What are the solutions to the equation (3x-2)(x^1)-(2x^3)(2-x)=182x ?

The solutions to the equation (3x-2)(x^1)-(2x^3)(2-x)=182x are x=0,x\approx 5.16202…

Find the zeros of (3x-2)(x^1)-(2x^3)(2-x)=182x

The zeros of (3x-2)(x^1)-(2x^3)(2-x)=182x are x=0,x\approx 5.16202…

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(3 x - 2) (x^{1}) - (2 x^{3}) (2 - x) = 182 x

x = 0, x \approx 5.16202 \dots

Solution steps

(3 x - 2) (x^{1}) - (2 x^{3}) (2 - x) = 182 x

Expand

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

3 x^{2} - 2 x - 4 x^{3} + 2 x^{4} = 182 x

Move

182 x

to the left side

2 x^{4} - 4 x^{3} + 3 x^{2} - 184 x = 0

Factor

2 x^{4} - 4 x^{3} + 3 x^{2} - 184 x : x (2 x^{3} - 4 x^{2} + 3 x - 184)

x (2 x^{3} - 4 x^{2} + 3 x - 184) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 2 x^{3} - 4 x^{2} + 3 x - 184 = 0

Solve

2 x^{3} - 4 x^{2} + 3 x - 184 = 0 : x \approx 5.16202 \dots

The solutions are

x = 0, x \approx 5.16202 \dots

(x^{2} - 3 x)^{2} - 16 (x^{2} - 3 x) - 36 = 0

x^{3} - x = 1

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

(x + 4)^{3} = - 729

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

What are the solutions to the equation (3x-2)(x^1)-(2x^3)(2-x)=182x ?
The solutions to the equation (3x-2)(x^1)-(2x^3)(2-x)=182x are x=0,x\approx 5.16202…
Find the zeros of (3x-2)(x^1)-(2x^3)(2-x)=182x
The zeros of (3x-2)(x^1)-(2x^3)(2-x)=182x are x=0,x\approx 5.16202…