What are the solutions to the equation 1=(sqrt(3))/4 a^2 ?

The solutions to the equation 1=(sqrt(3))/4 a^2 are a= 2/(sqrt(\sqrt{3))},a=-2/(sqrt(\sqrt{3))}

Find the zeros of 1=(sqrt(3))/4 a^2

The zeros of 1=(sqrt(3))/4 a^2 are a= 2/(sqrt(\sqrt{3))},a=-2/(sqrt(\sqrt{3))}

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1 = \frac{3}{4} a^{2}

a = \frac{2}{3}, a = - \frac{2}{3}

Decimal Notation

a = 1.51967 \dots, a = - 1.51967 \dots

Solution steps

1 = \frac{3}{4} a^{2}

Switch sides

\frac{3}{4} a^{2} = 1

Multiply both sides by

4

3 a^{2} = 4

Divide both sides by

3

a^{2} = \frac{4}{3}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

a = \frac{4}{3}, a = - \frac{4}{3}

\frac{4}{3} = \frac{2}{3}

- \frac{4}{3} = - \frac{2}{3}

a = \frac{2}{3}, a = - \frac{2}{3}

\frac{π}{4} + \frac{π}{2}

\frac{x - 6}{x + 5} \geq 0

\frac{( 2 n + 1 ) !}{( 2 n + 2 ) !}

81 + 30.25

- 24 \leq 4 x - 11 \leq 1

What are the solutions to the equation 1=(sqrt(3))/4 a^2 ?
The solutions to the equation 1=(sqrt(3))/4 a^2 are a= 2/(sqrt(\sqrt{3))},a=-2/(sqrt(\sqrt{3))}
Find the zeros of 1=(sqrt(3))/4 a^2
The zeros of 1=(sqrt(3))/4 a^2 are a= 2/(sqrt(\sqrt{3))},a=-2/(sqrt(\sqrt{3))}