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@ShittuMathematicsClass01 28 минут бұрын
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@b213videoz 14 сағат бұрын
I already solved it yesterday: + and - 2, + and - sqrt(5) * i
@ShittuMathematicsClass01 12 сағат бұрын
Great 👍
@ShittuMathematicsClass01 Күн бұрын
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@ShittuMathematicsClass01 2 күн бұрын
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@user-ny6jf9is3t 2 күн бұрын
α=0,α=0,α=ριζα2, α=-ριζα2
@nasrullahhusnan2289 2 күн бұрын
x⁴+x²-20=0 is a quadratic equation in x² which can be factored as (x²-4)(x²+5)=0 x²-4=0 --> x=±2 x²+5=0 --> x=±isqrt(5)
@ShittuMathematicsClass01 2 күн бұрын
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@samueladler9080 3 күн бұрын
Plug in the x valve and let see if it balances.
@ShittuMathematicsClass01 3 күн бұрын
This is not an equation that will need to do substitution
@devondevon4366 3 күн бұрын
9
@ShittuMathematicsClass01 4 күн бұрын
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@PrithwirajSen-nj6qq 4 күн бұрын
One more solution Let log x base3 =m Hence x =3^m Then the given equation will be (3^m)^m =81=3^4 >( 3)^m^2 = 3^4 >m^2 =4 > m = +2/- 2 So when m= 2 log x base 3= 2 then x =3^2=9 When m = - 2, then log x base 3= - 2 x = (3)^( - 2)=1/(3)^2=1/9 Comment please.
@ShittuMathematicsClass01 4 күн бұрын
That's great 👍
@ShittuMathematicsClass01 4 күн бұрын
That's great 👍
@PrithwirajSen-nj6qq 4 күн бұрын
x ^log x(base 3) =81 Taking log of both sides logx*logx=log3^4 (logx)^2=4(base is 3 in all cases) logx =+2 and logx=-2 Then x=3^2 and 3^( - 2) x = 9 and 1/3^2=1/9 Comment please.
@ShittuMathematicsClass01 4 күн бұрын
Great stuff
@kevinmadden1645 3 күн бұрын
Both 9 and 1/9 satisfy the original equation .Consequently both are solutions.
@ShittuMathematicsClass01 4 күн бұрын
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@PrithwirajSen-nj6qq 5 күн бұрын
Fine
@user-ee7nw2rx9s 5 күн бұрын
А почему не логарифмировать по базе 3?
@ShittuMathematicsClass01 5 күн бұрын
You're free to do that too Prof.
@markwinfield1679 5 күн бұрын
@@ShittuMathematicsClass01 log_{3}x^{log_{3}x} = log_{3}81 (log_{3}x)^2 = 4 log_{3}x = \pm 2 x = 3^{\pm 2} simple
@ShittuMathematicsClass01 5 күн бұрын
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@user-ee7nw2rx9s 6 күн бұрын
Решение в уме Ответ 2.5+sqrt 6
@ShittuMathematicsClass01 6 күн бұрын
Welldone prof 👍
@ShittuMathematicsClass01 7 күн бұрын
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@user-ee7nw2rx9s 7 күн бұрын
Если х^2-у^2=9, то (х-у) (х+у) =9 Возводим в квадрат и пусть а=(х+у)^2 (х-у)^2*(х+у)^2=81 (х^2+у^2-2ху+2ху-2ху) *а=81 ((х+у)^2-4ху)*а=81 (а-4*3)*а=81, потому что ху=3 Легко решить уравнение Решение от силы 2 минуты, а не 9 и хватит одно и тоже решать. Куча видео на один пример, и позорные решения
@ShittuMathematicsClass01 7 күн бұрын
I don't see your solution here, drop your solution that takes just 2 minutes so we can learn from you, Professor 😁.
@michaelcolbourn6719 7 күн бұрын
4^x + 4^x = 2(4^x) = 16 4^x = 8 2^2x = 2^3 2x = 3 x=1.5 4 lines of work. 15 seconds. No logarithms. Dont overcomplicate things. Using log base 2 would have made this method simpler too.
@ShittuMathematicsClass01 7 күн бұрын
Thank you 🙏🏻, thanks for watching.
@gislayllsondias7022 8 күн бұрын
Is the final negative value a valid solution to the question? Thank you, professor!
@ShittuMathematicsClass01 7 күн бұрын
The two solutions are valid. Since they are both reel numbers.
@khaledjubeh8818 8 күн бұрын
nice problem ✈️ I did the same but first i took root(2) and root(5) as a common factors which simplified the problem it became [root(2) + root(5)] / [root(2) - root(5)] then I did the same as the video
@ShittuMathematicsClass01 8 күн бұрын
That's beautiful ❤
@ShittuMathematicsClass01 8 күн бұрын
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@hassegreiner9675 9 күн бұрын
You move too slowly with the algebra to maintain my focus
@ShittuMathematicsClass01 9 күн бұрын
Good to hear
@RexxSchneider 9 күн бұрын
4^x + 4^x = 16. So 4^x = 8. 4 = 2^2, so 4^x = 2^2x = 8 = 2^3. Since 2^2x = 2^3, we must have 2x = 3 or x = 3/2. Mental arithmetic. That should not take five minutes.
@ShittuMathematicsClass01 10 күн бұрын
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@sumantchavan2753 10 күн бұрын
If it is multiplied and divided by denominator then what will happen?
@user-ee7nw2rx9s 10 күн бұрын
Два вопроса 1. Зачем вообще логарифмировать если 4=2^2, 8=2^3 2^(2х)=2^3 2х=3, х=3/2 Вопрос 2 Чем обусловлено выбор базы 10, почему не 150 или 7 или е. Если уже логарифмировать то по базе 2. Не позорьтесь решением
@ShittuMathematicsClass01 10 күн бұрын
Thank you 1. To do something differently 2. You can use any form of log (with any base you feel comfortable with) Thank you 🙏🏻 ❤️
@user-is9vx5ft2s 11 күн бұрын
Copy paste question
@ShittuMathematicsClass01 11 күн бұрын
May be copy and paste video 😂 Must you talk if you don't have anything to say?
@roger7341 12 күн бұрын
27 in the denominator requires that y-x=3 and 32 in the numerator requires that y-2x=5. Thus, x=-2 and y=1. Check: (4/3)^2*(2/3)=32/27 And it follows that 3y-2x=3+4=7.
@ShittuMathematicsClass01 12 күн бұрын
Beautifully done
@SuhbanIo 12 күн бұрын
This is called rationalising the denominator
@ShittuMathematicsClass01 12 күн бұрын
Exactly
@ShittuMathematicsClass01 12 күн бұрын
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@ShittuMathematicsClass01 13 күн бұрын
Like and follow my KZbin channel as it helps me alot ❤❤
@ShittuMathematicsClass01 14 күн бұрын
You're not aware that we introduce variables why solving equations? The x and y are only introduced to get the value of a.
@user-ee7nw2rx9s 14 күн бұрын
Подход не верный. Как я понял ввели обозначение х, у. А кто сказал что х, у обязательно целые числа? То есть, если например 39=2*19,5=(-2)*(-19,5) То х+у=19,5, а х-у=2 х=10,75 у=8,75 Единственное что можно сказать из условий что х,у положительные, но натуральными они быть не обязаны вовсе.
@user-ee7nw2rx9s 14 күн бұрын
Устно 39=64-25 64=8^2 25=5^2 Тогда очень легко (sqrt x) /2=2 Sqrt x=4 x=16 Вы хотите сказать что х=8 является одним из корней? По крайней мере так показано в видео 6:40. Подставьте и убедитесь что это не верно
@ceciliafreitas7856 14 күн бұрын
X=1.430
@ShittuMathematicsClass01 14 күн бұрын
Great
@ceciliafreitas7856 14 күн бұрын
@@ShittuMathematicsClass01 Thanks you. Hugs from 🇵🇹
@francescaamadeiaccardo363 14 күн бұрын
Nice job👍
@ShittuMathematicsClass01 14 күн бұрын
Thank you 🙏🏻
@ShittuMathematicsClass01 14 күн бұрын
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@culater 15 күн бұрын
Great 👍
@ShittuMathematicsClass01 15 күн бұрын
Thank you 🙏🏻
@tanveer_badar_ 15 күн бұрын
1 and 0 both satisfy this equation. Took 5 seconds to solve. 💪
@ngolokante4271 15 күн бұрын
and 4
@ShittuMathematicsClass01 14 күн бұрын
4 is also a solution
@culater 15 күн бұрын
NICE 👍
@ShittuMathematicsClass01 15 күн бұрын
Thank you 🙏🏻
@culater 15 күн бұрын
SWEET ! 👍
@ShittuMathematicsClass01 15 күн бұрын
Thank you 🙏🏻
@ShittuMathematicsClass01 16 күн бұрын
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@ShittuMathematicsClass01 16 күн бұрын
Drop your comments here 💧
@ShittuMathematicsClass01 16 күн бұрын
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@ShittuMathematicsClass01 18 күн бұрын
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@ShittuMathematicsClass01 18 күн бұрын
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@allozovsky 19 күн бұрын
And may the other solution *x = 2* be considered as a solution to this irrational equation over the field of complex numbers, where the square root is multivalued and can take on any of its two mutually opposite values?
@mr.d8747 19 күн бұрын
*The (principal) square root of a number is usually understood to be the nonnegative number that squares to the number itself. If you look at the graph of √(x+2) and the graph of -x (for example, in GeoGebra), then you can see that the √(x+2) graph is basically the x² graph rotated by 90 degrees (a horizontal half-parabola, to be precise) with the starting point (minimum) at the coordinates (-2: 0). You can see that the two graphs only meet at the coordinates (-1; 1), the point (2; -2) is where the other half of the parabola would have crossed the -x function. In this context, -2 isn't the square root of 4, 2 is. The number 2 is a solution to the quadratic equation x² - x - 2, but it is conidered a **_false root_** of the equation √(x+2) = -1. You can't just say that 4 = 4, take the square root on both sides and say -2 = 2.*
@allozovsky 19 күн бұрын
Thank you for your thorough, detailed and well-formateed answer. I appreciate it. I have carefully studied it and I have some follow-up questions.
@allozovsky 19 күн бұрын
You say that *"The (principal) square root of a number is **_usually_** understood to be the nonnegative number that squares to the number itself",* but what do mean "usually", are there any exclusions where this isn't the case?
@allozovsky 19 күн бұрын
Thank you for the reference to GeoGebra - never happened to use it before, very convenient, much more convenient than Desmos (in a mobile browser).
@ShittuMathematicsClass01 19 күн бұрын
Yes it will definitely be a solution over a complex plane.
@mr.d8747 19 күн бұрын
*Since the left hand side is a root, it must be greater than or equal to zero. So:* *√(x+2) ≥ 0 /()²* *x+2 ≥ 0 /-2* *x ≥ -2.* *But if the left hand side is greater than or equal to 0, that means the right hand side is also greater than or equal to 0:* *-x ≥ 0 /÷(-1)* *x ≤ 0.* *So if -2 ≤ x ≤ 0, that means x is in the interval [-2; 0]. Now we can solve the equation:* *√(x+2) = -x /()²* *x+2 = x² /-x²* *-x² + x + 2 = 0.* *This quadratic equation's solutions are 2 and -1, but only -1 is in the interval [-2; 0] and thus the only solution is x=-1.*
@ShittuMathematicsClass01 19 күн бұрын
Great 👍 👍