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
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
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.
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!
@@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.
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.
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.
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.
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.
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 😎
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.
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 💕
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.
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
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.
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!
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
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
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.
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
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
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.
√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
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 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?
@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.
@@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.
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)
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.
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
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
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 😎
sqrt(x²)=-x ±x=-x x=0??? what did I do wrong? edit: i forgor no complex. sqrt(x²)=-x we can assume x=|x| because otherwise the result would be a comolex number. because x=|x|, this doesnt matter at all since sqrt(|x|²) is still ±x edit2: plugging random shit into the equation. sqrt(-1²)=1 sqrt(-1)=1 is incorrect. sqrt(0²)=0 is correct sqrt(1²)=-1 sqrt(1)=±1, not -1, incorrect. edit3: apparently sqrt(x²)=|x|????? why??????? its ±x??? if sqrt(x²) were |x| sqrt(-7²)=7, -7=7, 0=14?? edit4: after reading the comments some more, i found out it depends on if your opinion on what sqrt(x²) is is wrong or not.
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
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.
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.
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.
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Did he really just give people over the internet homework?
Some front benchers already worked out the sum in their comment notebook and submitted it online 😅
I'm cool with this. His video shorts are like a gym for my brain.
Haha!😂
Yes
Yaa
I will post the answer tomorrow. 😊
The correct answer is B) Nonpositive only
Sqrt(x^2) = abs(x)
The left branch of abs(x) is y = -x
So answer is nonpositive
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
What but I studied √+ve number will always be positive and can't be negative 💀
But here👁️
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
But I already did 🙂
√(a)² = |a|
Definition of absolute value:
If a>0 or a=0, |a| = a
If a
absolute zero
but why is square root of (a)² = |a| ?
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
@@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|.
@@hetanshthakore5886square root result is always positive, by definition
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.
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.
B) Nonpositive only
Your channel has quickly become one of my favorites ! 🎉 Keep it up, please!
Thanks! Will do!
X = -1 works right?!
Excellent video wonderful 😊😊😊😊😊❤❤❤❤❤
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!
👍
Now also plug in a positive number.
@@pawebandulet5119 plugging in a positive number will not give you -x
B) non positive only, because the square root gives you non negative values by definition.
True
this comment on a t-shirt.
Why is aquare root always positive by definition?
What are you smoking? Sqrts give negative and positive answers. √4= {2,-2}
@@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.
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.
Netizens are stunned to see the negative zero.
yaaaa .....y zero negative kbse aane laga
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.
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.
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.
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 😎
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.
Wonderful!
sqrt(x^2)=|x|
If you graph |x| and -x you see two graphes intercept at negative numbers including zero
Answer choice is B
if it wasnt a mcq you would lose points
Simplify to x=-x. Solving this is 0 only
@@bsfighter4721That's not the only solution.
@@heinrich.hitzingerwhat else?
@@heinrich.hitzingerbro you’re supposed to say what else 😭
D
Since sqrt(x^2) is equal to absval(x), you can simply solve by graphing and you get (-inf, 0]
B
Option E). because complex no. "-i" or "1/i" , 0 can satisfy it.
Bro read the 1st line , x is a real number .......😅
A bit tricky!! But if you make the correct decision of where the negative really lies!!
People should memorize this property: √(x²) = |x|
So |x| = - x if and only if x is a negative real number
Or zero
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
left hand side is x or -x. The real domain is the correct answer imho.
@@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.
@@nizogos Not what I learned. Hey, where do we find the answers to all those questions this teacher poses ?
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 💕
I said the same thing 😊❤👏🏻
@@user-kj4pn5vy9l where are u from bro ?
@@bhaskarsaha6251 Kurdistan 🥰
@@bhaskarsaha6251 do you know anything about my region?
@@user-kj4pn5vy9l no
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.
👍
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.
A
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.
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
me resisting the urge to type an imaginary number
Hint: make graphics and watch the points where these functions intersect
P.D.: sqrt(x^2) = absolute value of x
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.
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!
Professor, to me the answer is C) only positives. Any number with even exponents will always be positive!
the left side of the equation is basically |x| which defined to be -x for non positives
Negatives only. The negative part of the left will automatically equal the right.
What about 0? He said 0 works.
So negatives and 0 which is nonpositive only
@@user-iu8uk5tc9s Yeah you're right I forgot the 0 :)
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
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
Thus is about making LHS=RHS. Not about solving for 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.
A and B.
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
Does non positive include 0?
@@finite17310 is neither positive nor negative so nonpositive does include 0 since it isn’t positive
The answer is All Reals
@@realmuru6365How is that?
Just think about it graphically.
The left part of abs(x) is -x
Left part is where x is nonpositive.
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
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
Square root of 4 cannot be -2. The definition of “square root” implies only the positive number.
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.
A and B
To solve this equation, you only need to find the domain of the equation which is x
Multiply both sides by -1. Then x always equals a non-positive number.
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.
agreed
√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
thats where im at, is this poorly written or do i need to involve square -1 @@joe-ib1wn
@@clmnyng It's not poorly written, and @joe-ib1wn describe the situation very well.
d because every number squared becomes posative then when you squareroot it it becomes itself again and its negative.
Non of the above the multiplication operations is a e log algorithm increment matlab responses from viewers
Bro literally😊
D)Non positive only
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.
None of that makes any sense at all. For one thing, the RHS is always "-x" ... not x and -x.
@@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?
@jash21222 ok. I need clarify the notations. It happens the same with "-"
@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.
@@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.
E is true
A is the correct answer
Because result of the square root should be positive or zero but if the number is imaginary number
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)
C positive
√x² = |x|
• |x| = -x
Therefore non-positive only ( B )
👍
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.
my ahh when I put "x is equal to plus or minus 0"
There he is again doing his wizardry, LoL
On one it works
Yep it's true I like math
Because x² is in a square root, -x has to be >=0 so x=
Guessing before looking at the answer:
x is equal to any real number
That's above my pay grade. I'd go with negetives but can't explain it well.
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
B! B! I choose B!
It is i mustly say positive
A option
√(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
People using calculus and stuff that I don't know to get the answer but I only know how to cancel and integers 😅😢
hi can you try the proof equation 13 in the paper "new highly anti-interference regularization method for ill-posed problems"
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
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
B.. i think.😊
Option B
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 😎
D is the anwser 👌
I think the option a and c are correct al negative numbers are valide to
Zero and -1.
Option B nonpositive only.
Yes~
I think people might struggle a bit with the - (-x) = postive x.
Rewrite that as x^(2/2) therefore x can only be nonpositive only to equal to =x
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
sqrt(x²)=-x
±x=-x
x=0???
what did I do wrong?
edit: i forgor no complex.
sqrt(x²)=-x
we can assume x=|x| because otherwise the result would be a comolex number.
because x=|x|, this doesnt matter at all since sqrt(|x|²) is still ±x
edit2: plugging random shit into the equation.
sqrt(-1²)=1
sqrt(-1)=1 is incorrect.
sqrt(0²)=0 is correct
sqrt(1²)=-1
sqrt(1)=±1, not -1, incorrect.
edit3: apparently sqrt(x²)=|x|????? why??????? its ±x??? if sqrt(x²) were |x| sqrt(-7²)=7, -7=7, 0=14??
edit4: after reading the comments some more, i found out it depends on if your opinion on what sqrt(x²) is is wrong or not.
B) non +ve only
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
A principle root can't be negative. Therefore the answer is x = 0.
I going with D
I want him as my tr fr
It is D because the root has two solutions
We can also cancel square with root
The square of a root, yes. but not necessarily the root of a square. You have to be careful.
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.
Positive numbers are strictly greater than zero, so zero is non-positive.
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.
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.