How do you solve x^4=2x^3 + 1?

Question

I'm taking pre-calculus in the summer, and this one just stumped me. I know it's easier than I think it is, but it's just not coming! Any help?

Answer 1

There are 2 real roots of that equation. The best way is to use Newton's method.

x_new = x_old - f(x_old) / f'(x_old)
f(x) = x^4 - 2x^3 - 1
f'(x) = 4x^3 - 6x^2

Starting with x = 2, the positive root is
x = 2.1069
Starting with x = -2, the negative root is
x = -0.7167

**EDIT**
If I'm reading the question correctly, it is
x^4 = 2x^3 + 1
NOT x^4 = 2x^3 x 1
x = 2 does not work

Answer 2

This is the "if you haven't got a fancy calculator" method.

Rearrange to : x^4 - 2x^3 - 1 = 0

This is a quartic equation of the form :
x^4 + ax^3 + bx^2 + cx + d = 0,
where, a = -2, b = 0, c = 0, d = -1.

First calculate coefficients of the cubic equation
given by : y^3 + ey^2 + fy + g = 0,
where, e = -b, f = ac - 4d, g = 4bd - c^2 - a^2d.

Thus : e = 0, f = 4, g = 4.

Now to find a real root of the cubic : y^3 + 4y + 4 = 0.

Calculate : Q = (3f - e^2) / 9 = 12 / 9 = 4 / 3.
Calculate: R = (9ef - 27g - 2e^3) / 54 = -108 / 54 = -2
Calculate: Q^3 + R^2 = 64 / 27 + 4 = 172 / 27
Calculate: S = [R + sqrt(Q^3 + R^2)]^(1/3) = 0.8061809063
Calculate: T = [R - sqrt(Q^3 + R^2)]^(1/3) = -1.6538885044
Calculate the real root: y = S + T - e/3 = -0.8477075981

Now to find the roots of the quadratic equation
given by : z^2 + mz + n = 0,
where, m = (1/2)[a ± sqrt(a^2 - 4b + 4y)]
and n = (1/2)[y ± sqrt(y^2 - 4d)].
The positive sqrt for m goes with the positive sqrt for n,
and the negative sqrt for m goes with the negative sqrt for n.

Thus :
(1) m = -0.6097534089 and n = 0.6622640780
(2) m = -1.3902465911 and n = -1.5099716762.

Solving (1) :
z^2 - 0.6097534089z + 0.6622640780 = 0
Solutions are : z = 0.30487670 ± 0.75452917*i

Solving (2) :
z^2 - 1.3902465911z - 1.5099716762 = 0
Solutions are : z = -0.71667275 or 2.10691934

The four roots of x^4 - 2x^3 - 1 = 0 are thus :

x1 = -0.71667275
x2 = 2.10691934
x3 = 0.30487670 - 0.75452917*i
x4 = 0.30487670 + 0.75452917*i

Answer 3

I dont know if there is a way to look for the root of any quartic equation, so I just type
evalf[10](solve(x^4=2*x^3+1,x))
in Maple console and it can search for the answer using a tecnique called numeric method. This is all the 4 roots of the equation.
2.106919340,
.3048767045+.7545291731*I,
-.7166727493,
.3048767045-.7545291731*I

Answer 4

x^4 / 2x^3=1 x/2=1 x=2
2^4=16 2(8) +1=17
I think you have an inconsistent equation.

Answer 5

the answer is two. you divide 2x^3 by both sides, therefore creating x^4/2x^3 = 1.

you subtract the exponents so you are left with x^1/2 = 1

since x^1 is the same as x, you are left with x/2 = 1. multiply 2 by both sides, therefore resulting in x=2

Answer 6

x^4 - 2x^3 - 1 = 0
By numeric method, the two real roots are
x = - 0.716672749, 2.10691934