What are the solutions to the equation 4a^4+28a^2+48=0 ?

The solutions to the equation 4a^4+28a^2+48=0 are a=sqrt(3)i,a=-sqrt(3)i,a=2i,a=-2i

Find the zeros of 4a^4+28a^2+48=0

The zeros of 4a^4+28a^2+48=0 are a=sqrt(3)i,a=-sqrt(3)i,a=2i,a=-2i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 a^{4} + 28 a^{2} + 48 = 0

a = 3 i, a = - 3 i, a = 2 i, a = - 2 i

Solution steps

4 a^{4} + 28 a^{2} + 48 = 0

Rewrite the equation with

u = a^{2}

and

u^{2} = a^{4}

4 u^{2} + 28 u + 48 = 0

Solve

4 u^{2} + 28 u + 48 = 0 : u = - 3, u = - 4

u = - 3, u = - 4

Substitute back

u = a^{2},

solve for

a

Solve

a^{2} = - 3 : a = 3 i, a = - 3 i

Solve

a^{2} = - 4 : a = 2 i, a = - 2 i

The solutions are

a = 3 i, a = - 3 i, a = 2 i, a = - 2 i

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

y^{3} + 2 y^{2} + 4 y + 2 = 0

18 x^{5} - 21 x^{4} - 9 x^{3} = 0

2 x^{4} + 8 x^{2} + 4 = 0

99 = c^{3}

What are the solutions to the equation 4a^4+28a^2+48=0 ?
The solutions to the equation 4a^4+28a^2+48=0 are a=sqrt(3)i,a=-sqrt(3)i,a=2i,a=-2i
Find the zeros of 4a^4+28a^2+48=0
The zeros of 4a^4+28a^2+48=0 are a=sqrt(3)i,a=-sqrt(3)i,a=2i,a=-2i