Alternative solution based on the knowledge of the depressed cubic formula (which is probably the way the problem was generated): -q/2=7 q=-14 (q/2)²+(p/3)³=50 49+p³/27=50 p³=27 p=3 x³+3x-14=0 (guess and check using rational root theorem) (x-2)(x²+2x+7)=0 Determinant = 2²-4*1*7 = -24 < 0 x=2 is only solution
So first you make the assumption that the expression under the cube root ( 7+√50) can be expressed as a cube (a+b)³. Is this always true? Or if not, how would you know to look for it?
Yes, in most cases you can express it as a cube, the sum or difference of two terms. If not, then the real root of the equation cannot be determined in the way the professor did, and he found the apparent solution x=2.@@musicsubicandcebu1774
Yes, in most cases you can express it as a cube, the sum or difference of two terms. If not, then the real root of the equation cannot be determined in the way the professor did, and he found the apparent solution x=2.@@musicsubicandcebu1774
Bro I'm serious no joke I did it without solving anything, I made it like an equation simplified it so 1=1 but then put the + that I changed before in = and the answer was 2 (I'm 15)
The easiest way is to notice that 7+√50=7+5√2=(1+√2)³ and likewise 7-√50=(1-√2)³ and from that ³√(7+√50)+³√(7-√50)=1+√2+1-√2=2
7+5√2 ≠ (1+√2)³ ?
Nvm it does. That's really smart how did you work that out
@@ImmortalityVenerable7+5√2=1+3√2+6+2√2=
(1)³+3(1)²(√2)+3(1)(√2)²+(√2)³ =(1+√2)³[(a+b)³=a³+3a²b+3ab²+b³]
7-5√2=1-3√2+6-2√2=(1-√2)³
@ناصريناصر-س4ب That's crazy👏👏
This is really good! Thanks for sharing :)
Alternative solution based on the knowledge of the depressed cubic formula (which is probably the way the problem was generated):
-q/2=7
q=-14
(q/2)²+(p/3)³=50
49+p³/27=50
p³=27
p=3
x³+3x-14=0 (guess and check using rational root theorem)
(x-2)(x²+2x+7)=0
Determinant = 2²-4*1*7 = -24 < 0
x=2 is only solution
my mind is really blown 🔥🤯 Nice one
Very good method
Damn, isn't there a shorter and easiar way?
The easiest way is to notice that 7+√50=7+5√2=(1+√2)³ and likewise 7-√50=(1-√2)³ and from that ³√(7+√50)+³√(7-√50)=1+√2+1-√2=2
@ناصريناصر-س4ب 🫡
So first you make the assumption that the expression under the cube root ( 7+√50) can be expressed as a cube (a+b)³. Is this always true? Or if not, how would you know to look for it?
Yes, in most cases you can express it as a cube, the sum or difference of two terms. If not, then the real root of the equation cannot be determined in the way the professor did, and he found the apparent solution x=2.@@musicsubicandcebu1774
Yes, in most cases you can express it as a cube, the sum or difference of two terms. If not, then the real root of the equation cannot be determined in the way the professor did, and he found the apparent solution x=2.@@musicsubicandcebu1774
I wish oneday i will be able to solve those.
Your chanel is really helping
You will, just keep practicing!
Bro I'm serious no joke I did it without solving anything, I made it like an equation simplified it so 1=1 but then put the + that I changed before in = and the answer was 2 (I'm 15)
Yep it's actually a question for 14-15 years old.
2
Looks like oversimplifying to me
👍👍👍