What are the solutions to the equation 2(x^2-9)^2(2x-1)=0 ?

The solutions to the equation 2(x^2-9)^2(2x-1)=0 are x=-3,x=3,x= 1/2

Find the zeros of 2(x^2-9)^2(2x-1)=0

The zeros of 2(x^2-9)^2(2x-1)=0 are x=-3,x=3,x= 1/2

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2 (x^{2} - 9)^{2} (2 x - 1) = 0

x = - 3, x = 3, x = \frac{1}{2}

Decimal Notation

x = - 3, x = 3, x = 0.5

Solution steps

2 (x^{2} - 9)^{2} (2 x - 1) = 0

Factor

2 (x^{2} - 9)^{2} (2 x - 1) : 2 (x + 3)^{2} (x - 3)^{2} (2 x - 1)

2 (x + 3)^{2} (x - 3)^{2} (2 x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 3 = 0 or x - 3 = 0 or 2 x - 1 = 0

Solve

x + 3 = 0 : x = - 3

Solve

x - 3 = 0 : x = 3

Solve

2 x - 1 = 0 : x = \frac{1}{2}

The solutions are

x = - 3, x = 3, x = \frac{1}{2}

x, y^{3} = C x^{4} + 2

4 x^{4} - 40 x^{2} + 36 = 0

m^{3} + 5 m^{2} + 4 m + 4 = 0

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

z^{4} - 4 = 0

What are the solutions to the equation 2(x^2-9)^2(2x-1)=0 ?
The solutions to the equation 2(x^2-9)^2(2x-1)=0 are x=-3,x=3,x= 1/2
Find the zeros of 2(x^2-9)^2(2x-1)=0
The zeros of 2(x^2-9)^2(2x-1)=0 are x=-3,x=3,x= 1/2