What are the solutions to the equation ax^4+bx^2+c=0 ?

The solutions to the equation ax^4+bx^2+c=0 are

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ax^4+bx^2+c=0

a x^{4} + b x^{2} + c = 0

x = \frac{- b + b ^{2} - 4 a c}{2 a}, x = - \frac{- b + b ^{2} - 4 a c}{2 a}, x = \frac{- b - b ^{2} - 4 a c}{2 a}, x = - \frac{- b - b ^{2} - 4 a c}{2 a}

Solution steps

a x^{4} + b x^{2} + c = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

a u^{2} + b u + c = 0

Solve

a u^{2} + b u + c = 0 : u = \frac{- b + b ^{2} - 4 a c}{2 a}, u = \frac{- b - b ^{2} - 4 a c}{2 a}; a \neq = 0

u = \frac{- b + b ^{2} - 4 a c}{2 a}, u = \frac{- b - b ^{2} - 4 a c}{2 a}; a \neq = 0

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = \frac{- b + b ^{2} - 4 a c}{2 a} : x = \frac{- b + b ^{2} - 4 a c}{2 a}, x = - \frac{- b + b ^{2} - 4 a c}{2 a}

Solve

x^{2} = \frac{- b - b ^{2} - 4 a c}{2 a} : x = \frac{- b - b ^{2} - 4 a c}{2 a}, x = - \frac{- b - b ^{2} - 4 a c}{2 a}

The solutions are

x = \frac{- b + b ^{2} - 4 a c}{2 a}, x = - \frac{- b + b ^{2} - 4 a c}{2 a}, x = \frac{- b - b ^{2} - 4 a c}{2 a}, x = - \frac{- b - b ^{2} - 4 a c}{2 a}

y, 2 (x^{2} + y^{2})^{2} = 25 (x^{2} - y^{2})

y, y^{3} = x

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

(x - 3)^{4} + (x - 5)^{4} = 20

5 y^{3} + 6 y^{2} + 25 y + 30 = 0

What are the solutions to the equation ax^4+bx^2+c=0 ?
The solutions to the equation ax^4+bx^2+c=0 are
Find the zeros of ax^4+bx^2+c=0
The zeros of ax^4+bx^2+c=0 are