Ah. I came here to point that out.

Imaginary numbers...

I was just going to say that it was |x+4| instead of x+4

​@@legitjimmyjaylight8409That analysis works if we were talking about (√(x+4))² instead of √( (x+4)² )

I dont think you need absolute function since x will always be positive

Right. I have added "x+4 is greater than or equal to 0" in the comments above.

@@mrhtutoring uhm... Nope, it's not the same thing. The square root of the square of x+4 is | x + 4 | . Defined on R. If for some reason you want to restrict the exercise to x+4 geq 0, this must be part of the assignment. If a different domain is not specified, all functions are automatically defined in their natural domain. In this case, the natural domain of the function is R.

​@@DaveJ6515bro didn't you hear him 💀

Thank you.

@@mrhtutoring your growing fast and deserve it

@@AmericanVRisReal I appreciate it!

Presta atención en case.

Thank you. I appreciate it

x²+2² is not equal to (x+2)²

(x+2)² = x²+4x+4 ≠ (x²+2²)

Glad you like it. Some don't unfortunately.

Thanks

Well yes. That’s why he called it a mistake.

It is a quadratic equation

@@1saxyboi191 Thank you👍

It's not an equation at all, since an equation has an equal sign. But yeah, it's a quadratic expression.

@@chessandmathguyThanks👍

@@55Quirll 🙂

It isn't.

@@mrhtutoring so final answer is?

I think if root of x²+4.. using trigono method... Right?

Isn't it + 14x?

It cannot be simplified further.

No solution.

That's as far as you can go.

​@@mrhtutoringbinomial theorem could help you can insert υ = 1/2 in the expression (a+b)^υ It gives us √b(a/b +1)^υ

no child the formula is xpower2-2power2

also this topic is a bit more advanced than 8th grade it involves domain and range

Foil. Y^2 from y times y. Then y times 3 = 3y. And y times 3 again = 3y. Then 3 times 3 = 9. Should now be y^2 + 3y +3y + 9. We can add the 3y’s because they are like terms so final answer is y^2 + 6y + 9

(y+3)(y+3)=y²+3y+3y+9. The 3y+3y results in 6y.

Given √(a²+b²), it doesn't equal a+b because it's added. Only when you have √(a²b²), it's equal to ab.

Thank you for the comment. I appreciate it.

This example will work if the characteristic is 2 or 3 but not for 5, 7 etc

Think of the binomial expansion. All the coefficients are 0 mod p except for the first and last terms, leaving only a^p + b^p. Look up Frobenius endomorphism.

@@rileyweisman3789 The question is when is (y+3)^2 = y^2 + 3^2 ? In any commutative ring (y+3)^2 = y*2 +2*3*y + 3^2. In Z/2Z we get 2*3*y = 0. In Z/3Z y+3 = y. However in Z/5Z 1+3 = 4 so (1+3)^2 = 1 in Z/5Z but 1^2 + 3^2 = 0 in Z/5Z In other words, if p and q are distinct primes do not expect (a+b)^P = a^p + b^p in Z/qZ.

What you’re noticing is that the (a+b)^p we must have char p. Not any p, obviously. Of course this can’t work in rings of different characteristics. In our case, we would work in Z2 for your example. *must have char p for the Frobenius mapping. second note: with regards to your last line: of course we couldn’t expect that if p neq q. But the whole point is to have char p, so that when we raise a+b to the power of p, we will have lots of cancellation with the binomial expansion.

That doesn't work because the original contains (x^2 plus 4), not times. If you started with square root of (4x^2) then you could take out the x^2 to get x * square root of 4

Basically you are trying to turn an addition into a multiplication.

I'm quite sure it's correct.

Right. If we had √(x+2)² then it would equal to x+2.

(y+3)(y+3)=y²+3y+3y+9. The 3y+3y results in 6y.