What are the solutions to the equation (x+1)(x+2)^2(x+4)-2x^2=0 ?

The solutions to the equation (x+1)(x+2)^2(x+4)-2x^2=0 are x=sqrt(5)-3,x=-sqrt(5)-3

Find the zeros of (x+1)(x+2)^2(x+4)-2x^2=0

The zeros of (x+1)(x+2)^2(x+4)-2x^2=0 are x=sqrt(5)-3,x=-sqrt(5)-3

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(x + 1) (x + 2)^{2} (x + 4) - 2 x^{2} = 0

x = 5 - 3, x = - 5 - 3

Decimal Notation

x = - 0.76393 \dots, x = - 5.23606 \dots

Solution steps

(x + 1) (x + 2)^{2} (x + 4) - 2 x^{2} = 0

Expand

(x + 1) (x + 2)^{2} (x + 4) - 2 x^{2} : x^{4} + 9 x^{3} + 26 x^{2} + 36 x + 16

x^{4} + 9 x^{3} + 26 x^{2} + 36 x + 16 = 0

Find one solution for

x^{4} + 9 x^{3} + 26 x^{2} + 36 x + 16 = 0

using Newton-Raphson:

x \approx - 0.76393 \dots

Apply long division:

\frac{x ^{4} + 9 x ^{3} + 26 x ^{2} + 36 x + 16}{x + 0.76393 \dots} = x^{3} + 8.23606 \dots x^{2} + 19.70820 \dots x + 20.94427 \dots

x^{3} + 8.23606 \dots x^{2} + 19.70820 \dots x + 20.94427 \dots \approx 0

Find one solution for

x^{3} + 8.23606 \dots x^{2} + 19.70820 \dots x + 20.94427 \dots = 0

using Newton-Raphson:

x \approx - 5.23606 \dots

Apply long division:

\frac{x ^{3} + 8.23606 \dots x ^{2} + 19.70820 \dots x + 20.94427 \dots}{x + 5.23606 \dots} = x^{2} + 3 x + 4

x^{2} + 3 x + 4 \approx 0

Find one solution for

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

using Newton-Raphson:

No Solution for

x \in R

Approximate to algebraic form

x \approx - 0.76393 \dots : x = 5 - 3

x \approx - 5.23606 \dots : x = - 5 - 3

x = 5 - 3, x = - 5 - 3

The solutions are

x = 5 - 3, x = - 5 - 3

4 x^{4} - 17 x^{2} + 4 = 0

z^{3} = i

x^{6} + 1 = 0

x^{3} - 7 x + 6 = 0

2 5^{150} = 25 x^{50}

What are the solutions to the equation (x+1)(x+2)^2(x+4)-2x^2=0 ?
The solutions to the equation (x+1)(x+2)^2(x+4)-2x^2=0 are x=sqrt(5)-3,x=-sqrt(5)-3
Find the zeros of (x+1)(x+2)^2(x+4)-2x^2=0
The zeros of (x+1)(x+2)^2(x+4)-2x^2=0 are x=sqrt(5)-3,x=-sqrt(5)-3