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.
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
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.
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
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'.
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.
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.
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, ...
@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 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
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 . }
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 .
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 .
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 . }
