Find all of the real and imaginary solutions of x^3+8=0

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SkiffMackinaw SkiffMackinaw · Answer 1 · 2021-01-16T03:41:02+01:00

Answer:

[tex]x=-2, \:1+\sqrt{3}i,\: 1-\sqrt{3} i[/tex]

Step-by-step explanation:

[tex]x^3+8=0,\\x^3=-8,\\\mathrm{For\:}x\in \mathrm{R}, x=-2[/tex]

Real solutions ([tex]x \in \mathrm{R}[/tex]): [tex]x=-2[/tex]

However, recall that [tex](\frac{-1\pm\sqrt{3}i}{2})^3=1[/tex].

Therefore:

[tex]\frac{-1\pm\sqrt{3}i}{2}\cdot -2=-2[/tex].

Thus, for [tex]x^3=-8,\: x\notin \mathrm{R}[/tex]:

[tex]x=1\pm\sqrt{3}i[/tex].

Therefore, the real and nonreal solutions of [tex]x^3+8=0[/tex] are [tex]\fbox{$-2, \:1+\sqrt{3}i,\: 1-\sqrt{3}i$}[/tex].

lacampanella lacampanella · Answer 2 · 2021-01-16T03:41:17+01:00

Answer:

-2, 1 +/- i*sqrt(3)

(3 solutions)

Step-by-step explanation:

Because the left side is in the form a^3+b^3, we can factor it as (a+b)(a^2-ab+b^2) (good formula to remember)

So we have

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

We can see one value already

x+2 = 0

x = -2

now we solve for

x^2-2x+4 = 0

We can use the quadratic formula, and see that the determinant is negative (2^2-4*1*4 < 0)

Therefore we have two imaginary solutions

(-(-2) +/- sqrt(2^2-4*1*4))/)2*1) = (2 +/- sqrt(-12))/2 = (2 +/- 2sqrt(-3))/2 = 1+/- sqrt(-3) = 1 +/- i*sqrt(3)