Solving x^3+x^2+4=0 in Two Ways

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[laurentthais]

Please do not say "Cardano's method", this is giving credit to a cheat. Cardano copied and published the original method due to Niccolo Tartaglia who was the first to solve cubic equations.
In the special case where the 3 roots are real (4p^3+27q^2 < 0), the method of the French mathematician François Viette is the most efficient. It is based on a change of variable involving the cosine of the triple angle, cos(3x) = 4 cos^3(x) - 3 cos x.
This avoids the clumsiness of general Tartaglia's formulii.

[depressedguy]

Use rational root and synthetic division

[quite_unknown_]

Easy method:
x^3 + x^2 =-4
x^2(x+1) = -4
(x+1) must be negative since x^2 always positive
Notice x^2(x+1) is increasing from - infinity to -1 for all x <-1
x=-2 is a solution and therefore the only solution

[mintentha]

I found x = -2 by inspection (easier to find by inspection by looking at x^3 + x^2 = -4) and then used polynomial long division to divide x^3+x^2+4 / x+2 to factor it and solve it from there

[darklordbgextrachannel]

The second method is pure power and beauty

[rakenzarnsworld]

Difference: 3x^2+2x
x(3x+2)=0
x = 0 or -2/3
So f(x) slightly decreases on -2/3 < x < 0 and increases on x < -2/3 and x > 0
x^3+x^2 = -4, so real root must be negative.
When x = -2, f(x) = -8+4=-4, the real root is x = -2.
We already know that two roots depend on quadratic formula.
(x+2)(x^2-x+2)=0
{1+-(1-8)^(1/2)}/2 = {1+-7^(1/2)i}/2

[riccardofroz]

Similarly to the second method:

x^3 + x^2 + 4 = 0
add and subtract 4x
x^3 + x^2 - 4x + 4x + 4 = 0
regroup and factor x from (x^3 - 4x)
x(x^2 - 4) + (x^2 + 4x + 4) = 0
since (x^2 - 4)=(x + 2)(x - 2) and (x^2 + 4x + 4)=(x + 2)^2
x(x - 2)(x + 2) + (x + 2)^2 = 0
factor (x+2)

(x + 2)(x(x - 2) + x + 2) = 0
(x + 2)(x^2 - 2x + x + 2) = 0
(x + 2)(x^2 - x + 2) = 0
x=-2
x=(1+-sqrt(-7))/2

[angelmendez-rivera]

With these types of equations, the best way to start is using the rational root theorem. From the theorem, x^3 + x^2 + 4 = 0 ==> x = –2, and (x^3 + x^2 + 4)/(x + 2) = (x^3 + 2·x^2 – x^2 + 4) = x^2 – (x^2 – 4)/(x + 2) = x^2 – (x – 2) = x^2 – x + 2 = x^2 – x + 1/4 + 7/4 = (x – 1/2)^2 + 7/4 = [x – 1/2 + sqrt(7)·i/2]·[x – 1/2 – sqrt(7)·i/2] ==> x = [1 + sqrt(7)·i]/2 or x = [1 – sqrt(7)·i]/2. Therefore, x^3 + x^2 + 4 = 0 <==> x = –2 or x = [1 + sqrt(7)·i]/2 or x = [1 – sqrt(7)·i]/2

[vansf]

x^3 + x^2 + 4 = 0
x = - 2 ----> (- 2)^3 + (-2)^2 + 4 = 0
( -2) x (-2)x (-2) + (-2) x(-2) + 4 = 0
you can never prove (-2)^2 = ( -2 x - 2) = -2 -2 to be the same as (+2)^2 = (+2 x +2) = + 2 + 2 = + 4 or 4
because the negative sign " - 2 " stands for decreasing in value by 2, whereas +2 stands for an increasing value by 2, and negative sign - merely indicates decreasing in value, and positive sign merely indicates increasing in value, but they are not any actual value
If x = + 2, x^2 = +2^2 = ( + 2 ) x (+2) = + 2 + 2 = 4
1 2 = 2 times of + 2 = + 4 = 4
But If x = - 2, x^2 = (-2)^2 = negative (-2) x (-2) = negative 2 times of negative - 2
negative 1 2 = negative 2 times of -2 = the number of the negative 2 is negative 2, which is nonsense. The number of negative 2 can be 2, but can never be negative 2
Let the number negative two be a female. If you say the number of females in the room is 2, it will means that there are 2 females. It can mathematically expressed as 2 times of a female = 2 x 1 female = + 1 female + 1 female = 2 females
. But if you say that the number of females in the room is - 2 ( negative two), it will mean that there are negative 2 females, which is nonsensical, and can be mathematically expressed as negative 2 times of a female = - 2 x 1female = - (1 female + 1 female) = negative 2 females, which is obviously nonsense.
Let's take another example with a decreasing quantity value for you guys to understand why taking even power of a negative number can never turn it into a positive numberr as fundamentally misunderstood by ignorant human mathematicians
The population of the city of Rome has decreased by 1 million, which can be mathematically expressed as follows:
P = p - 1 millions, where p is the previous population of Rome. The negative number - 1million is the decreasing number of people, in which the negative sign indicates a decreasing in population, but it does not represents and actual value of the population, and the number 1 million is the quantity of people to be taken away from the previous population, the previous population P is purely the number of people of the previous population, but not any increasing nor decreasing number of people to be added to or taken away from anywhere nor any other population of Italy. Hence, P can never ever have any positive nor negative sign
P current = P previous - 1million people
Similarly, if the population has increased by 2 million people, the increasing number of people can be mathematically expressed as: + 2 million people, where the positive sign merely indicates an increasing number of people, but it does not represents any actual number of people, and the number of 2 million people is the number to be added to the previous population
P current = P previous + 2 million people
Now if the population has increased 10 times of the previous population, then we will have a multiplication as
P current = 10 x P previous . Let the previous population be 10 millions--->
P current = 10 x ( 10 m) = 10^2 = 10 x 10 =
10 m + 10m +10m+10m+10m +10m +10m+10m +10m +10m = 100 millions
1 2 3 4 5 6 7 8 9 10 = 10 times of P previous
We can see that the number of the current population is 10 times of P previous = 10X P previous, but not negative 10 times
Now if the population has decreased by 1/5 ---> P current = (4/5) P previous x =( 1/5) 10m= 2 millions. since 2 millions are the decreasing number of people to be taken away from 10 millions, it must have a negative sign - as - 2 m. The number of population now is 1/5 times of P previous, and 1/5 times can never be negative.
If the population keeps decreasing by 3 times of the previous decreasing number of people, which was - 2 millions, the current decreasing number of people now will be: 3 times of - 2 million = 3 times of -2 m= -2m - 2m -2 m = -6 millions.
1 2 3 = 3 times of -2 millions = - 6 millions, which is the current decreasing number of people to be taken away from the previous population of 10m-2m = 8 millions
So, the current decreasing number of people is - 6 millions where the negative sign - indicates 6 million is the decreasing value, and the number of 6 millions is the number of people to be taken away from the pervious population of 8 millions
----> P current = 10 millions - 2 millions - 6millions= 2 millions . Again, the number of decreasing population, which is 3 times of the previous decreasing number of people is not negative either

Now adding a negative sign to 3 times of - 2 millions will give:
- 3 times of - 2 millions = -3 x - 2 millions = negative - 2m - 2m - 2m
negative 1 2 3 = negative 3 times of - 2 millions, which is nonsensical
Similarly, earlier, we had - 2 times of - 2 = -2 x -2 = (-2)^2 = negative 2 times of -2 is obviously so nonsensical as this latest example

if you use merely human-invented defective concepts of mathematics full of flaws, without applying logos nor any form of matter in the real physical world to examining the mathematical baseless notion that taking power of 2 of a negative number will make it become positive as if you had a magic cane to turn negative into positive, you will obviously never ever be able to see such irrationalities or illogicalities as what i am showing you

[tontonbeber]

Another easy method ...
It's obvious by trying that -2 is solution, then to demonstrate it's unique in R, of course you can divide the polynome and solve the quadratic, which have complex conjugate solutions.
But there is much faster ...
If you derivate to check local min and max ... f' = 3x2+2x, whose solutions are -2/3 and 0.
As the cubic order term of the original equation is positive, the cubic is almost always increasing, so the lowest -2/3 is a maximum and 0 is a minimum.
You can also prove that with 2nd derivative, f'' = 6x+2 and f''(0) = 2 > 0, so 0 is the minimum.
As f(0)=4 >0 it is clear that the cubic has only one intersection with x axis, so only one real solution.

[andrasdekoos]

I must be getting better at these. I saw immediately that x=-2 is a solution. The rest is straightforward. Not a very systematic approach, but I'll take it.

[NadiehFan]

With the first method, you don't have to solve a quadratic to find the other two roots once you have

y₁ = a + b = −⁵⁄₃

and

a − b = ⅓√21

since the other roots are

y₂ = −½(a + b) + i·½√3(a − b) = ⅚ + i·⅙√63 = ⅚ + i·½√7
y₃ = −½(a + b) − i·½√3(a − b) = ⅚ − i·⅙√63 = ⅚ − i·½√7

and so

x₂ = y₂ − ⅓ = ½ + i·½√7
x₃ = y₃ − ⅓ = ½ − i·½√7

[Chhatrapatiraj]

I am able to find solution as i am in 10th standard, that is
x³+x²+4=0
x²(x+1)= -4
x²(x+1) = (-2)² (-2+1)
On comparison x = -2
So (x+2) is a factor of x³+x²+4
By long division we get x²-x+2
And now by quadratic equation we can easily find answers

[chaparral]

3rd method: just looking: one root is -2 so divide by x+2, gives x² - x + 2. with complex roots 1/2 + i*sqrt(7)/2 and 1/2 - i*sqrt(7)/2 . ligget se.

[stevenlitvintchouk]

I always like to graph these functions so I can see what we're dealing with. The graph of x^3 + x^2 + 4 = 0 suggested the root x = -2, which you then verify by substitution.Then by the Factor Theorem, divide the cubic by (x+2) to get a quadratic, from which you get the other two (complex) roots.

[__Junioor__]

-2 is an obvious solution so you can factor by (x+2). You find a degree 2 polynomial that you can easily solve with the quadratic formula

[vansf]

Factoring the eq will give (x+2)(x^2-x+2) = 0---> x + 2 = 0----> x = -2
x^2 - x + 2 = 0 is nonsense because
x^2 -x = -2, but x^2 is always non-negative, either = or > x, and thus x^2 - x will always be non-negative. Hence, the non-negative LHS can never ever = the negative RHS
Therefore x^2 - x + 2 cannot = 0. In other words, no solution for this nonsensical equation
Another way:
x^3 - x^2 + 4 = 0 ---> x^3 - x^2= -4 ---->( x^3/4 ) - (x^2/4) = -1
Since x^2/4 is always non-negative, x^3/4 must be negative for (x^3/4) - (x^2) to yield a negative result, and x^3/4 - x^2/4 must be -1. Notice that x^2/4 must be > 0 or >=1
It means that the smallest value of x^2/4/=1----> x^3/4 must be -2
x^3/4 = -2----> x = sqrt(-8) = -2

[ostorjulien]

Thanks for the video !

Simpler version of the first method : divide the equation by x^3 and set y = 1/x.

The new equation is 4y^3+y+1=0, you can directly apply Cardano's formula and the coefficients are nicer.

Bonus : multiply the equation by 2 and set u=2y, the equation becomes u^3+u+2=0.

[christianthomas]

same stuff: looking for an evident solution in that case, you find x= -2; then you have (x+2)(x**2-x+4)
as the second factor has no real solution, the only one is x=2. easy.

[pageegap]

We can easily find -2 as integer Root of x³+x²+4=0. Because x²+4 is positive the root must be negative . The trick is: if there is an integer root then it must divide the constant 4. So let's try with -1, -2, -4 and 1, 2, 4. Finding root -2 we divide by (x+2) to find the quadratic term.