Olympiad Mathematics | How many real solutions can be here?

Рет қаралды 33,179

Күн бұрын

Пікірлер: 25

@berndkru 2 ай бұрын

There is no calculation necessary, because within the reals no quadrat can be negative and so x^4 cannot be negative for reals.

@yiutungwong315 2 ай бұрын

All Complex Number Solution...

@thebasisti2482 Ай бұрын

There are no real solutions.

@ЛидийКлещельский-ь3х 2 ай бұрын

Thank you. But it’s possible a little differently . (1) X^4=-4=4*{cos[pi+2*pi*n]+i*sin[pi+2*pi*n] }=4*e^[i*pi*(1+2*n) , n=0,1,2,3 ; (2) Xn=sqrt(2)*e^(i*[pi/4]*[1+2*n] ]=sqrt(2)*{ cos( [pi/4]*[1+2*n ])+i*sin( [pi/4]*[1+2*n] ) ; (3.0) Xo=sqrt(2)*{cos(pi/4)+i*sin(pi/4) }=1+i ; (3.1) X1=sqrt(2)*{cos(3*pi/4)+i*sin(3*pi/4) }=-1+i ; (3.2) X2= sqrt(2)*{cos(5*pi/4)+i*sin(5*pi/4) }=-1-i ; (3.3) X3=sqrt(2)*{cos(7*pi/4)+i*sin(7*pi/4)}=1-i !!!!!!😊)!!!! With respect, Lidiy

@PerthScienceClinic Ай бұрын

"How many real solutions can be here?" Pretty sure none of those solutions are real...

@GillesF31 25 күн бұрын

Yes, or (via exponential and trigonometric forms) ... x⁴ = -4 x⁴ + 4 = 0 (x²)² + 2² = 0 (x² - 2·i)·(x² + 2·i) = 0 --- /// case: (x² - 2·i) = 0 x² - 2·i = 0 x² = 2·i || || set 2·i on exponential form: || || z = 0 + 2·i || || |z| = √(0² + 2²) = √4 = 2 || || θ = π/2 rad || || z = 2·e^(i·(π/2)) || x² = 2·e^(i·(π/2)) x = ±(2·e^(i·(π/2)))^(1/2) x = ±2^(1/2)·e^(i·(π/2))^(1/2) x = ±2^(1/2)·e^(i·(π/2)^(1/2)) x = ±2^(1/2)·e^(i·(π/4)) x = ±√2·e^(i·(π/4)) x = ±√2·(cos(π/4) + i·sin(π/4)) x = ±√2·(√2/2 + i·√2/2) x = ±(√2·√2/2 + i·√2·√2/2) x = 1 + i and -1 - i --- /// case: (x² + 2·i) = 0 x² + 2·i = 0 x² = -2·i || || set -2·i on exponential form:: || || z = 0 - 2·i || || |z| = √(0² + 2²) = √4 = 2 || || θ = -π/2 rad || || z = 2·e^(i·(-π/2)) || x² = 2·e^(i·(-π/2)) x = ±(2·e^(i·(-π/2)))^(1/2) x = ±2^(1/2)·e^(i·(-π/2))^(1/2) x = ±2^(1/2)·e^(i·(-π/2)^(1/2)) x = ±2^(1/2)·e^(i·(-π/4)) x = ±√2·e^(i·(-π/4)) x = ±√2·(cos(-π/4) + i·sin(-π/4)) x = ±√2·(√2/2 - i·√2/2) x = ±(√2·√2/2 - i·√2·√2/2) x = 1 - i and -1 + i --- /// final results: ■ root #1: x = 1 + i ■ root #2: x = -1 - i ■ root #3: x = 1 - i ■ root #4: x = -1 + i 🙂

@ssalmero Ай бұрын

x=4^(1/4).e^î((pi+2npi)/4) for n=0, 1, 2 and 3. All of them complex numbers and thus no real solutions.

@hec-hd8861 Ай бұрын

Depuis le début vous pouvez constater qu'il n y a pas de solutions dans IR

@БелАлекс Ай бұрын

Так х^4 больше либо равно нулю то есть сразу реальных корней нет. Ну а так как есть х^4 то комплексных корня четыре. Так ведь в задаче требовалось только количество корней определить, а не искать их конкреиные значения.

@ressouguerrier6043 2 ай бұрын

X⁴=-4;X⁴+4=0;(X²)²+2²=0 :La somme de deux carrée n'est jamais nul.(X²+2)(X²+2)=0:X²++2=>X²=-2,X²+2=0,X²+(√2)²=0;(X+√2)(X+√2)=0;X+√2=0;X=-√2; (-√2)⁴=-4;(-√2)2(-√2)²=-4; 2×2=-4; 4≠-4; S={0}.