x√x = (3√3)² = 9√9. x = 9 That is the obvious real solution. But there are two complex ones as well: x√x = (3√3)² √x³ = (3√3)² x³ = (3√3)⁴ x³ = 729 x³ − 729 = 0 Since we already know that x − 9 = 0 for x₁ = 9 is a solution, we can factor that out: (x − 9) (x² + 9x + 81) = 0 According to the rule of the zero product, the equation is zero when one of the factors is zero. We already know that for x₁ = 9, the first factor gets zero. So, let's solve for the second: x² + 9x + 81 = 0 x₂,₃ = −9/2 ± √(81/4 − 81) = −9/2 ± √(−243/4) = −9/2 ± i√(243/4) = −9/2 ± i√243 / 2 = −9/2 ± 9i√3 / 2 x₂ = (−9 − 9i√3)/2 x₃ = (−9 + 9i√3)/2 And now we have our three solutions.

Convert the exponents to fractions on both sides. It’s quicker. X^3/4=3^3/2 Now power 4/3 on both sides X=(3^3/2)^4/3=3^2=9

What do the hats or carpors on numbers mean?

Woww... that was actually sooo long to get a simple solution. X^3 = 27*9 = 729 --> x = +9 for real values (non imaginary).

In 7 sec, I have solved x=9

9.

There is only 1 solution, really: x = 9 € N. All your clever manipulations are not necessary. Instead of writing x^3 = 9^3, you could have equally written x^3 = 3^6, and taking the cube root of both sides would have given x = 3^2 = 9. Going the way of x^3 - 9^3 = 0 led you to complex and irrational roots that are not required, since doing it differently (x^3 = 3^6) would have resulted in only x = 9.

@Nikioko

24 күн бұрын

The problem is that x³ = 27 is a third degree polynomial. And a third degree polynomial of course has three solutions. One of them is x = 9. But there are two more: x³ = 729 x³ − 729 = 0 Since you already know that x − 9 = 0 for x₁ = 9 is a solution, you can factor that out: (x − 9) (x² + 9x + 81) = 0 According to the rule of the zero product, the equation is zero when one of the factors is zero. We already know that for x₁ = 9, the first factor gets zero. So, lets solve for the second: x² + 9x + 81 = 0 x₂,₃ = −9/2 ± √(81/4 − 81) = −9/2 ± √(−243/4) The root of a negative radicand is not defined for real numbers. However, for complex numbers, the definition i² = −1 means that √(−x) = i√x. Or in this case: x₂,₃ = −9/2 ± √(−243/4) = −9/2 ± i√(243/4) = −9/2 ± i√243 / 2 x₂ = (−9 − i√243)/2 x₃ = (−9 + i√243)/2 And now we have our three solutions. Ah, you can still replace √243 by 9√9, but that doesn't change the result.

Simplification? Really?

X³=27 X = 3, or -3

@Nikioko

24 күн бұрын

A third degree polynomial has three solutions. So, your answer is incomplete. And −3 is wrong.

@guitC

23 күн бұрын

not 27 but 81

I cannot see that the last 2 'solutions' are actually solutions.

@Nikioko

24 күн бұрын

They are complex solutions. A third degree polynomial always has three solutions.

@ericzacher509

24 күн бұрын

True that but they are not really of any use (at least I think so)

@Nikioko

24 күн бұрын

@@ericzacher509 In this equation or in general? In this equation, they fulfil the rule that every polynomial function has as many solutions as the highest exponent. In real life, complex numbers have a role in physics, like alternating current or the Schrödinger Equation.

@ericzacher509

24 күн бұрын

Yeah you are right with that. I meant that you cant really the non-real solutions im any practical way (besides in higher maths and the likes)

@Nikioko

24 күн бұрын

@@ericzacher509 As I said, complex numbers are important in quantum physics and electrical engineering.

you can find solution earlier, if x^3=9^3, then simply take root 3 from.both sides and you.will have solution that x = 9😊

Silly

@Nikioko

24 күн бұрын

What is silly?

@vics8873

24 күн бұрын

@@Nikioko too easy...

@Nikioko

24 күн бұрын

@@vics8873 Too easy to get all three solutions?

@vics8873

24 күн бұрын

@@Nikioko for most of us who went to high school...

@Nikioko

24 күн бұрын

@@vics8873 You went to high school? In the US? And there you had complex numbers?

задача решается в уме в два действия.

Too much unnecessary writing, many nonsense