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can also use synthetic division to determine x=3 and then use quadratic equation to find the remaining two complex roots.
That's a great way to solve the problem!
Thank you sir l hope to become a professer in mathimatic like you
Most welcome
x³-x²=18x²(x-1)=18 → x-1>0 → x>1x²(x-1)=9*2x²(x-1)=3²(3-1)x=3
x=3, solved according to the Horner's scheme.
❤️❤️
Linda solução, (-18 = -27 + 9) daqui para frente tudo tornou fácil.
Sim, essa parte foi fundamental!
x^3-3x^2+2x^2-6x+6x-18=0=>(x-3)(x^2+2x+6)=0=>x={3,two conjugate complex roots exist as discriminent of quadratic part
That's a great way to solve the equation!
X = 3
can also use synthetic division to determine x=3 and then use quadratic equation to find the remaining two complex roots.
That's a great way to solve the problem!
Thank you sir l hope to become a professer in mathimatic like you
Most welcome
x³-x²=18
x²(x-1)=18 → x-1>0 → x>1
x²(x-1)=9*2
x²(x-1)=3²(3-1)
x=3
x=3, solved according to the Horner's scheme.
❤️❤️
Linda solução, (-18 = -27 + 9) daqui para frente tudo tornou fácil.
Sim, essa parte foi fundamental!
x^3-3x^2+2x^2-6x+6x-18=0=>(x-3)(x^2+2x+6)=0=>x={3,two conjugate complex roots exist as discriminent of quadratic part
That's a great way to solve the equation!
X = 3