What are the solutions to the equation y^4-5y^2+4y=0 ?

The solutions to the equation y^4-5y^2+4y=0 are y=0,y=1,y=(-1+sqrt(17))/2 ,y=(-1-sqrt(17))/2

Find the zeros of y^4-5y^2+4y=0

The zeros of y^4-5y^2+4y=0 are y=0,y=1,y=(-1+sqrt(17))/2 ,y=(-1-sqrt(17))/2

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y^{4} - 5 y^{2} + 4 y = 0

y = 0, y = 1, y = \frac{- 1 + 17}{2}, y = \frac{- 1 - 17}{2}

Decimal Notation

y = 0, y = 1, y = 1.56155 \dots, y = - 2.56155 \dots

Solution steps

y^{4} - 5 y^{2} + 4 y = 0

Factor

y^{4} - 5 y^{2} + 4 y : y (y - 1) (y^{2} + y - 4)

y (y - 1) (y^{2} + y - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y = 0 or y - 1 = 0 or y^{2} + y - 4 = 0

Solve

y - 1 = 0 : y = 1

Solve

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

The solutions are

y = 0, y = 1, y = \frac{- 1 + 17}{2}, y = \frac{- 1 - 17}{2}

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

(x - 1)^{4} + (x - 7)^{4} = 272

x^{4} + x^{2} - 1 = 0

x^{4} = - 1

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

What are the solutions to the equation y^4-5y^2+4y=0 ?
The solutions to the equation y^4-5y^2+4y=0 are y=0,y=1,y=(-1+sqrt(17))/2 ,y=(-1-sqrt(17))/2
Find the zeros of y^4-5y^2+4y=0
The zeros of y^4-5y^2+4y=0 are y=0,y=1,y=(-1+sqrt(17))/2 ,y=(-1-sqrt(17))/2