Solve the binomial equation x^3 + 8 = 0

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

x^3 + 8 = 0

First let us subtact 8 from both sies:

==> x^3 = -8

Now take the cubic root for both sides:

==> (x^3)^1/3 = (-8)^1/3

==> x = -2

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To solve the binomial equation, we'll apply the formula of the sum of cubes:

a^3 + b^3 = (a+b)(a^2 - ab + b^2)

a^3 = x^3

a = x

b^3 = 2^3 = 8

b = 2

x^3 + 8 = (x+2)(x^2 - 2x + 4)

If x^3 + 8 = 0, then (x+2)(x^2 - 2x + 4) = 0

If a product is zero, then each factor could be zero.

x + 2 = 0

We'll subtract 2 both sides:

x1 = -2

x^2 - 2x + 4 = 0

We'll apply the quadratic formula:

x2 = [2 + sqrt(4-16)]/2

x2 = (2+2isqrt3)/2

We'll factorize by 2:

x2 = 2(1+isqrt3)/2

x2 = 1+isqrt3

x3 = 1- isqrt3

The roots of the equation are: {-2 , 1+isqrt3, 1- isqrt3 }.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

x^3+8 = 0.

Factorise  LHS:

(x+2)(x^2-2x+4) = 0

Therefore x+2 = 0. Or x = -2.

Or

x^2-2x+4 = 0. Solve by quadratic formula:

x2 = {- -2 +sqrt{(-2)^2 -4*1*4]}/2 .

x2 = {2+2sqrt(-3)}/2

x2 = {1+sqrt-3)

x3 = 1-sqrt(-3)

x1 = -2

 

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