What are the solutions to the equation x^4+x^3+x+1=4x^2 ?

The solutions to the equation x^4+x^3+x+1=4x^2 are x=1,x=(-3+sqrt(5))/2 ,x=(-3-sqrt(5))/2

Find the zeros of x^4+x^3+x+1=4x^2

The zeros of x^4+x^3+x+1=4x^2 are x=1,x=(-3+sqrt(5))/2 ,x=(-3-sqrt(5))/2

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x^{4} + x^{3} + x + 1 = 4 x^{2}

x = 1, x = \frac{- 3 + 5}{2}, x = \frac{- 3 - 5}{2}

Decimal Notation

x = 1, x = - 0.38196 \dots, x = - 2.61803 \dots

Solution steps

x^{4} + x^{3} + x + 1 = 4 x^{2}

Move

4 x^{2}

to the left side

x^{4} + x^{3} + x + 1 - 4 x^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

x^{4} + x^{3} - 4 x^{2} + x + 1 = 0

Factor

x^{4} + x^{3} - 4 x^{2} + x + 1 : (x - 1)^{2} (x^{2} + 3 x + 1)

(x - 1)^{2} (x^{2} + 3 x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x^{2} + 3 x + 1 = 0

Solve

x - 1 = 0 : x = 1

Solve

x^{2} + 3 x + 1 = 0 : x = \frac{- 3 + 5}{2}, x = \frac{- 3 - 5}{2}

The solutions are

x = 1, x = \frac{- 3 + 5}{2}, x = \frac{- 3 - 5}{2}

x^{4} - 18 x^{2} + 32 = 0

x^{3} - 5 x^{2} + 6 x - 1 = 0

x^{3} + x^{2} - 21 x - 45 = 0

x^{4} - x^{3} - 7 x^{2} - 14 x - 24 = 0

9 = d^{7} d

What are the solutions to the equation x^4+x^3+x+1=4x^2 ?
The solutions to the equation x^4+x^3+x+1=4x^2 are x=1,x=(-3+sqrt(5))/2 ,x=(-3-sqrt(5))/2
Find the zeros of x^4+x^3+x+1=4x^2
The zeros of x^4+x^3+x+1=4x^2 are x=1,x=(-3+sqrt(5))/2 ,x=(-3-sqrt(5))/2