Let y=x^2+1. The given equation then reads y-x = sqrt(4/3xy) > x^2+y^2-10/3xy=0 > y = x/3, 3x. The first does not yield any real x as solutions. If y=3x, x^2-3x+1=0 > x = 1/2[3+/-sqrt(5)], both of which are valid solutions.

Quartic nicely avoided! Clever move - thanks!

(3+√5)/2,(3-√5)/2, (1+√35 i)/2,(1-√35 i)/2

X= (3+ -√5)/2 only real solns. Rest is complex.

Two real solutions x=(3+or-sqrt5)/2 and two complex ones x=(1+or-isqrt35)/6. The other real solution (double) x=-1 is rejected.

I have got the correct answer, happy.

χ=(3+ -ριζα5)/2 στο R

3(x² - x + 1)² = 4(x³ + x) = 4x(x² + 1) x² + 1 = u 3(u - x)² = 4ux 3(u + x)² = 16ux ux = (3/16)(u + x)² u + x = b ux = (3/16)b² t² - bt + (3/16)b² = 0 t = (b ± b/2)/2 t = 3b/4 or t = b/4 => u = 3x or u = x/3 x² + 1 = u x² + 1 = 3x x² - 3x + 1 = 0 *x = (3 ± √5)/2* x² + 1 = u x² + 1 = x/3 x² - x/3 + 1 = 0 x = [1/3 ± √(1/9 - 4)]/2

3x^9/3x^9 1x^1 (x ➖ 1x+1) 4x^3+4x^3 ➖4x+4x ➖ = {8x^6+8x^2}=18x^8 18x^8/3=6x2^6.2 3^2x2^3^2^2.2 1^1x1^3^1^.2 x^3^.2 (x ➖ 3x+2)

1.Почему без проверки может ли х=0, делится на х. А если х=0, является корнем исходного уравнения? 2. Почему в одних примерах находятся комплексные корни, а в других нет. Какой критерий? Разумеется если это не определяется условиями примеров.

@moeberry8226

20 күн бұрын

There is no criteria for him to ask for real solutions or non real solutions, that is at his discretion. The reason you can divide by x because he’s saying we know obviously x=0 is not a solution because you can put it into the original equation and you will not get a true statement. However even if you didn’t know if x=0 was a solution you can still divide by x stating x cannot be 0 and then handle that case separately at the end.