What are the solutions to the equation solvefor x,x^2(5-2x)=4 ?

The solutions to the equation solvefor x,x^2(5-2x)=4 are x=2,x=(1+sqrt(17))/4 ,x=(1-sqrt(17))/4

Find the zeros of solvefor x,x^2(5-2x)=4

The zeros of solvefor x,x^2(5-2x)=4 are x=2,x=(1+sqrt(17))/4 ,x=(1-sqrt(17))/4

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solve for

x, x^{2} (5 - 2 x) = 4

x = 2, x = \frac{1 + 17}{4}, x = \frac{1 - 17}{4}

Decimal Notation

x = 2, x = 1.28077 \dots, x = - 0.78077 \dots

Solution steps

x^{2} (5 - 2 x) = 4

Expand

x^{2} (5 - 2 x) : 5 x^{2} - 2 x^{3}

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

Move

4

to the left side

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

Write in the standard form

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

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

Factor

- 2 x^{3} + 5 x^{2} - 4 : - (x - 2) (2 x^{2} - x - 2)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 2 = 0 : x = 2

Solve

2 x^{2} - x - 2 = 0 : x = \frac{1 + 17}{4}, x = \frac{1 - 17}{4}

The solutions are

x = 2, x = \frac{1 + 17}{4}, x = \frac{1 - 17}{4}

x^{5} = (2 - x)^{3}

5 x^{4} = 3, x^{2} = 8

2 x^{5} - 6 x^{3} + 8 x^{2} - 5 = 0

(x^{3} - 4 x^{2}) = (16 x - 64)

15 x^{4} + 15 x^{2} = 0

What are the solutions to the equation solvefor x,x^2(5-2x)=4 ?
The solutions to the equation solvefor x,x^2(5-2x)=4 are x=2,x=(1+sqrt(17))/4 ,x=(1-sqrt(17))/4
Find the zeros of solvefor x,x^2(5-2x)=4
The zeros of solvefor x,x^2(5-2x)=4 are x=2,x=(1+sqrt(17))/4 ,x=(1-sqrt(17))/4