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

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

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

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

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

x = 1, x = 1 + 3, x = 1 - 3

Decimal

x = 1, x = 2.73205 \dots, x = - 0.73205 \dots

Solution steps

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

Expand

(3 - x) x^{2} : 3 x^{2} - x^{3}

2 = 3 x^{2} - x^{3}

Switch sides

3 x^{2} - x^{3} = 2

Move

2

to the left side

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

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x^{2} - 2 x - 2 = 0

Solve

x - 1 = 0 : x = 1

Solve

x^{2} - 2 x - 2 = 0 : x = 1 + 3, x = 1 - 3

The solutions are

x = 1, x = 1 + 3, x = 1 - 3

4 x^{5} - 5 x^{4} + 5 x^{3} + 11 x^{2} - 24 x + 12 = 0

y, x^{4} + y^{4} = x^{2} y^{2}

x^{3} + 6 x^{2} + 8 x + 10

2 x^{3} - 6 x^{2} - 3 x - 10 = 0

z^{3} - 8 i = 0

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