What are the solutions to the equation (x^2-4)^2=(x+4)^2+32 ?

The solutions to the equation (x^2-4)^2=(x+4)^2+32 are x\approx-3.11851…,x\approx 3.67940…

Find the zeros of (x^2-4)^2=(x+4)^2+32

The zeros of (x^2-4)^2=(x+4)^2+32 are x\approx-3.11851…,x\approx 3.67940…

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(x^{2} - 4)^{2} = (x + 4)^{2} + 32

x \approx - 3.11851 \dots, x \approx 3.67940 \dots

Solution steps

(x^{2} - 4)^{2} = (x + 4)^{2} + 32

Expand

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

Expand

(x + 4)^{2} + 32 : x^{2} + 8 x + 48

x^{4} - 8 x^{2} + 16 = x^{2} + 8 x + 48

Move

48

to the left side

x^{4} - 8 x^{2} - 32 = x^{2} + 8 x

Move

8 x

to the left side

x^{4} - 8 x^{2} - 32 - 8 x = x^{2}

Move

x^{2}

to the left side

x^{4} - 9 x^{2} - 8 x - 32 = 0

Find one solution for

x^{4} - 9 x^{2} - 8 x - 32 = 0

using Newton-Raphson:

x \approx - 3.11851 \dots

Apply long division:

\frac{x ^{4} - 9 x ^{2} - 8 x - 32}{x + 3.11851 \dots} = x^{3} - 3.11851 \dots x^{2} + 0.72512 \dots x - 10.26130 \dots

x^{3} - 3.11851 \dots x^{2} + 0.72512 \dots x - 10.26130 \dots \approx 0

Find one solution for

x^{3} - 3.11851 \dots x^{2} + 0.72512 \dots x - 10.26130 \dots = 0

using Newton-Raphson:

x \approx 3.67940 \dots

Apply long division:

\frac{x ^{3} - 3.11851 \dots x ^{2} + 0.72512 \dots x - 10.26130 \dots}{x - 3.67940 \dots} = x^{2} + 0.56088 \dots x + 2.78885 \dots

x^{2} + 0.56088 \dots x + 2.78885 \dots \approx 0

Find one solution for

x^{2} + 0.56088 \dots x + 2.78885 \dots = 0

using Newton-Raphson:

No Solution for

x \in R

The solutions are

x \approx - 3.11851 \dots, x \approx 3.67940 \dots

2 x^{3} + 7 x^{2} = 15 x

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

x, z = (3 x + 2 y)^{5}

x^{4} - 25 x^{2} = - 100

2 x^{3} + 8 x^{2} - 24 x = 0

What are the solutions to the equation (x^2-4)^2=(x+4)^2+32 ?
The solutions to the equation (x^2-4)^2=(x+4)^2+32 are x\approx-3.11851…,x\approx 3.67940…
Find the zeros of (x^2-4)^2=(x+4)^2+32
The zeros of (x^2-4)^2=(x+4)^2+32 are x\approx-3.11851…,x\approx 3.67940…