What are the solutions to the equation (x+2)^4-4(x+2)^2-5=0 ?

The solutions to the equation (x+2)^4-4(x+2)^2-5=0 are x=sqrt(5)-2,x=-sqrt(5)-2

Find the zeros of (x+2)^4-4(x+2)^2-5=0

The zeros of (x+2)^4-4(x+2)^2-5=0 are x=sqrt(5)-2,x=-sqrt(5)-2

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(x + 2)^{4} - 4 (x + 2)^{2} - 5 = 0

x = 5 - 2, x = - 5 - 2

Decimal Notation

x = 0.23606 \dots, x = - 4.23606 \dots

Solution steps

(x + 2)^{4} - 4 (x + 2)^{2} - 5 = 0

Expand

(x + 2)^{4} - 4 (x + 2)^{2} - 5 : x^{4} + 8 x^{3} + 20 x^{2} + 16 x - 5

x^{4} + 8 x^{3} + 20 x^{2} + 16 x - 5 = 0

Find one solution for

x^{4} + 8 x^{3} + 20 x^{2} + 16 x - 5 = 0

using Newton-Raphson:

x \approx 0.23606 \dots

Apply long division:

\frac{x ^{4} + 8 x ^{3} + 20 x ^{2} + 16 x - 5}{x - 0.23606 \dots} = x^{3} + 8.23606 \dots x^{2} + 21.94427 \dots x + 21.18033 \dots

x^{3} + 8.23606 \dots x^{2} + 21.94427 \dots x + 21.18033 \dots \approx 0

Find one solution for

x^{3} + 8.23606 \dots x^{2} + 21.94427 \dots x + 21.18033 \dots = 0

using Newton-Raphson:

x \approx - 4.23606 \dots

Apply long division:

\frac{x ^{3} + 8.23606 \dots x ^{2} + 21.94427 \dots x + 21.18033 \dots}{x + 4.23606 \dots} = x^{2} + 4 x + 5

x^{2} + 4 x + 5 \approx 0

Find one solution for

x^{2} + 4 x + 5 = 0

using Newton-Raphson:

No Solution for

x \in R

Approximate to algebraic form

x \approx 0.23606 \dots : x = 5 - 2

x \approx - 4.23606 \dots : x = - 5 - 2

x = 5 - 2, x = - 5 - 2

The solutions are

x = 5 - 2, x = - 5 - 2

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

12 x^{3} - 30 x^{2} + 12 x = 0

6 x^{3} + 4 x^{2} - 7 x + 2

r^{3} = 0

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

What are the solutions to the equation (x+2)^4-4(x+2)^2-5=0 ?
The solutions to the equation (x+2)^4-4(x+2)^2-5=0 are x=sqrt(5)-2,x=-sqrt(5)-2
Find the zeros of (x+2)^4-4(x+2)^2-5=0
The zeros of (x+2)^4-4(x+2)^2-5=0 are x=sqrt(5)-2,x=-sqrt(5)-2