What are the solutions to the equation 4x^4+19x^2-5=0 ?

The solutions to the equation 4x^4+19x^2-5=0 are x= 1/2 ,x=-1/2 ,x=sqrt(5)i,x=-sqrt(5)i

Find the zeros of 4x^4+19x^2-5=0

The zeros of 4x^4+19x^2-5=0 are x= 1/2 ,x=-1/2 ,x=sqrt(5)i,x=-sqrt(5)i

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4 x^{4} + 19 x^{2} - 5 = 0

x = \frac{1}{2}, x = - \frac{1}{2}, x = 5 i, x = - 5 i

Solution steps

4 x^{4} + 19 x^{2} - 5 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

4 u^{2} + 19 u - 5 = 0

Solve

4 u^{2} + 19 u - 5 = 0 : u = \frac{1}{4}, u = - 5

u = \frac{1}{4}, u = - 5

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = \frac{1}{4} : x = \frac{1}{2}, x = - \frac{1}{2}

Solve

x^{2} = - 5 : x = 5 i, x = - 5 i

The solutions are

x = \frac{1}{2}, x = - \frac{1}{2}, x = 5 i, x = - 5 i

x^{3} + x^{2} + x + 1 = 0

x^{4} - x^{3} - 30 x^{2}

2 x^{4} + 5 x^{3} - 23 x^{2} + 10 x = 0

x^{3} + 9 x^{2} + 20 x = 0

12 x^{3} - 36 x = 0

What are the solutions to the equation 4x^4+19x^2-5=0 ?
The solutions to the equation 4x^4+19x^2-5=0 are x= 1/2 ,x=-1/2 ,x=sqrt(5)i,x=-sqrt(5)i
Find the zeros of 4x^4+19x^2-5=0
The zeros of 4x^4+19x^2-5=0 are x= 1/2 ,x=-1/2 ,x=sqrt(5)i,x=-sqrt(5)i