That part where m^2 = -10. I actually solved for this using the quadratic formula to get -5+-isqrt(10) and tried to plug in -5+isqrt(10) into the x values to solve for it but it didn't work. And like that you rejected it right off the spot. It was fun to verify it anyways. Nice stuff!
This video is not good. If the viewers are only students who have not yet learned about imaginary and complex numbers, it is okay to reject these because they are not real solutions, but if not, the problem needs to include the condition that x is a real number. Another thing, m²=4 ⇒m=√4 ⇒m=±2, is incorrect. You must write m²=4 ⇒m=±√4 ⇒m=±2.
Simple. -3 is an obvious root.
Very nice! ❤
That part where m^2 = -10. I actually solved for this using the quadratic formula to get -5+-isqrt(10) and tried to plug in -5+isqrt(10) into the x values to solve for it but it didn't work. And like that you rejected it right off the spot. It was fun to verify it anyways. Nice stuff!
Thanks for liking! ❤
Thailand, Math Olympiad: (x + 6)⁴ + (x + 4)⁴ = 82; x =?
First method:
82 > (x + 6)⁴ > (x + 4)⁴ > 0; 0 > x > - 6 or 82 > (x + 4)⁴ > (x + 6)⁴ > 0; - 6 > x > - 8
(x + 6)⁴ + (x + 4)⁴ = 81 + 1 = 3⁴ + 1⁴ = (- 3 + 6)⁴ + (- 3 + 4)⁴; x = - 3
(x + 6)⁴ + (x + 4)⁴ = 1 + 81 = 1⁴ + 3⁴ = (- 7 + 6)⁴ + (- 7 + 4)⁴; x = - 7
Missing two complex value roots
Second method:
Solve the equation directly to find the two complex value roots
Let: y = x + 5; (x + 6)⁴ + (x + 4)⁴ = (y + 1)⁴ + (y - 1)⁴ = 82
(y ± 1)⁴ = y⁴ ± 4y³ + 6y² ± 4y + 1, (y + 1)⁴ + (y - 1)⁴ = 2(y⁴ + 6y² + 1) = 82
y⁴ + 6y² + 1 = 41, y⁴ + 6y² - 40 = (y² - 4)(y² + 10) = 0, y² - 4 = 0 or y² + 10 = 0
y² = 4, y = ± 2 = x + 5, x = - 3 or x = - 7; y² = - 10, y = ± i√10, x = - 5 ± i√10
Answer check:
x = - 3 or x = - 7: (x + 6)⁴ + (x + 4)⁴ = 82; Confirmed as shown in First method
x = - 5 ± i√10, x + 5 = ± i√10 = y: (x + 6)⁴ + (x + 4)⁴ = 2(y⁴ + 6y² + 1)
2(y⁴ + 6y² + 1) = 2[(± i√10)⁴ + 6(± i√10)² + 1] = 2(100 - 60 + 1) = 82; Confirmed
Final answer:
x = - 3; x = - 7; Two complex value roots, if acceptable;
x = - 5 + i√10 or x = - 5 - 5i√10
Why you don't search complex roots of this equation?
Then the video will be very lengthy ❤
(x+6)^4+(x+4)^4=3^4+1^4.....ecc ecc facciamolo facile
-3
Is there a real solution besides -3 and -7? ❤
Oh, I understand, Thanks a lot! ❤
This video is not good. If the viewers are only students who have not yet learned about imaginary and complex numbers, it is okay to reject these because they are not real solutions, but if not, the problem needs to include the condition that x is a real number.
Another thing, m²=4 ⇒m=√4 ⇒m=±2, is incorrect. You must write m²=4 ⇒m=±√4 ⇒m=±2.
Yes, you are right! ❤
x+4=y => (y+2)⁴+y⁴=82 =>
=> y⁴+4y³+12y²+16y-33=0 =>
=> (y-1)(y³+5y²+17y+33)=0 =>
=> (y-1)(y+3)(y²+2y+11)=0 =>
=> y1=1; y2=-3; y3,4=-1±ivʼ10 =>
=> x1=-3; x2=-7; x3,4=-5±ivʼ10 😁
y⁴+4y³+12y²+16y-33=0
y=±1,±3,±11,±33
y=1; 1⁴+4•1³+12•1²+16•1-33=0
1 4 12 16 - 33
0 1 5 17 33
1 5 17 33 0
y³+5y²+17y+33=0
y=±1,±3,±11,±33
y=-3; (-3)³+5(-3)²+17(-3)+33=0
1 5 17 33
0 -3 - 6 -33
1 2 11 0
y²+2y+11=0
y=-1±vʼ(1²-11)=-1±ivʼ10
y⁴+4y³+12y²+16y-33=0
y=±1,±3,±11,±33
y=1; 1⁴+4•1³+12•1²+16•1-33=0
1. 4 12 16 -33
0. 1 5 17 33
1 5 17 33 0
y³+5y²+17y+33=0
y=±1,±3,±11,±33
y=-3; (-3)³+5(-3)²+17(-3)+33=0
1 5 17 33
0 -3 -6 -33
1 2 11 0
y²+2y+11=0
y=-1±vʼ(1²-11)=-1±ivʼ10
Very nice! ❤
Very nice! ❤