What are the solutions to the equation 1200=600x^2(3-x) ?

The solutions to the equation 1200=600x^2(3-x) are x=1,x=1+sqrt(3),x=1-sqrt(3)

Find the zeros of 1200=600x^2(3-x)

The zeros of 1200=600x^2(3-x) are x=1,x=1+sqrt(3),x=1-sqrt(3)

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1200 = 600 x^{2} (3 - x)

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

Decimal Notation

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

Solution steps

1200 = 600 x^{2} (3 - x)

Expand

600 x^{2} (3 - 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

x^{4} - 3 x^{2} = 10

w^{3} = 5

2 x^{4} + x^{3} - 7 x^{2} - 3 x + 3 = 0

7 x^{9} = (7 x^{10}) - (7 x)

x, 27 x^{4} + 60 x^{2} = 32

What are the solutions to the equation 1200=600x^2(3-x) ?
The solutions to the equation 1200=600x^2(3-x) are x=1,x=1+sqrt(3),x=1-sqrt(3)
Find the zeros of 1200=600x^2(3-x)
The zeros of 1200=600x^2(3-x) are x=1,x=1+sqrt(3),x=1-sqrt(3)