itself again and its negative.
@georgewatters8441
7 ай бұрын
That's above my pay grade. I'd go with negetives but can't explain it well.
@user-id6gk4bj2v
10 ай бұрын
√(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
@gibson2623
7 ай бұрын
There he is again doing his wizardry, LoL
@Gonnaberelaxed
10 ай бұрын
That's a tough question
@moeberry8226
7 ай бұрын
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
@ZantierTasa
6 ай бұрын
I sometimes confuse myself by seeing "-x" as a negative number 😓
@ofekn
10 ай бұрын
the left side of the equation is basically |x| which defined to be -x for non positives
@muhammadsaud5696
9 ай бұрын
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.
@rnseby
8 ай бұрын
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.
@marcusgloder8755
9 ай бұрын
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 😎
@wyboo2019
9 ай бұрын
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
@colinjava8447
9 ай бұрын
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.
@Nikioko
5 ай бұрын
A principle root can't be negative. Therefore the answer is x = 0.
@youtubepro5932
10 ай бұрын
Non of the above the multiplication operations is a e log algorithm increment matlab responses from viewers
@slavazinko
10 ай бұрын
The answer is only non-positive numbers
@RonnyS.1900
7 ай бұрын
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.
@Le8h0
3 ай бұрын
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
@f4achamp436
6 ай бұрын
Take x=-a so the equation becomes (sqrt (-a))^2 = -(-a) thus it becomes -a= a which is only possible when a= 0.. thus x =0 is the only solution.. correct me if I'm wrong
@valerimetev5628
6 ай бұрын
B is the right answer Sqrt(-2)² = - (-2)
@DoxxTheMathGeek
4 ай бұрын
sqrt(x²) = |x| = -x Since abs for real x is always a positive number or 0, -x has to be positive or 0 and x negative or 0. So all the negative real numbers and 0.
@kpopalitfonzelitaclide2147
5 ай бұрын
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
@420sakura1
5 ай бұрын
Thus is about making LHS=RHS. Not about solving for X.
@RaquelSantos-hj1mq
10 ай бұрын
I didn't know that negative zero was a thing.
@TheNetkrot
9 ай бұрын
it isn't ... hes bullshitting
@joe-ib1wn
8 ай бұрын
In this context -0 = (-1)*0 = 0 In other contexts, such as calculating limits, -0 may be used to represent an infinitesimal negative number.
@RaquelSantos-hj1mq
8 ай бұрын
@@joe-ib1wn Makes sense that way. Thanks!
@StandDont
7 ай бұрын
People using calculus and stuff that I don't know to get the answer but I only know how to cancel and integers 😅😢
@MichaelZankel
7 ай бұрын
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).
@Kale164
6 ай бұрын
its B since (-3)^2 is 9, √9 is 3, while -(-3) is also 3, so its B.
@leodavinci9682
10 ай бұрын
I think people might struggle a bit with the - (-x) = postive x.
@user-qv2dd8ex8k
6 ай бұрын
Square root of negative numbers results in imaginary numbers. Which we shortened to square root of √(-1) So "A" is the only answer. Another way - You need to know the rules of math here. When we take the square root of a number the rule is we only look at the absolute value of that #. So zero is the only answer. No other solutions. Answer choice "A". Remember|-3| = 3 Means it can NEVER BE NEGATIVE.
@idontmine7215
9 ай бұрын
Rewrite that as x^(2/2) therefore x can only be nonpositive only to equal to =x
@BlackSakura33
9 ай бұрын
8th grade! This is 6th grade introductory stuff.
@hasanalmonem5713
9 ай бұрын
root(x^2)=x^2^1/2=x so, x=-x 2x=0 so x=0
@tarekh123
8 ай бұрын
I'm not completely sure now, if x=-3, is it √((-3)²) or √(-3²) ?
@GiovanniSanguedolce
10 ай бұрын
Nonpositive only
@oscarmartinpico5369
6 ай бұрын
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.
@davidbroadfoot1864
6 ай бұрын
None of that makes any sense at all. For one thing, the RHS is always "-x" ... not x and -x.
@oscarmartinpico5369
6 ай бұрын
@@davidbroadfoot1864 root_2(2^2) = 2 and -2, as root_2(2^2) = -2 , x still equals 2. root((-3)^2) = 3 and -3, as root_2((-3)^2) = 3, x still equals -3. root_k(a^k) = a·e^(i·2·pi·n/2), n = 0,1, ... k-1. But what does RHS mean?
@oscarmartinpico5369
6 ай бұрын
@jash21222 ok. I need clarify the notations. It happens the same with "-"
@oscarmartinpico5369
6 ай бұрын
@jash21222 So, what does it output root_5(32)? I wold say 2, 2(cos(2/5·pi) + i·sin(2/5·pi)), 2(cos(4/5·pi) + i·sin(4/5·pi)), 2(cos(6/5·pi) + i·sin(6/5·pi)) and 2(cos(8/5·pi) + i·sin(8/5·pi)). So, I do not not how to set +/- in this operation, but if you tell me I have to choose the main value, which is 2, what would be the main value of root_5(32+i·64), for example. That is the reason I say that the convenctions has to be clarify first and we shouldn't take for acepted the one you use to take. root_2(4) = 2 and -2 because 2^2=4 and (-2)^2 = 4. The same happens for the order of the opertations in a large mathematical expresion.
@davidbroadfoot1864
6 ай бұрын
@@oscarmartinpico5369 That is not a clarification. Re "It happens the same with '-'" ... What is "it"? What happens the same? Putting "-" where? You are being totally unclear. Please do maths ... not word salad.
@ashwanibeohar8172
9 ай бұрын
It is last choice(D) non of above, as the ans is zero and all non positive or n-ve numbers.
@user-cv9ov4fu9c
3 ай бұрын
D)Non positive only
@anantswaroop.college
10 ай бұрын
Brain stopped functioning. 😂
@theLaboratoryofScience
7 ай бұрын
If X²½ = -x , X²½ = -1•x ; 2x= -1 ; ∴ x=-0.5 is that true?, if I have done mistake, please teach me kindly.
@schockmetamorphose7729
6 ай бұрын
It is D because the root has two solutions
@davidguy9197
10 ай бұрын
le seul nombre qui mes semble pouvoir donner une racine négative est un nombre complexe, mais pour la résolution d'une équation racine (x)= x 1 pourrais correspondre mais les solution de racine de x = 1 donnerais 1 et -1 mais par convention il me semble que la racine est toujours dans R positive
@marcossanchezburgardt9253
4 ай бұрын
I think the option a and c are correct al negative numbers are valide to
@somebody_33
8 ай бұрын
The square root of x^2 is |x|. |x| is -x if x is less than zero, and x if x is greater than zero. In the equation |x| = -x, only negative values of x (and zero) will satisfy this equation. Therefore, *B* is the correct answer.
@joe-ib1wn
8 ай бұрын
0 also works, the answer is x ∈ (−∞ ; 0]
@a.r.9822
5 ай бұрын
Just a quick question When you square root something isn’t the answer positive or negative. For example the square root of 4 isn’t just positive 2 but also negative 2
@mrhtutoring
5 ай бұрын
By definition, square roots of 4 are ±2. But when we use the radical sign, √, we only use the positive root or the principal root. Meaning √4 is only equal to +2.
@a.r.9822
5 ай бұрын
@@mrhtutoring thanks a lot
@agnidas5816
7 ай бұрын
square root of a number is plus minus x
@IddoKeren-yo3tm
8 ай бұрын
Sqrt(x^2) = |x|. So -x = |x|. |x| = -x when x 《 0. So the answer is B.
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