Find the number of real zeros of f(x) = 4x3 + 2

Select one:
a. 2
b. 3
c. 4
d. 1

Question

Find the number of real zeros of f(x) = 4x3 + 2

Select one:
a. 2
b. 3
c. 4
d. 1

accessdenied · Answer

You understand that real zeros are solutions to the equation f(x) = 0  such that x is a real number?

1645323 · Answer

yea so how do you solve this?

accessdenied · Answer

$4x^3 + 2 = 0$
This is deceptively painless.  We can isolate the $x^3 $ by subtracting the 2 from both sides then dividing by 4,  and then we take the cube root of both sides for x.

accessdenied · Answer

$x^3 = a  \iff x = \sqrt[3]{a}$

1645323 · Answer

ok so cubedroot(2)?

accessdenied · Answer

$ 4x^3 + 2 = 0 $
$ 4x^3 =  -2 $    Subtract 2 from both sides
$ x^3 = \dfrac{-2}{4} $    Divide both ides by 4.
Then we can use the cube root.  :)

accessdenied · Answer

It may not seem evident at first why we have a polynomial of degree 3, which we would predict has 3 complex roots, only has this one real value solution.  However, we are only using the piece of the cube root that gives us real values (there are two other complex cube roots of a number which we intentionally ignore).

1645323 · Answer

ok so is the answer 2 then?

accessdenied · Answer

The answer is the quantity of real solutions we found.
There could only be one real solution for x^3 = -2/4 <==>  x = cuberoot(-2/4)