High School Math Practice Test | Be Careful!
Вставка
- Опубліковано 12 січ 2025
- A great and tricky equation today, be careful. What do you think about this question? If you're reading this ❤️.
My second math channel : / @relaxingmath
Thank you for your support💖
✅️Check out my latest videos:
🔴 Can you solve this Cambridge Entrance Exam Question?
• Can you solve this Cam...
🔴 Can You Pass Harvard University Entrance Exam?
• Can You Pass Harvard U...
🔴 Everything is possible in math: • Everything is possible...
🔴 Can You Pass Harvard's Entrance Exam?: • Can You Pass Harvard's...
Hello My Friend ! Welcome to my channel. I really appreciate it!
@higher_mathematics
#maths #math
OK, that's a cubic equation in disguise whose real solution x1=2
Division by (x-2) yields x²-8x+32 which has the complex solutions x3,4 = 4 +/- 4i, if I did not miscalculate.
Not sure where you went wrong, but the x² - 8x + 32 is not correct. It should be x² - 4x + 8. Perhaps your cubic was incorrect? That should have been
x⁴ = x⁴ - 16x³ + 96x² - 256x + 256
which simplifies to
x³ - 6x² + 16x - 16 = 0
At 1:44 transform this to ((x-4)/x)^4 =1
Now take 4th root on both sides.
((x-4)/x) = 1, -1, i, -i
so x =4 +(+x, -x, ix, -ix)
first case is spurious the other three give the three sokutions
x=4-x .... x=2
x= 4+ix
x= 4/(1+i) = 4(1-i)/(1²-i²)
x= 2-2i
similarly replacing i with -i
gives
x=2+2i
lntroduce y=x-2 to get (y+2)^4-(y-2)^4=16y^3+64y=0, or
y(y^2+4)=0 => y=0, ±2i => x=2, 2±2i
The complex roots can also be expressed as 8^(1/2)*e^(i*pi/4) and 8^(1/2)*e^(-i*pi/4).
4 is not solution, the given equation is equivalent to (x/(x - 4))^4 = 1, so is equivalent to x/(x - 4) = u, with u among the solutions of z^4 = 1.
It gives x = (4.u)/(u -1), so u = 1 is eliminated, we have 3 solutions: x = (4.i)/(i-1) = 2 - 2.i or x = (4.-1)/(-1-1) = 2 or x = (4.-i)/(-i-1) = 2 + 2.i
@TheMathManPrifundities did all correct. In summing up, the expression x⁴ - (x - 4)⁴ reduces to a cubic equation on simplification by cancelling x⁴ terms on either side when
(x-2)(x² - 4x + 8) gives 01 real root & 02 complex conjugate roots. This is an equation in which the highest degree of x is unnecessarily put in either side as it is unretainable on simplification; hence, it should not add a 4th. root.
Thanks for the analysis! I need some advice: I have a SafePal wallet with USDT, and I have the seed phrase. (alarm fetch churn bridge exercise tape speak race clerk couch crater letter). How should I go about transferring them to Binance?
What about the fourth solution?
If x is on the fourth power, then it should have four answers, right?
the fourth degree disappear if you write it out, (x-4)^4-x^4 is a third degree polynomial
@yuezie I see, but on a test, wouldn't they say that "you didn't find the fourth answer" ?
@@tamasstrengher6407There is no fourth solution to find.. When put in standard polynomial form the coefficient of x⁴ is 0 so it is a cubic equation and has three roots.
@@tamasstrengher6407 No, because there isn't a fourth root.
The equation x^4 = (x - 4)^4 is not a fourth degree equation as the term x^4 simplifies, it is only a third degree equation.
Already can tell the i is coming back.
Real solutions: x=±(x-4) ⇒ x=2.
Complex solutions:
x⁴ = (x-4)⁴ ⇒ x⁴ - (x-4)⁴ = 0 ⇒ x⁴ - {x⁴ - 4x³(4) + 6x²(4²) - 4x(4³) + 4⁴} = 0 ⇒ x³ - 6x² + 16x - 16 = 0 ⇒ (x-2)(x² - 4x + 8) ⇒ x² - 4x + 8 = 0 ⇒ x=2±2i.
So x=2 or x=2±2i.
More Easy
8x - 16 does not equal 0 in that work up
Hello,
In the video the answer includes 3 roots but if I use the the roots to reproduce the starting polinom, I will have X^3, but the starting polinom is X^4. Is there a root missing?
Right, I see. It seems the starting polinom is X^4 but it is not, in fact the starting polinom is X^3. Really tricky 😅
no because when you expand (x-4)^4 you get x^4 plus + cubic. but subtracting x^4 from the lhs leaves just the cubic with 3 roots.
what's the differance between / log / ln and lambert w
log is the decimal logarithm (base 10): log(x) = log₁₀(x)
ln is the natural logarithm (base e): ln(x) = logₑ(x)
W is the inverse function of x∙eˣ: W(x∙eˣ) = x or generally W(■∙e^■) = ■
Log is log with base 10
Ln is log with base e
(e is Euler's number)
Lambert W is the product log function such that:
W(xe^x)=x
Where is the fourth solution? You correctly say that this 4th degree equation and has 4 solution. One disappeared???
Pls could you solve that?
(x^2)-(y^2)+2x-6y-25=0 {x;y € Z} (x and y are integers)
The given equation is equivalent to (x + 1)^2 - (y + 3)^2 = 17, or X^2 - Y^2 = 17 with X = x + 1 and Y = y + 3, or (X - Y).(X + Y) = 17
As X and Y are in Z: X - Y and X + Y are divisors of 17. We have 4 solutions:
1) X - Y = 1 and X + Y = 17, so X = 9 and Y = 8, so x = 8 and y = 5. 2) X - Y = 17 and X + Y = 1, so X = 9 and Y = -8, so x = 8 and y = -11
3) X - Y = -1 and X + Y = -17, so X = -9 and Y = -8, so x = -10 and y = -11. 4) X -Y = -17 and X + Y = -1, so X = -9 and Y = 8, so x = -10 and y = 5. Finally the couples (x, y) solutions are: (8, 5) (8, -11) (-10, -11) and (-10, 5).
@@marcgriselhubert3915thank you :)
(x^2)^2-((x-4)^2)^2=0 , (x^2-(x^2-8x+16))(x^2+(x^2-8x+16))=0 , (8x-16)(2x^2-8x+16)=0 ,
(x-2)(x^2-4x+8)=0 , x=2 , x=(4+/-V(16-32))/2 , x= 2+2i , 2-2i ,
Where is the 4th root
x⁴s cancel so it is a cubic equation.
X = 2 for real solution
x^2^2(x ➖ 2x+2). (x^4 ➖ 256)={x^0+x^0 ➖ x^0+x^0 ➖ x^0+x^0 ➖ x^0+x^0 ➖}={x^1+x^1+x^1+x^1}=x^4 x^2^2(x ➖ 2x+2).
I have a question for you 5^x = 26x.
I got this question in my olympiad exam and i didn't know how to solve it : [2^(x-3)] × [3^(2x-8) ]=36 ; find x
x=2
Already the second step is incorrect
Wrong, Wrong, Wrong. You need 4 solutions, not 3
Notice that (wrong)^4 = wrong ;-)
What is wrong was to say (now deleted) that this is a 4th degree equation. It is not, it is a 3rd degree equation as the terms in x^4 disappear. It is simply a degree 3, with 3 solutions.
Really bad, think about real number, complex numbers with argument...... perhaps just one minute to solve this.....
Nachdenken ist noch kein Beweis. Er hat wohl den klarsten Weg gezeigt!