What are the solutions to the equation solvefor a,y=9a^4-a^2b^2 ?

The solutions to the equation solvefor a,y=9a^4-a^2b^2 are

Find the zeros of solvefor a,y=9a^4-a^2b^2

The zeros of solvefor a,y=9a^4-a^2b^2 are

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solve for

a, y = 9 a^{4} - a^{2} b^{2}

a = \frac{b ^{2} + b ^{4} + 36 y}{3 2}, a = - \frac{b ^{2} + b ^{4} + 36 y}{3 2}, a = \frac{b ^{2} - b ^{4} + 36 y}{3 2}, a = - \frac{b ^{2} - b ^{4} + 36 y}{3 2}

Solution steps

y = 9 a^{4} - a^{2} b^{2}

Switch sides

9 a^{4} - a^{2} b^{2} = y

Move

y

to the left side

9 a^{4} - a^{2} b^{2} - y = 0

Write in the standard form

a_{n} x^{n} + \dots + b x + y = 0

9 a^{4} - b^{2} a^{2} - y = 0

Rewrite the equation with

u = a^{2}

and

u^{2} = a^{4}

9 u^{2} - b^{2} u - y = 0

Solve

9 u^{2} - b^{2} u - y = 0 : u = \frac{b ^{2} + b ^{4} + 36 y}{18}, u = \frac{b ^{2} - b ^{4} + 36 y}{18}

u = \frac{b ^{2} + b ^{4} + 36 y}{18}, u = \frac{b ^{2} - b ^{4} + 36 y}{18}

Substitute back

u = a^{2},

solve for

a

Solve

a^{2} = \frac{b ^{2} + b ^{4} + 36 y}{18} : a = \frac{b ^{2} + b ^{4} + 36 y}{3 2}, a = - \frac{b ^{2} + b ^{4} + 36 y}{3 2}

Solve

a^{2} = \frac{b ^{2} - b ^{4} + 36 y}{18} : a = \frac{b ^{2} - b ^{4} + 36 y}{3 2}, a = - \frac{b ^{2} - b ^{4} + 36 y}{3 2}

The solutions are

a = \frac{b ^{2} + b ^{4} + 36 y}{3 2}, a = - \frac{b ^{2} + b ^{4} + 36 y}{3 2}, a = \frac{b ^{2} - b ^{4} + 36 y}{3 2}, a = - \frac{b ^{2} - b ^{4} + 36 y}{3 2}

r, f = - 1 r^{8}

1 - 4 x^{2} - 4 x^{4} = 0

8 = x^{5}

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

(y^{2} - 5)^{2} + 2 (y^{2} - 5) - 24 = 0

What are the solutions to the equation solvefor a,y=9a^4-a^2b^2 ?
The solutions to the equation solvefor a,y=9a^4-a^2b^2 are
Find the zeros of solvefor a,y=9a^4-a^2b^2
The zeros of solvefor a,y=9a^4-a^2b^2 are