What are the solutions to the equation 2x^4-12x^2+16=0 ?

The solutions to the equation 2x^4-12x^2+16=0 are x=2,x=-2,x=sqrt(2),x=-sqrt(2)

Find the zeros of 2x^4-12x^2+16=0

The zeros of 2x^4-12x^2+16=0 are x=2,x=-2,x=sqrt(2),x=-sqrt(2)

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2 x^{4} - 12 x^{2} + 16 = 0

x = 2, x = - 2, x = 2, x = - 2

Decimal Notation

x = 2, x = - 2, x = 1.41421 \dots, x = - 1.41421 \dots

Solution steps

2 x^{4} - 12 x^{2} + 16 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

2 u^{2} - 12 u + 16 = 0

Solve

2 u^{2} - 12 u + 16 = 0 : u = 4, u = 2

u = 4, u = 2

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = 4 : x = 2, x = - 2

Solve

x^{2} = 2 : x = 2, x = - 2

The solutions are

x = 2, x = - 2, x = 2, x = - 2

- 40 + 6 x^{3} = 0

3 x^{4} + 5 x^{3} + 3 x^{2} + x = 20

- 5.3 t^{3} + 21 t^{2} + 16 t = 0

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

(4 m^{2} + 2)^{2} = 0

What are the solutions to the equation 2x^4-12x^2+16=0 ?
The solutions to the equation 2x^4-12x^2+16=0 are x=2,x=-2,x=sqrt(2),x=-sqrt(2)
Find the zeros of 2x^4-12x^2+16=0
The zeros of 2x^4-12x^2+16=0 are x=2,x=-2,x=sqrt(2),x=-sqrt(2)