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This was so fucking confusing

x^5 - 1 = 0 x^5 = 1 x = penta root of of 1 x = 1

cant u apply quadratic equations in solving x^2+4x+12=0 .....up there on that step

While me noticing him say serd instead of third😅

simplify but tssss aahahah

I used synthetic division to break down the quartic, don't forget to have the zero place holder where the x^3 value should be and your golden. Once down to the quadratic, complete the square to find the last two possible solutions. As P and Q are small integer values , finding the factors (1,-2) in short order shouldn't be an issue with the factor theorem. You can do a quick check with Descartes rule of signs to find all solutions are real. Then plug in your values to check solutions. Apologies for getting the order all jumbled up, do Descartes rule of signs first, then the synthetic division.

Boring

x^(4+3+2+1)=-1 x^(10)=-1 x=-1

X=3 😂

Nice

This isnt even what the cambridge entrance exam questions are like

I think 0 is a answer to most of the questions

Fucking maths X= horse power Y= donkey power So what is man power

Good solution a=0 a=3^1/2 a=-3^1/2

4 complex solutions placed at the 4 of 5 vertices (except 1+0i)of the perfect pentagon

Yes

1 LOL

1

The question is solution and no need to waste time more .

😂😂😂 in the age of AI we need to use calculator. Requires only 20 seconds

1.705

It is profound wrong to leave the final solution with an irrational number as denominator of the fraction.

Mathematics subject in a high school in Turkey is taught at a university in America Turkey math is very very hard you 've ever seen😞

I enjoyed following along but it seems like you simply rewrote the problem instead of solving it. Your “answer” still contains the awkward fractional exponent. In fact, one could argue that your “answer” is more complex than the original problem.

This is a geometric series with common factor x, and initial term equal to 1. Doing this we obtain (x^5-1)/(1-x)=0. We know x=/=1, so this equation reduces to x^5=1. There are complex roots. Simple.

I remember doing expansions that wouldn't fit the page 😅

x^4+x^3+x^2+x+1 = (x^5 - 1)/(x - 1)

I can do that too much easy form.

I dont see how make it into ‘x’ can solve the problem faster

Here's how you do it: x⁴ + x³ + x² + x + 1 = 0 ....(I) x⁴ + x³ + x² + x = -1 .....(II) Multiply (I) by X to get x⁵ + x⁴ + x³ + x² + x = 0 .....(III) Put (II) in (III) to get x⁵ - 1 = 0 x⁵ = 1 So, X = 1

I've watched this earlier.As far as i can remember,i came in international math Olympiad as (x^5-1=0,solve for x).Where the first part is to get (x-1)(this video's equation)=0

Since when did i tell you that math problems are part of this competition?

Just compute complex square root of e^i2πn for n=0,1,2,3,4

Расписать по треугольнику Паскаля и привести подобные

I am proud i can answer one fifth of the answer correctly.

x=0.5

Olympiad question?! I was able to solve it in around 5 sec in my brain. And I came here thinking the solution should not be that easy that requires a utub video...

In my earlier comment.

Correction :the second equation in the last line should be X^2-x+1=0.

At x = 1, 1.5 , 2 ; f(x) = 2^x + x is kind of linear behaviour with f(x) changing by 0.3 at incremental change of 0.1 in x. Hence, at x= 1.5 , f(x) ~ 4.3. For f(x) to be 5, x should increase by 0.2 from 1.5 i.e 1.7

One way of factorisation of x^4 +x^3 +x^2+x+1=0 after dividing by x^2 writing it as (x+1/x+1)(x+1/x-1)=0 and proceeding with solving two quadratic equations x^2+x+1=0 and x^2+x-1=0

After 4✓(3/4)^3, multiply inside by (3/4)*(4/3) We get 4✓((3/4)^4*(4/3) Hence (3/4)*4√(4/3) Much simpler

Stirling’s approximation is good tool to deal with factorial

58.4 ans

Mattez ma tique pas ma b..e :L'inconnu est par quatre, l'inconnu est par trois, l'inconnu est par deux, l'inconnu est , puis un et toi .

You really like the smell of your own farts don’t you

0, ±(3)^½ but I'm pretty sure, I would forget 0 while I'm in the exam hall😅

-1

We need to solve the equation: \[ 2^x + x = 11 \] This is a **transcendental equation**, meaning it cannot be solved algebraically using standard methods. Instead, we can use numerical methods or trial and error to approximate the solution. ### Step 1: Trial and error Let's test some values of \(x\): - If \(x = 2\): \[ 2^2 + 2 = 4 + 2 = 6 \quad (\text{too small}) \] - If \(x = 3\): \[ 2^3 + 3 = 8 + 3 = 11 \quad (\text{exact solution}) \] - If \(x = 4\): \[ 2^4 + 4 = 16 + 4 = 20 \quad (\text{too large}) \] From this, we see that \(x = 3\) is an exact solution. ### Step 2: Check for other solutions We can also check if there are other solutions. For example: - If \(x = 0\): \[ 2^0 + 0 = 1 + 0 = 1 \quad (\text{too small}) \] - If \(x = 1\): \[ 2^1 + 1 = 2 + 1 = 3 \quad (\text{too small}) \] - If \(x = 2.5\): \[ 2^{2.5} + 2.5 \approx 5.656 + 2.5 = 8.156 \quad (\text{too small}) \] - If \(x = 3.5\): \[ 2^{3.5} + 3.5 \approx 11.313 + 3.5 = 14.813 \quad (\text{too large}) \] From this, it appears that \(x = 3\) is the only real solution. ### Final Answer: \[ x = 3 \]