What are the solutions to the equation 1200=30x^2(60-20x) ?

The solutions to the equation 1200=30x^2(60-20x) are x=1,x=1+sqrt(3),x=1-sqrt(3)

Find the zeros of 1200=30x^2(60-20x)

The zeros of 1200=30x^2(60-20x) are x=1,x=1+sqrt(3),x=1-sqrt(3)

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1200 = 30 x^{2} (60 - 20 x)

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

Decimal Notation

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

Solution steps

1200 = 30 x^{2} (60 - 20 x)

Expand

30 x^{2} (60 - 20 x) : 1800 x^{2} - 600 x^{3}

1200 = 1800 x^{2} - 600 x^{3}

Switch sides

1800 x^{2} - 600 x^{3} = 1200

Move

1200

to the left side

1800 x^{2} - 600 x^{3} - 1200 = 0

Write in the standard form

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

- 600 x^{3} + 1800 x^{2} - 1200 = 0

Factor

- 600 x^{3} + 1800 x^{2} - 1200 : - 600 (x - 1) (x^{2} - 2 x - 2)

- 600 (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

y^{4} = 16

x^{4} - 13 x^{2} + 30 = 0

16 x^{3} = 2

4 x^{3} - 24 x^{2} - 9 x + 54 = 0

x^{5} = 3

What are the solutions to the equation 1200=30x^2(60-20x) ?
The solutions to the equation 1200=30x^2(60-20x) are x=1,x=1+sqrt(3),x=1-sqrt(3)
Find the zeros of 1200=30x^2(60-20x)
The zeros of 1200=30x^2(60-20x) are x=1,x=1+sqrt(3),x=1-sqrt(3)