it is normal cubic equation, x to the degree of 3 + 2 x to the degree of 2 + x - 100 = 0, you can find the discriminant (q/2) to the degree of 2 + (p/3) to the degree of 3, and then proceed accordingly, depends if discriminant is bigger or less than zero and what value of p is
You would do your students a much better service if at 1:28 you would show them how to use the Rational Root Theorem to identify m=2 as a root. Then show them how to do polynomial division to find m^2 + 2m + 5 as the other factor. The magic you are doing in rewriting -10 as -8 - 2 makes sense only because you know the answer in advance. This teaches them nothing about how really to solve these problems.
Is it possible that the two complex solutions x2 and x3 are not correct ? When I substitute these values in the left side of your original equation, I got -10 not 10 !
or x√x +√x=10 signifie √x(x+1)=10 signife (√x(x+1))exp2=100 siginfie x(x2+2x+1)-100 =0 signifie x3+2x2+x-100=0 on remarque 4 estune racine apparente de l'equaion donc E =(x-4)(ax2+bx+c)=0 1ére méthode on développe puis identification des coefficients pour trouver une équation de sécond dégre or on 2ieme méthode on fait la division eucludiennhe de E par x-4 on trouve E =(x-4)(x2+6x+25)=0 aprés calcule delta on trouve tous calcul fait x =-3-4i et x=-3+4i conclusion x=( 4;-3+4i;-3-4i)
it is normal cubic equation, x to the degree of 3 + 2 x to the degree of 2 + x - 100 = 0, you can find the discriminant (q/2) to the degree of 2 + (p/3) to the degree of 3, and then proceed accordingly, depends if discriminant is bigger or less than zero and what value of p is
Very nice! ❤
You would do your students a much better service if at 1:28 you would show them how to use the Rational Root Theorem to identify m=2 as a root. Then show them how to do polynomial division to find m^2 + 2m + 5 as the other factor. The magic you are doing in rewriting -10 as -8 - 2 makes sense only because you know the answer in advance. This teaches them nothing about how really to solve these problems.
Noted! very nice suggestion! ❤
Correct
Very nice! ❤
4
Very nice! ❤
Is it possible that the two complex solutions x2 and x3 are not correct ? When I substitute these values in the left side of your original equation, I got -10 not 10 !
m = {2, -1+2i, -1-2i}, so x = {4, -3-4i, -3+4i}.
I will check it! ❤
It took my 10 secs answer is 4.
Very nice! ❤
That's just a guess. How would you solve it if it were, for example, 11, or 68?
It can be solved in the same way! ❤
@@SALogics How?
I thing no. For example x√x+√x=11. You will get the equation m^3 - m = 11. Please show how you would solve this?
@SALogics Sorry for the mistake in the sign, not m^3 - m = 11, but m^3 + m = 11. How would you solve it please?
@@paveltubobviously he can't do this via his method. Please tell how will you do this without cubic formula.
O bhqi sidha question tha x ki value 4 put kr do😊
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let u=Vx , u^3+/-u^2+u-10=0 , (u-2)(u^2+2u+5)=0 , u=2 , Vx=2 , x=4 , test , 4*V4+V4=8+2 , --> , 10 , OK ,
1 -2
2 -4
5 -10
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x=4
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or x√x +√x=10 signifie √x(x+1)=10 signife (√x(x+1))exp2=100
siginfie x(x2+2x+1)-100 =0
signifie x3+2x2+x-100=0
on remarque 4 estune racine apparente de l'equaion
donc E =(x-4)(ax2+bx+c)=0 1ére méthode
on développe puis identification des coefficients pour trouver une équation de sécond dégre
or on 2ieme méthode on fait la division eucludiennhe de E par x-4
on trouve E =(x-4)(x2+6x+25)=0 aprés calcule delta on trouve tous calcul fait x =-3-4i et x=-3+4i
conclusion x=( 4;-3+4i;-3-4i)
Very nice! ❤
...ou en divisant le polynôme x3+2x2+x-100 par x-4...