If had to go through expansion why not just expand the original equation and solve it directly? But here is how I did it. (x+2)^4 - 1 + (x+1)^4 - 4^2 =0 …. Will get x = -3 and 0 pretty quickly, then use the formula to get the two solutions of complex numbers
That's a nice way to factor the quartic equation! (x+2)⁴ + (x+1)⁴ = 17 = 2⁴ + 1⁴ (x+2)⁴-1⁴ + (x+1)⁴-2⁴ = 0 ((x+2)²-1²)*((x+2)²+1²) + ((x+1)²-2²)*((x+1)²+2²) = 0 (x+1)*(x+3)*(x²+4*x+5) + (x-1)*(x+3)*(x²+2*x+5) = 0 (x+3)*[(x+1)*(x²+4*x+5) + (x-1)*(x²+2*x+5)] = 0 (x+3)*[ x*Σ + 1*∆ ] = 0 (x+3)*[ x*(2*x²+6*x+10) + 1*(2*x) ] = 0 (x+3)*[ 2*x*(x²+3*x+5 + 1) ] = 0 2*x*(x+3)*(x²+3*x+6) = 0 x = 0, -3, (-3 ± i*√15)/2 Most folks would the sum of differences of 4th powers by pairing like terms: (x+2)⁴-2⁴ + (x+1)⁴-1⁴ = 0 . And the next step will reveal x can be factored out easily. But that leaves a cubic polynomial -- it may be hard to find x+3 as a factor to reduce that down. By pairing the counterintuitive way [(x+2)⁴-1⁴ + (x+1)⁴-2⁴ = 0], factor x+3 is removed first, leaving the cubic polynomial that's easily reduced by factoring out x. This is useful when someone creates a challenge with the sum of the 4th power of consecutive integers: 0⁴+1⁴=1, 1⁴+2⁴=17, 2⁴+3⁴=97, 3⁴+4⁴=337, 881, 1921, 3697, 6497, 10657, 16561, ...
It might be helpful not to the use the same letters (a, b and c) with different meanings (first a = x + 2 and b = x + 1, later they are used in the equation ax^2 + bx + c = 0). Furthermore the non-real (complex) solutions have been skipped
since x=0 is a solution we can already factor that out, and for polynomes of degree 3 there are formulas. if there is another obvious solution the problem can be further simplied to a 2nd degree polynomial.
Yes, there's another obvious solution. The exponent is 4, an even number, so ± bases yield the same value. (2)⁴ + (1)⁴ = 17 , for x = 0 (-2)⁴ + (-1)⁴ = 17 But the terms must be swapped to match the equation: (-1)⁴ + (-2)⁴ = 17 , for x = -3 Thus, x and x+3 are factors of the quartic equation. The missing factor is a quadratic polynomial that yields a complex conjugate pair of roots, (-3 ± i*√15)/2 .
His obfuscated way is unnecessarily complicated. It's faster to do the binomial expansion and solve: (x⁴ + 4*2*x³ + 6*2²*x² + 4*2³*x + 2⁴) + (x⁴ + 4*x³ + 6*x² + 4*x + 1) = 17 2*x⁴ + 12*x³ + 30*x² + 36*x + 17 = 17 x⁴ + 6*x³ + 15*x² + 18*x = 0 ‡ x*((x + 3)*x² + (x + 3)*3*x + (x + 3)*6) = 0 x*(x + 3)*(x² + 3*x + 6) = 0 x = 0, -3, (-3 ± i*√15)/2 { ‡ In the original equation, the values in parentheses work for 2⁴+1⁴ and (-1)⁴+(-2)⁴ . Thus, we know x = 0 and -3 . We know x and x+3 are factors. } In general, a² - b² isn't a perfect square. ∆ is called the "discriminant". At 07:20, he can complete the square: z² + 2*z + 1 = 9 , z = a*b z + 1 = ±3 z = -1 ± 3 = 2, -4 There's no reason to reject the complex solutions.
It's faster to do the binomial expansion and solve. Your obfuscated way is unnecessarily complicated. In general, a² - b² isn't a perfect square. ∆ is called the "discriminant". At 07:20, you can complete the square: Z² + 2*Z + 1 = 9 , Z = a*b Z + 1 = ±3 Z = -1 ± 3 = 2, -4 There's no reason to reject the complex solutions.
Note the special contion here. As 2⁴+1⁴=17 and the power is even then the sign of 2 can be + or -. Similarly the sign og 1. Both can be the same sign, or not. And x is found accordingly.
His obfuscated way is unnecessarily complicated. It's faster to do the binomial expansion and solve. It turns out that no substitution is faster and less error-prone. (x⁴ + 4*2*x³ + 6*2²*x² + 4*2³*x + 2⁴) + (x⁴ + 4*x³ + 6*x² + 4*x + 1) = 17 2*x⁴ + 12*x³ + 30*x² + 36*x + 17 = 17 x⁴ + 6*x³ + 15*x² + 18*x = 0 ‡ x*((x + 3)*x² + (x + 3)*3*x + (x + 3)*6) = 0 x*(x + 3)*(x² + 3*x + 6) = 0 x = 0, -3, (-3 ± i*√15)/2 { ‡ Upon inspection of the original equation, the values in parentheses work for (2)⁴+(1)⁴ and (-1)⁴+(-2)⁴ . Thus, we know x = 0 and x = -3 , and that x and x+3 are factors of the quartic equation. Factor them out to leave a quadratic equation for the complex conjugate pair of roots, x = (-3 ± i*√15)/2 . }
Its a 4th degree equation. Theres 4 solutions. You missed the two complex solutions; x= (-3/2) +or- (radical(15)i/2). This is an OLYMPIAD question. Those solutions belong there
Is here anybody, who automatically calculated that x=0😂😂 Being honest, I didn't watch till the end, because I already thought that 2^4+1^4 equals to 17 And second x equals to -3
The goal is to learn how to apply several rules of algebra to find the solutions. This channel is the best there is and PLEASE stop offering guess and check solutions. He's connecting the problem to so many profound concepts. I truly like his methods and if they are too long, then at least offer a better way.
@is7728: It turns out that no substitution is faster and less error-prone. Upon inspection of the original equation, the values in parentheses work for 2⁴+1⁴ and (-1)⁴+(-2)⁴ . Thus, we know x = 0 and x = -3 . We know x and x+3 are factors of the quartic equation. Factor them out to leave a quadratic equation for the complex conjugate pair of roots, (-3 ± i*√15)/2 . (x⁴ + 4*2*x³ + 6*2²*x² + 4*2³*x + 2⁴) + (x⁴ + 4*x³ + 6*x² + 4*x + 1) = 17 2*x⁴ + 12*x³ + 30*x² + 36*x + 17 = 17 x⁴ + 6*x³ + 15*x² + 18*x = 0 x*((x + 3)*x² + (x + 3)*3*x + (x + 3)*6) = 0 x*(x + 3)*(x² + 3*x + 6) = 0 x = 0, -3, (-3 ± i*√15)/2
I made the same variable change. You could have developped faster by using the binomial expansion method: (a+b)^4=a⁴+4a³b+6a²b²+4ab³+b⁴ It was even possible to skip the calculation of the terms in y and y³ by noticing that if y is solution then -y is also a solution, so only the even powers of y would remain. Best method in my opinion.
His obfuscated way is unnecessarily complicated. It's faster to do the binomial expansion and solve. It turns out that no substitution is faster and less error-prone: (x⁴ + 4*2*x³ + 6*2²*x² + 4*2³*x + 2⁴) + (x⁴ + 4*x³ + 6*x² + 4*x + 1) = 17 2*x⁴ + 12*x³ + 30*x² + 36*x + 17 = 17 x⁴ + 6*x³ + 15*x² + 18*x = 0 ‡ x*((x + 3)*x² + (x + 3)*3*x + (x + 3)*6) = 0 x*(x + 3)*(x² + 3*x + 6) = 0 x = 0, -3, (-3 ± i*√15)/2 { ‡ Upon inspection of the original equation, the values in parentheses work for (2)⁴+(1)⁴ and (-1)⁴+(-2)⁴ . Thus, we know x = 0 and x = -3 , and that x and x+3 are factors of the quartic equation. Factor them out to leave a quadratic equation for the complex conjugate pair of roots, x = (-3 ± i*√15)/2 . }
To be honest....videos like this tend to put students off mathematics by making it seem more complicated than it actually is....the 2 real solutions are easily derived by inspection....given the length of the video, I thought the 2 imaginary roots were going to be derived. We should be trying to teach mathematical instinct rather than complicated 'recipes'.
Upon inspection of the original equation, the values in parentheses work for 2⁴+1⁴ and (-1)⁴+(-2)⁴ . Thus, we know x = 0 and x = -3 . We know x and x+3 are factors to the quartic equation. Factor them out to leave a quadratic equation for the complex conjugate pair of roots, (-3 ± i*√15)/2 .
I tried this 1st, got the answer in5 seconds, my god his procedure is very confusing, don’t understand a thing he’s doing, kmt 🤦♀️, y this long procedure?. One math problem should never take so long, very confusing, u turn off ppl from math
Let h be x + 1.
(h + 1)^4 + h^4 = 17
h^4 + 4h^3 + 6h^2 + 4h + 1 + h^4 = 17
2h^4 + 4h^3 + 6h^2 + 4h - 16 = 0
h^4 + 2h^3 + 3h^2 + 2h - 8 = 0
When h = 1, the eqn is satisfied.
∴ x = 0
h^3 + 3h^2 + 6h + 8 = 0
When h = -2, the eqn is satisfied.
∴ x = -3
h^2 + h + 4 = 0
h = ( -1 +- √(1 - 16) ) / 2
= -0.5 +- √15 i / 2
∴ x = -1.5 +- √15 i / 2
If had to go through expansion why not just expand the original equation and solve it directly? But here is how I did it. (x+2)^4 - 1 + (x+1)^4 - 4^2 =0 …. Will get x = -3 and 0 pretty quickly, then use the formula to get the two solutions of complex numbers
*complex numbers
You are correct. Edited
That's a nice way to factor the quartic equation!
(x+2)⁴ + (x+1)⁴ = 17 = 2⁴ + 1⁴
(x+2)⁴-1⁴ + (x+1)⁴-2⁴ = 0
((x+2)²-1²)*((x+2)²+1²) + ((x+1)²-2²)*((x+1)²+2²) = 0
(x+1)*(x+3)*(x²+4*x+5) + (x-1)*(x+3)*(x²+2*x+5) = 0
(x+3)*[(x+1)*(x²+4*x+5) + (x-1)*(x²+2*x+5)] = 0
(x+3)*[ x*Σ + 1*∆ ] = 0
(x+3)*[ x*(2*x²+6*x+10) + 1*(2*x) ] = 0
(x+3)*[ 2*x*(x²+3*x+5 + 1) ] = 0
2*x*(x+3)*(x²+3*x+6) = 0
x = 0, -3, (-3 ± i*√15)/2
Most folks would the sum of differences of 4th powers by pairing like terms: (x+2)⁴-2⁴ + (x+1)⁴-1⁴ = 0 . And the next step will reveal x can be factored out easily. But that leaves a cubic polynomial -- it may be hard to find x+3 as a factor to reduce that down. By pairing the counterintuitive way [(x+2)⁴-1⁴ + (x+1)⁴-2⁴ = 0], factor x+3 is removed first, leaving the cubic polynomial that's easily reduced by factoring out x. This is useful when someone creates a challenge with the sum of the 4th power of consecutive integers: 0⁴+1⁴=1, 1⁴+2⁴=17, 2⁴+3⁴=97, 3⁴+4⁴=337, 881, 1921, 3697, 6497, 10657, 16561, ...
It might be helpful not to the use the same letters (a, b and c) with different meanings (first a = x + 2 and b = x + 1, later they are used in the equation ax^2 + bx + c = 0). Furthermore the non-real (complex) solutions have been skipped
since x=0 is a solution we can already factor that out, and for polynomes of degree 3 there are formulas. if there is another obvious solution the problem can be further simplied to a 2nd degree polynomial.
Yes, there's another obvious solution. The exponent is 4, an even number, so ± bases yield the same value.
(2)⁴ + (1)⁴ = 17 , for x = 0
(-2)⁴ + (-1)⁴ = 17
But the terms must be swapped to match the equation:
(-1)⁴ + (-2)⁴ = 17 , for x = -3
Thus, x and x+3 are factors of the quartic equation. The missing factor is a quadratic polynomial that yields a complex conjugate pair of roots, (-3 ± i*√15)/2 .
17 = 1**4 +2**4 or (-1)**4+(-2)**4
х+2=2, х+1=1 => x=0
x+2=-1, x+1=-2 = x=-3
Explain the solving strategy and ideas rather than a long confusing mechanical calculation
His obfuscated way is unnecessarily complicated. It's faster to do the binomial expansion and solve:
(x⁴ + 4*2*x³ + 6*2²*x² + 4*2³*x + 2⁴) +
(x⁴ + 4*x³ + 6*x² + 4*x + 1) = 17
2*x⁴ + 12*x³ + 30*x² + 36*x + 17 = 17
x⁴ + 6*x³ + 15*x² + 18*x = 0 ‡
x*((x + 3)*x² + (x + 3)*3*x + (x + 3)*6) = 0
x*(x + 3)*(x² + 3*x + 6) = 0
x = 0, -3, (-3 ± i*√15)/2
{ ‡ In the original equation, the values in parentheses work for 2⁴+1⁴ and (-1)⁴+(-2)⁴ . Thus, we know x = 0 and -3 . We know x and x+3 are factors. }
In general, a² - b² isn't a perfect square.
∆ is called the "discriminant".
At 07:20, he can complete the square:
z² + 2*z + 1 = 9 , z = a*b
z + 1 = ±3
z = -1 ± 3 = 2, -4
There's no reason to reject the complex solutions.
It's faster to do the binomial expansion and solve.
Your obfuscated way is unnecessarily complicated.
In general, a² - b² isn't a perfect square.
∆ is called the "discriminant".
At 07:20, you can complete the square:
Z² + 2*Z + 1 = 9 , Z = a*b
Z + 1 = ±3
Z = -1 ± 3 = 2, -4
There's no reason to reject the complex solutions.
(x⁴ + 4*2*x³ + 6*2²*x² + 4*2³*x + 2⁴) +
(x⁴ + 4*x³ + 6*x² + 4*x + 1) = 17
2*x⁴ + 12*x³ + 30*x² + 36*x + 17 = 17
x⁴ + 6*x³ + 15*x² + 18*x = 0
x*(x³ + 6*x² + 15*x + 18) = 0
x*((x + 3)*x² + (x + 3)*3*x + (x + 3)*6) = 0
x*(x + 3)*(x² + 3*x + 6) = 0
x = 0, -3, (-3 ± i*√15)/2
Note the special contion here. As 2⁴+1⁴=17 and the power is even then the sign of 2 can be + or -. Similarly the sign og 1. Both can be the same sign, or not. And x is found accordingly.
His obfuscated way is unnecessarily complicated. It's faster to do the binomial expansion and solve. It turns out that no substitution is faster and less error-prone.
(x⁴ + 4*2*x³ + 6*2²*x² + 4*2³*x + 2⁴) +
(x⁴ + 4*x³ + 6*x² + 4*x + 1) = 17
2*x⁴ + 12*x³ + 30*x² + 36*x + 17 = 17
x⁴ + 6*x³ + 15*x² + 18*x = 0 ‡
x*((x + 3)*x² + (x + 3)*3*x + (x + 3)*6) = 0
x*(x + 3)*(x² + 3*x + 6) = 0
x = 0, -3, (-3 ± i*√15)/2
{ ‡ Upon inspection of the original equation, the values in parentheses work for (2)⁴+(1)⁴ and (-1)⁴+(-2)⁴ . Thus, we know x = 0 and x = -3 , and that x and x+3 are factors of the quartic equation. Factor them out to leave a quadratic equation for the complex conjugate pair of roots, x = (-3 ± i*√15)/2 . }
Mine is more simple.
(x+2)⁴ + (x+1)⁴ = 17
[(x+1)+1]⁴ + (x+1)⁴ = 17
example ; y = (x+1)
(y+1)⁴ + y⁴ = 17
y⁴ + 4y³ + 6y² + 4y + 1 + y⁴ - 17 = 0
2y⁴ + 4y³ + 6y² + 4y - 16 = 0
2(y⁴ + 2y³ + 3y² +2y - 8) = 0
Lets find the factors of y
Using polinomial
Factors of y⁴ + 2y³ + 3y² + 2y - 8 = 0
(y - 1) → y = 1 → 1+2+3+2-8 = 0 (ok!)
(y + 2) → y = -2 → -8+12-12+8 = 0 (ok.!)
The others y = i
So, we get y = 1 and y = -2
(y = x+1)
y = 1
x+1 = 1 → x = 0 (check ok.!)
y = -2
x+1 = -2 → x = -3 (check ok.!)
So, x = 0, -3
Factor out x and x+3 to leave a quadratic equation. You can easily get the complex conjugate pair of roots, (-3 ± i*√15)/2 .
Let y=x+1.5, then (y+0.5)^4+(y-0.5)^4=17 (4y^2-9)(4y^2+15)=0; y=+/-3/2 and y= +/-isqrt(15)/2. If y=x+1.5, then x=-3, x=0, , x=-1.5 +/- isqrt(15)/2
Its a 4th degree equation. Theres 4 solutions. You missed the two complex solutions; x= (-3/2) +or- (radical(15)i/2). This is an OLYMPIAD question. Those solutions belong there
x = 0, -3, (-3 ± i*√15)/2
Is here anybody, who automatically calculated that x=0😂😂
Being honest, I didn't watch till the end, because I already thought that 2^4+1^4 equals to 17
And second x equals to -3
Yes, at the first glance one can get the two results, since both parentheses must be 1/-1 and 2/-2.
The goal is to learn how to apply several rules of algebra to find the solutions. This channel is the best there is and PLEASE stop offering guess and check solutions. He's connecting the problem to so many profound concepts. I truly like his methods and if they are too long, then at least offer a better way.
Did you also find the complex solutions?
I did. Had the number NOT been -1, 0, or +1, I would have worked it out. But a quick inspection worked this time.
Natürlich!
Diese Lösung springt einem ja mit 'nem nackten Ar... ins Auge!!!!
Very confusing procedure.
He's going for obfuscation. The direct solution is faster, and less prone to mistakes.
What did we learn?
2^4 + 1^4 = 17
(-1)^4 + (-2)^4 = 17
Math Olympiad: (x + 2)⁴ + (x + 1)⁴ = 17; x = ?
Second method;
Let: y = x + 3/2, a = 1/2; (x + 2)⁴ + (x + 1)⁴ = (y + a)⁴ + (y - a)⁴ = 17
(y ± a)⁴ = y⁴ ± 4ay³ + 6a²y² ± 4a³y + a⁴, (y + a)⁴ + (y - a)⁴ = 2y⁴ + 12a²y² + 2a⁴ = 17
2y⁴ + 12(1/4)y² + 2(1/16) = 17, 2y⁴ + 3y² + 1/8 = 17, 16y⁴ + 24y² + 1 = 136
16y⁴ + 24y² - 135 = 0, (2y)⁴ + 6(2y)² - 135 = [(2y)² - 9][(2y)² + 15] = 0
(2y)² - 9 = 0, (2y)² = 9; 2y = ± 3 or (2y)² + 15 = 0, 2y = ± i√15; y = x + 3/2
2y = 2(x + 3/2) = 2x + 3 = ± 3; x = 0 or x = - 3
2y = 2(x + 3/2) = 2x + 3 = ± i√15; x = (- 3 ± i√15)/2
And about the unreal solutions ?
Too complex for him perhaps?
thanks bravo!
一秒钟做出来，0和-3.还有其他解么？
no
x = 0, -3, (-3 ± i*√15)/2
Math Olympiad: (x + 2)⁴ + (x + 1)⁴ = 17; x = ?
Let: y = x + 2; y⁴ + (y - 1)⁴ = 2y⁴ - 4y³ + 6y² - 4y + 1 = 17
2y⁴ - 4y³ + 6y² - 4y = 16, y⁴ - 2y³ + 3y² - 2y - 8 = 0
(y⁴ - 2y³ + y²) + (2y² - 2y) - 8 = 0, (y² - y)² + 2(y² - y) - 8 = 0
(y² - y - 2)(y² - y + 4) = (y + 1)(y - 2)(y² - y + 4) = 0
y + 1 = 0, y = - 1; y - 2 = 0, y = 2 or y² - y + 4 = 0, y = (1 ± i√15)/2
x = y - 2: x = - 3; x = 0 or x = (1 ± i√15)/2 - 2 = (- 3 ± i√15)/2
Answer check:
x = - 3: (x + 2)⁴ + (x + 1)⁴ = (- 1)⁴ + (- 2)⁴ = 1 + 16 = 17; Confirmed
x = 0: 2⁴ + 1⁴ = 16 + 1 = 17; Confirmed
x = (- 3 ± i√15)/2: y = x + 2, y² - y + 4 = 0, y² - y = - 4
(x + 2)⁴ + (x + 1)⁴ = 2(y⁴ - 2y³ + 3y² - 2y) + 1 = 2[(y² - y)² + 2(y² - y)] + 1
= 2[(- 4)² + 2(- 4)] + 1 = 2(16 - 8) + 1 = 17; Confirmed
Final answer:
x = - 3, x = 0, Two complex value roots, if acceptable;
x = (- 3 + i√15)/2 or x = (- 3 - i√15)/2
Nice explanation
@naweenraaj
Maybe this better?
Let h be x + 1.
(h + 1)^4 + h^4 = 17
h^4 + 4h^3 + 6h^2 + 4h + 1 + h^4 = 17
2h^4 + 4h^3 + 6h^2 + 4h - 16 = 0
h^4 + 2h^3 + 3h^2 + 2h - 8 = 0
When h = 1, the eqn is satisfied.
∴ x = 0
h^3 + 3h^2 + 6h + 8 = 0
When h = -2, the eqn is satisfied.
∴ x = -3
h^2 + h + 4 = 0
h = ( -1 +- √(1 - 16) ) / 2
= -0.5 +- √15 i / 2
∴ x = -1.5 +- √15 i / 2
@is7728: It turns out that no substitution is faster and less error-prone. Upon inspection of the original equation, the values in parentheses work for 2⁴+1⁴ and (-1)⁴+(-2)⁴ . Thus, we know x = 0 and x = -3 . We know x and x+3 are factors of the quartic equation. Factor them out to leave a quadratic equation for the complex conjugate pair of roots, (-3 ± i*√15)/2 .
(x⁴ + 4*2*x³ + 6*2²*x² + 4*2³*x + 2⁴) +
(x⁴ + 4*x³ + 6*x² + 4*x + 1) = 17
2*x⁴ + 12*x³ + 30*x² + 36*x + 17 = 17
x⁴ + 6*x³ + 15*x² + 18*x = 0
x*((x + 3)*x² + (x + 3)*3*x + (x + 3)*6) = 0
x*(x + 3)*(x² + 3*x + 6) = 0
x = 0, -3, (-3 ± i*√15)/2
@@oahuhawaii2141 But your factorization from cubic to quadratic eqns is quite subtle to notice
The polynomial coefficients used in synthetic division, with annotations:
1 6 15 18 / 1 3 = 1 3 6
1 3 { 1 3 * 1 _ _ }
0 3 15 18
0 3 9 { 1 3 * 0 3 _ }
0 0 6 18
0 0 6 18 { 1 3 * 0 0 6 }
{ 1 _ _ + 0 3 _ + 0 0 6 = 1 3 6 }
Let y = x + 3/2
(y + 1/2)^4 + (y - 1/2)^4 = 17
(y^2 + y + 1/4)^2 + (y^2 - y + 1/4)^2 = 17
[ (y^2 + 1/4) + y ]^2 + [ (y^2 + 1/4) - y ]^2 = 17
[ (y^2 + 1/4)^2 + 2 y (y^2 + 1/4) + y^2 ] + [ (y^2 + 1/4)^2 - 2 y (y^2 + 1/4) + y^2 ]=17
2((y^2 + 1/4)^2 + y^2) = 17
2(y^4 + 1/2 y^2 + 1/16 + y^2) = 17
2 y^4 + y^2 + 1/8 + 2 y^2 - 17 = 0
2 y^4 + 3 y^2 - 135/8 = 0
y^2 = (-3 +/- √(9 - 4 * 2 * (- 135/8)))/(2*2)
y^2 = (-3 +/- √(9 + 135))/4
y^2 = (-3 +/- √144)/4
y^2 = (-3 +/- 12)/4
y^2 = (-3 + 12)/4 = 9/4 //y^2 = (-3 - 12)/4 < 0 so it is rejected
y = +/- 3/2
x = y - 3/2 = +/- 3/2 - 3/2
x = (0, -3)
I made the same variable change. You could have developped faster by using the binomial expansion method:
(a+b)^4=a⁴+4a³b+6a²b²+4ab³+b⁴
It was even possible to skip the calculation of the terms in y and y³ by noticing that if y is solution then -y is also a solution, so only the even powers of y would remain.
Best method in my opinion.
17 = 1^4 + 2^4, so (x+1)^4=2^4, (x+2)^4=1^4 then x = -3.
what is he actually doing?
His obfuscated way is unnecessarily complicated. It's faster to do the binomial expansion and solve. It turns out that no substitution is faster and less error-prone:
(x⁴ + 4*2*x³ + 6*2²*x² + 4*2³*x + 2⁴) +
(x⁴ + 4*x³ + 6*x² + 4*x + 1) = 17
2*x⁴ + 12*x³ + 30*x² + 36*x + 17 = 17
x⁴ + 6*x³ + 15*x² + 18*x = 0 ‡
x*((x + 3)*x² + (x + 3)*3*x + (x + 3)*6) = 0
x*(x + 3)*(x² + 3*x + 6) = 0
x = 0, -3, (-3 ± i*√15)/2
{ ‡ Upon inspection of the original equation, the values in parentheses work for (2)⁴+(1)⁴ and (-1)⁴+(-2)⁴ . Thus, we know x = 0 and x = -3 , and that x and x+3 are factors of the quartic equation. Factor them out to leave a quadratic equation for the complex conjugate pair of roots, x = (-3 ± i*√15)/2 . }
Why are you making this so bloody complicated. It's crap like this that made me fail Algebra 2 in HS. I find your videos maddening.
(x + 2)⁴ + (x + 1)⁴ = 17
We see x = 0, -3 are 2 of 4 solutions.
We'll use them for factoring.
Let y = x + 1 to ease the work.
(y + 1)⁴ + y⁴ = 17
y⁴ + 4*y³ + 6*y² + 4*y + 1 + y⁴ = 17
2*y⁴ + 4*y³ + 6*y² + 4*y - 16 = 0
y⁴ + 2*y³ + 3*y² + 2*y - 8 = 0
(y - 1)*(y³ + 3*y² - 6*y - 8) = 0
(y - 1)*(y + 2)*(y² + y + 4) = 0
y = 1, -2, (-1 ± i*√15)/2
x = y - 1 = 0, -3, (-3 ± √15)/2
2^4=16だからX+2=2
X=0
(-2)^2=16だからX+1=-2
X=-3
∴X=0,-3
x = 0, -3, (-3 ± i*√15)/2
I think there are no imaginary numbers in our would.
@稲次将人: Complex numbers are needed in STEM fields, such as physics, electronics, and math.
X=-3
(-3+2)⁴+(-3+1)⁴=17
x = 0, -3, (-3 ± i*√15)/2
Real Number
X = 0 and (-3)
16 + 1 = 17
To be honest....videos like this tend to put students off mathematics by making it seem more complicated than it actually is....the 2 real solutions are easily derived by inspection....given the length of the video, I thought the 2 imaginary roots were going to be derived. We should be trying to teach mathematical instinct rather than complicated 'recipes'.
用二项式定理啦
പ്ളീസ്കോറെഡ് utharm
So cool thanks
Thank you very much!!
Your welcome😀
Just know I got lost😂😂😂😂
x=0,x=-3
식을 보면 3초만에 답이 나오는데?
X=0 et.....
x = 0, -3, (-3 ± i*√15)/2
1
X=0; Don't do it stupidly ?!
2 power 4 = 16
1 power 4 = 1
Thus 16+ 1= 17
Upon inspection of the original equation, the values in parentheses work for 2⁴+1⁴ and (-1)⁴+(-2)⁴ . Thus, we know x = 0 and x = -3 . We know x and x+3 are factors to the quartic equation. Factor them out to leave a quadratic equation for the complex conjugate pair of roots, (-3 ± i*√15)/2 .
Immediately guessing, x=0.
x = 0, -3, (-3 ± i*√15)/2
I solve it by other method.
Only 2 secands x=0 and -3
x = 0, -3, (-3 ± i*√15)/2
@@oahuhawaii2141 forr x€R 👍
0; -3.
x = 0, -3, (-3 ± i*√15)/2
0 / -3
X = 0
ОТВЕТ: Х=-3
Equation: (x+2)⁴ + (x+1)⁴ = 17
... is equivalent to: 2*x*(x+3)*(x²+3*x+6) = 0
... which means: x = 0, -3, (-3 ± i*√15)/2 .
I tried this 1st, got the answer in5 seconds, my god his procedure is very confusing, don’t understand a thing he’s doing, kmt 🤦♀️, y this long procedure?. One math problem should never take so long, very confusing, u turn off ppl from math
You didn't find all 4 roots in 5 seconds.
Ardışık 2 sayı toplamı 17. 9+8= 17 uğraşmayın bu kadar😊 şaka yapıyorum
Teacher passed over imaginary number. There are 4 answers
0
Equation: (x+2)⁴ + (x+1)⁴ = 17
... is equivalent to: 2*x*(x+3)*(x²+3*x+6) = 0
... which means: x = 0, -3, (-3 ± i*√15)/2 .
0。
x = 0, -3, (-3 ± i*√15)/2
0