What are the solutions to the equation (2/3)^2+(x)^2=-(1/4)^2 ?

The solutions to the equation (2/3)^2+(x)^2=-(1/4)^2 are x=(sqrt(73))/(12)i,x=-(sqrt(73))/(12)i

Find the zeros of (2/3)^2+(x)^2=-(1/4)^2

The zeros of (2/3)^2+(x)^2=-(1/4)^2 are x=(sqrt(73))/(12)i,x=-(sqrt(73))/(12)i

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(\frac{2}{3})^{2} + (x)^{2} = - (\frac{1}{4})^{2}

x = \frac{73}{12} i, x = - \frac{73}{12} i

Solution steps

(\frac{2}{3})^{2} + x^{2} = - (\frac{1}{4})^{2}

Expand

(\frac{2}{3})^{2} + x^{2} : \frac{4}{9} + x^{2}

Expand

- (\frac{1}{4})^{2} : - \frac{1}{16}

\frac{4}{9} + x^{2} = - \frac{1}{16}

Move

\frac{4}{9}

to the right side

x^{2} = - \frac{1}{16} - \frac{4}{9}

Simplify

- \frac{1}{16} - \frac{4}{9} : - \frac{73}{144}

For

x^{2} = f (a)

the solutions are

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

x = - \frac{73}{144}, x = - - \frac{73}{144}

- \frac{73}{144} = \frac{73}{12} i

- - \frac{73}{144} = - \frac{73}{12} i

x = \frac{73}{12} i, x = - \frac{73}{12} i

b^{2} - 4 b - 5

5 v^{2} - 26 v = 24

\frac{ln ( 2 )}{0.5}

∣ x + 2 ∣ \geq ∣ x - 3 ∣ - ∣ x - 2 ∣

\frac{1}{7} - \frac{1}{2}

What are the solutions to the equation (2/3)^2+(x)^2=-(1/4)^2 ?
The solutions to the equation (2/3)^2+(x)^2=-(1/4)^2 are x=(sqrt(73))/(12)i,x=-(sqrt(73))/(12)i
Find the zeros of (2/3)^2+(x)^2=-(1/4)^2
The zeros of (2/3)^2+(x)^2=-(1/4)^2 are x=(sqrt(73))/(12)i,x=-(sqrt(73))/(12)i