What are the solutions to the equation 3x^4-5x^2+2=0 ?

The solutions to the equation 3x^4-5x^2+2=0 are x=1,x=-1,x=sqrt(2/3),x=-sqrt(2/3)

Find the zeros of 3x^4-5x^2+2=0

The zeros of 3x^4-5x^2+2=0 are x=1,x=-1,x=sqrt(2/3),x=-sqrt(2/3)

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

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

Decimal Notation

x = 1, x = - 1, x = 0.81649 \dots, x = - 0.81649 \dots

Solution steps

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

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

3 u^{2} - 5 u + 2 = 0

Solve

3 u^{2} - 5 u + 2 = 0 : u = 1, u = \frac{2}{3}

u = 1, u = \frac{2}{3}

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = 1 : x = 1, x = - 1

Solve

x^{2} = \frac{2}{3} : x = \frac{2}{3}, x = - \frac{2}{3}

The solutions are

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

24 x^{3} - 303 x^{2} + 918 x - 331 = 0

x, 4 x^{3} - 4 x = 0

- 2 (2) + b^{9} = 59

55 mod 3 = (1 x^{5}) mod 3

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

What are the solutions to the equation 3x^4-5x^2+2=0 ?
The solutions to the equation 3x^4-5x^2+2=0 are x=1,x=-1,x=sqrt(2/3),x=-sqrt(2/3)
Find the zeros of 3x^4-5x^2+2=0
The zeros of 3x^4-5x^2+2=0 are x=1,x=-1,x=sqrt(2/3),x=-sqrt(2/3)