What are the solutions to the equation 4x^4-19x^2+12=0 ?

The solutions to the equation 4x^4-19x^2+12=0 are x=2,x=-2,x=(sqrt(3))/2 ,x=-(sqrt(3))/2

Find the zeros of 4x^4-19x^2+12=0

The zeros of 4x^4-19x^2+12=0 are x=2,x=-2,x=(sqrt(3))/2 ,x=-(sqrt(3))/2

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4 x^{4} - 19 x^{2} + 12 = 0

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

Decimal Notation

x = 2, x = - 2, x = 0.86602 \dots, x = - 0.86602 \dots

Solution steps

4 x^{4} - 19 x^{2} + 12 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

4 u^{2} - 19 u + 12 = 0

Solve

4 u^{2} - 19 u + 12 = 0 : u = 4, u = \frac{3}{4}

u = 4, u = \frac{3}{4}

Substitute back

u = x^{2},

solve for

x

Solve

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

Solve

x^{2} = \frac{3}{4} : x = \frac{3}{2}, x = - \frac{3}{2}

The solutions are

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

0 = x^{3} - 3 x^{2} + 4 x - 12

2 x^{3} + 6 x^{2} + 5 x + 15 = 0

- 9 m^{3} + 18 m^{2} - 8 m = 0

x, (x - 3)^{3} = 8 (y - 2)

x^{3} + 3 x^{2} + 3 x + 9 = 0

What are the solutions to the equation 4x^4-19x^2+12=0 ?
The solutions to the equation 4x^4-19x^2+12=0 are x=2,x=-2,x=(sqrt(3))/2 ,x=-(sqrt(3))/2
Find the zeros of 4x^4-19x^2+12=0
The zeros of 4x^4-19x^2+12=0 are x=2,x=-2,x=(sqrt(3))/2 ,x=-(sqrt(3))/2