Find all x ( square-root equation)

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@m.h.6470 3 ай бұрын

Solution: √(x²) = |x| As such, the equation can be written as: |x + 3| + |x - 2| + |2x - 8| = 9 The critical points of this equation are: x < -3, because all terms will be negative x < 2, because the second and third term will be negative x < 4, because the third term will be negative as soon as x ≥ 4, all terms will be positive case x < -3: -(x + 3) + -(x - 2) + -(2x - 8) = 9 -x - 3 - x + 2 - 2x + 8 = 9 -4x + 7 = 9 |-7 -4x = 2 |:-4 x = -1/2 → not in valid region, therefore no solutions case -3 ≤ x < 2: x + 3 + -(x - 2) + -(2x - 8) = 9 x + 3 - x + 2 - 2x + 8 = 9 -2x + 13 = 9 |-13 -2x = -4 |:-2 x = 2 → not in valid region, therefore no solutions case 2 ≤ x < 4: x + 3 + x - 2 + -(2x - 8) = 9 2x + 1 - 2x + 8 = 9 9 = 9 → valid, so any value 2 ≤ x < 4 is a solution case x ≥ 4: x + 3 + x - 2 + 2x - 8 = 9 4x - 7 = 9 |+7 4x = 16 |:4 x = 4 → valid So the valid solutions are x = [2, 4]

@platypi_otbs 3 ай бұрын

I set the equation up wrong so I watched the video instead of finding the answer.

@fredesch3158 3 ай бұрын

How's your handwriting so beautiful?? Dude, congrats, I find it so cool when people have good handwriting in blackboards, it's so hard, and you make it seam so easy!

@MikeGz92 3 ай бұрын

In A region, there is a sign error. The reduced equation is not -4x-7=9, but -4x+7=9 that gives x=-2/4=-1/2, which don't belong to the interval x

@shmuelzehavi4940 2 ай бұрын

That's right. There are actualy two errors and the second error cancels the first one (x = - 4 would be a valid value if it were arithmetically correct).

@nothingbutmathproofs7150 3 ай бұрын

Once you get a possible answer for x, if it is in region that you are working in then that x value will be a solution. If the x value you got is not in the the interval that you are working in, then that x-value is not a solution. For region A, you got x=-4. Since x is region A, then x=-4 will work. Since it didn't work you need to go back and find your mistake. Remember, doing arithmetic in public is not a good move!

@tanguc007 3 ай бұрын

THere is a typo in Testing A solution. Can we check it again..

@keithrobinson2941 3 ай бұрын

+7, I think, on the top line for region A. So we test -½, which comes to 14, not 9; so it doesn't work anyway.

@Mrcasgoldfinch 3 ай бұрын

I made the same error too before watching the video, there must be some active spots on the Sun today:-) Or a quantum entaglement...

@davidgagen9856 3 ай бұрын

It's +7

@Grecks75 3 ай бұрын

Solution: The equation with radicals can be equivalently written in terms of absolute values: |x + 3| + |x - 2| + 2|x - 4| = 9. I suppose we're solving for real numbers x. Then we have 4 cases to consider in total, related to the branches of the absolute value function: (A) x < -3, (B) -3

@larswilms8275 3 ай бұрын

If we suppose that x is a complex number, that would mean that x = a + bi. For the real part of the equation the critical points are the same and the solution are the same. for the imaginary part (+bi) for all the radicals the critical point is b = 0. there for either all the radicals are positive, negative or 0. So they either all ad up subtract or do not exist. In the first two cases we will always be left with a imaginary residual. Therefore only the case where b = 0 will give valid solutions and we can reduce the problem to the real numbers.

@Grecks75 3 ай бұрын

@@larswilms8275 Sorry, but this is not correct. I was pretty sure there exist complex solutions with a non-zero imaginary part, and it turned out to be true on further analysis. Here's one: x = 2 + i. If we take the usual definition of the single-valued complex square root function (as the principal branch with non-negative real part), we have: sqrt((x + 3)^2) = x + 3 = 5 + i, sqrt((x - 2)^2) = x - 2 = i, sqrt((2x - 8)^2) = -(2x - 8) = 4 - 2i. Their sum is exactly 9 + 0i as needed for the equation to hold. Of course, there are many other solutions with non-zero imaginary part, and what makes it worse is that the solution set isn't even bounded, contrary to the real numbers case! (Because the imaginary part of a solution can grow without bounds.) The situation in the complex plane is a lot more complicated: The sqrt() function takes on complex values (by the way, it is well-defined for _all_ complex numbers), we have to decide on a branch to make it single-valued, the relation sqrt(x^2) = |x| does not hold anymore (for the complex modulus), we cannot make a case-by-case analysis as easy as in the real numbers situation, because complex numbers cannot be ordered, etc. etc.

@Grecks75 3 ай бұрын

@@larswilms8275 I now have a description of the full solution set in the complex plane: Solutions have the form x = a + ib with the real part a taking values from the closed interval [2, 4] (as in the real numbers case) and the imaginary part b being any real number if 2 < a < 4, or any non-negative real number (b >= 0) if a = 2, or any non-positive real number (b

@StaR-uw3dc 3 ай бұрын

While Testing A we get -4x+7=9 i.e. x=-1/2 which is out of region A (x

@georgesbv1 3 ай бұрын

actually the absolute value is contiguous in 0 as well. So he can bend the rules on those intervals.

@StaR-uw3dc 3 ай бұрын

@@georgesbv1 I agree, but it would be more clear when the regions be disjoint but covering the whole axis.

@Psykolord1989 3 ай бұрын

Before watching: So, we have to test 6 cases. See, √(a+b)^2 can be either a+b or -a-b. It's probably easiest for us to write this using piecewise functions. First, let's find the spots where any term equals 0. We do this because these will be our critical points. For √(x+3)^2, this point is at x=-3. We will have -x-3, x< -3, or x+3, x≥-3. We will see similar piecewise functions for √(x-2)^2, giving us (-x+2, x

@pierreneau5869 2 ай бұрын

Thanks to share such exercise. I'm surprised by the writing x = [2;4]. Should be more correct x € [2;4]

@shmuelzehavi4940 2 ай бұрын

That's right. Or, {x} = [2 , 4].

@0lympy 3 ай бұрын

Normally you don't need to validate solution by putting it into the original equation. As all transformation were equivalent within the regions given, there is no way to get any side roots. So you only need to check, if the solution fits the current region. Thus, solution A is wrong as it doesn't fit (-INF; -3) region due to sign mistake (+7 / -7). Or else, for consistency, you had to check the whole [2;4] region. It was a challenge for me to get, where did the side root -4 in region A came from :D

@bhagyashrigadekar8618 3 ай бұрын

I liked and commented video first 😎 And that point 2.0000001 made me laugh 😂😂

@jay_13875 3 ай бұрын

-x-3-x+2-2x+8 is -4x+7, not -4x-7. The solution of -4x+7 = 9 is x=-1/2, which is outside the interval A (x

@carterhyde908 3 ай бұрын

As nice as it is watching a more edited channel's flow, it feels more impactful having all the little mistakes left it. Makes my errors seem that much more reasonable.

@ibrahimkonefilsdiarrassoub5736 3 ай бұрын

❤ you make me fall in love with math, everyday I watch your videos ❤...Thanks ❤

@BartBuzz 3 ай бұрын

Such a straight forward solution.

@hacerkayal1740 3 ай бұрын

Sir, your solution style is so cool❤ thanks❤

@srisaishravan5512 3 ай бұрын

Sir could you please start uploading videos on integration, your videos got me into calculus and i am unable to find such amazibg videos so please start uploading calculus

@alansun70 3 ай бұрын

It cancels if the square is outside the radical.

@tezeralorisso4823 3 ай бұрын

The solution A is wrong twice. 1st: instead of -7 you used +7 2nd: the solution x=-4 fits the condition . ie, you were testing the region where x≤-3 Hence -4≤-3 so that it was a good sign to check.

@Pigemerzyt 3 ай бұрын

There is no need to testing after drawing line just place iniquities on A,B,C,D, after this coman area is answered

@tcmxiyw 2 ай бұрын

The solution when solving over region A is x=-1/2, which you can throw out without testing because -1/2 is not in region A.

@Metaverse-d9f 3 ай бұрын

Use triangle inequality. abs(x)+abs(y)+abs(z)≧abs(x+y+z),when taking the equal sign, every term has the same sign. The answer will appear very quickly.

@surendrakverma555 3 ай бұрын

Thanks Sir 🙏🙏🙏🙏

@ruchirgupta610 3 ай бұрын

I am a big fan of you, brother!

@allanmarder456 3 ай бұрын

Rewrite the equation as abs(x+3) +abs(x-2) +2*abs(x-4) - 9=0. For x=>4 the equation becomes x+3 +x-2 +2x-8 -9 =0 or 4x=16 So x=4 is the only solution and no solution for x>4. For x>=0 and x==0 and x=-3 and x

@ThePayner11 3 ай бұрын

Am I right in thinking the solutions are an inequality where 2 ≤ x ≤ 4?

@maxvangulik1988 3 ай бұрын

|x+3|+|x-2|+2|x-4|=9 case 1: x>=4 4x-7=9 4x=16 x=4 case 2: 2

@dan-florinchereches4892 3 ай бұрын

A modulus question. Not bad. I X

@jakehobrath7721 3 ай бұрын

This would be simpler and easier to follow if you took the pivot points and arranged them as such: x

@ssaalktbi2970 3 ай бұрын

You teach algebra?

@ryhorabramovich1457 3 ай бұрын

I have shorter solution. |x+3|+|x-2|+|2x-8|=9 Since |x|=|-x|, then |x+3|+|x-2|+|8-2x|=9 Further, |a|+|b|+|c|=a+b+c a>=0, b>=0 and c>=0 (vectors' property). Because x+3+x-2+8-2x=9, then x+3>=0 AND x-2>=0 AND 8-2x>=0, and answer is x in [2; 4]

@shmuelzehavi4940 2 ай бұрын

You still have to prove uniqueness.

@hervesergegbeto3352 3 ай бұрын

Merci professeur

@mdashrafulahmed2820 3 ай бұрын

isnt this a reupload

@georgesbv1 3 ай бұрын

problem yes, solution is different since he ordered the switch points and obtained fewer intervals.

@eowmob 3 ай бұрын

Like your videos, your explanations and presentation. But, do me as a mathematician a favor: Don't write x=[2,4] x is a scalar, a real number and [2,4] an interval or a set. Write sth. x element in [2,4] (does this work: x ∈ [2,4]) or the set of the solutions is [2,4]. Thx ;-)