What are the solutions to the equation 20^2=0^2+(2a^{40}) ?

The solutions to the equation 20^2=0^2+(2a^{40}) are a=\sqrt[40]{200},a=-\sqrt[40]{200}

Find the zeros of 20^2=0^2+(2a^{40})

The zeros of 20^2=0^2+(2a^{40}) are a=\sqrt[40]{200},a=-\sqrt[40]{200}

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2 0^{2} = 0^{2} + (2 a^{40})

Decimal Notation

a = 1.14163 \dots, a = - 1.14163 \dots

Solution steps

2 0^{2} = 0^{2} + (2 a^{40})

Expand

2 0^{2} : 400

Expand

0^{2} + (2 a^{40}) : 2 a^{40}

400 = 2 a^{40}

Switch sides

2 a^{40} = 400

Divide both sides by

2

a^{40} = 200

For

x^{n} = f (a)

, n is even, the solutions are

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

2 x^{3} - 3 x^{2} - 6 = 0

a^{8} - 47 a^{4} + 1 = 0

(2 x + 1)^{8} = 5. 9^{7} + 4. 9^{7}

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

What are the solutions to the equation 20^2=0^2+(2a^{40}) ?
The solutions to the equation 20^2=0^2+(2a^{40}) are a=\sqrt[40]{200},a=-\sqrt[40]{200}
Find the zeros of 20^2=0^2+(2a^{40})
The zeros of 20^2=0^2+(2a^{40}) are a=\sqrt[40]{200},a=-\sqrt[40]{200}