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

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

Find the zeros of x^2(x+1)^2+x^2=3(x+1)^2

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

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Popular >

x^2(x+1)^2+x^2=3(x+1)^2

Solution

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

Solution

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

Decimal Notation

x = - 0.61803 \dots, x = 1.61803 \dots

Solution steps

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

Expand

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

Expand

3 (x + 1)^{2} : 3 x^{2} + 6 x + 3

x^{4} + 2 x^{3} + 2 x^{2} = 3 x^{2} + 6 x + 3

Move

3

to the left side

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

Move

6 x

to the left side

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

Move

3 x^{2}

to the left side

x^{4} + 2 x^{3} - x^{2} - 6 x - 3 = 0

Find one solution for

x^{4} + 2 x^{3} - x^{2} - 6 x - 3 = 0

using Newton-Raphson:

x \approx - 0.61803 \dots

Apply long division:

\frac{x ^{4} + 2 x ^{3} - x ^{2} - 6 x - 3}{x + 0.61803 \dots} = x^{3} + 1.38196 \dots x^{2} - 1.85410 \dots x - 4.85410 \dots

x^{3} + 1.38196 \dots x^{2} - 1.85410 \dots x - 4.85410 \dots \approx 0

Find one solution for

x^{3} + 1.38196 \dots x^{2} - 1.85410 \dots x - 4.85410 \dots = 0

using Newton-Raphson:

x \approx 1.61803 \dots

Apply long division:

\frac{x ^{3} + 1.38196 \dots x ^{2} - 1.85410 \dots x - 4.85410 \dots}{x - 1.61803 \dots} = x^{2} + 3 x + 3

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

Find one solution for

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

using Newton-Raphson:

No Solution for

x \in R

Approximate to algebraic form

x \approx - 0.61803 \dots : x = - \frac{5 - 1}{2}

x \approx 1.61803 \dots : x = \frac{5 + 1}{2}

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

The solutions are

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

Popular Examples

x^3-x=x^2+5x-6

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

roots x^4-81 roots

x^{4} - 81

x^6-8x^5+12x^4=0

x^{6} - 8 x^{5} + 12 x^{4} = 0

x^3-4x^2-3x+6=0

x^{3} - 4 x^{2} - 3 x + 6 = 0

2x^4-x^3-14x^2+19x-6=0

2 x^{4} - x^{3} - 14 x^{2} + 19 x - 6 = 0

Frequently Asked Questions (FAQ)

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