Show that x^3 + 4x - 1 = 0 has exactly one real root?

Question

Please show all processes involved to solve it please, i'm having difficulty in solving it

Answer 1

In order to solve this equation or show that it has exactly one real root, we need to state the Fundamental Theorem of Algebra. It states that a Polynomial of degree n has at least one root, real or complex. With this theorem, we can elaborate that our Polynomial P(x) of degree n has exactly n roots, real or complex.

Now, let P(x) = x^3 + 4x - 1. Based on the aforementioned theorem and its subsequent corollary, our P(x) has 3 roots, real or complex.

Using the Descartes' rule of signs, there is one shift from + to -. This means that it has at most one positive real root. If we replace it with the -x, we get 0 sign changes. This means there are no negative real roots.

In order to determine for the real root, we can run an equation solver (Matematica) and get that the root is approximately 0.26. I know this is a bit weird to end in this manner but the equation does not have an integer root if we check it using the Integer Root Theorem.

The other method you can try is the Newton's Approximation Method but it involves Calculus.

Answer 2

this equation can have:
2 imaginary roots (since they come in pairs), 1 real root
or
3 real roots.

The rational root theorem says that this equation can have only 1 as its roots (being that the root is rational). This number is obtained by dividing the constant (1) with the coefficient of the highest degree (1). However, if this equation is graphed, one is not the root. Therefore, this polynomial does not have a rational root.
This problem is unique in that 1 is a number that can be obtained only if it is multiplied with another 1. Thus, since this equation has 3 possible roots, if the root is not divisible by 1, then there must not be 3 real roots. Because no cube root of any number can give 1 anything other than 1. This means that, by using the graph, this equation must have imaginary roots in order for the real root to be anything other than a 1. Therefore, if this problem has 2 imaginary roots, the third one must be real.

Sorry it's too long. Hard to explain. Hope it's not confusing but helpful.

Answer 3

Let's use Rolle's theorem.
That says that between any 2 roots of this function,
there must be at least 1 zero of the derivative.
But the derivative of f(x) = x³ + 4x -1 is 3x² + 4, which
has no real zeros, so f(x) has at most 1 real zero.
To see that has 1, note that f(-2) = -17
and f(1) = 4. Since f is continuous(it is a polynomial),
the intermediate value theorem tells us that there
must be a zero of f(x) between -2 and 1.

Answer 4

Let's put f(x)=x^3+4x-1
f is continous in R and
f(0)=-1<0
f(3)=14>0
So (Bolzano's theorem)
has 1 at least root in (0,3)
Let's suppose it has 2 roots
r1 and r2 (r1 Then f(r1)=f(r2)=0
f is continous in R and
has derivative in R
f'(x)=3x^2+4
Then(Roll's Theorem)
exists mER such as
f'(m)=0
3m^2+4=0
That's not true so f
has not more than 1 root
Finally f has exactly 1