What are the solutions to the equation 5^2=-x^2+4^2 ?

The solutions to the equation 5^2=-x^2+4^2 are x=3i,x=-3i

Find the zeros of 5^2=-x^2+4^2

The zeros of 5^2=-x^2+4^2 are x=3i,x=-3i

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5^{2} = - x^{2} + 4^{2}

x = 3 i, x = - 3 i

Solution steps

5^{2} = - x^{2} + 4^{2}

Switch sides

- x^{2} + 4^{2} = 5^{2}

Move

4^{2}

to the right side

- x^{2} = 9

Divide both sides by

- 1

x^{2} = - 9

For

x^{2} = f (a)

the solutions are

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

x = - 9, x = - - 9

- 9 = 3 i

- - 9 = - 3 i

x = 3 i, x = - 3 i

y - 6 > - 13

47 \cdot 5

3 - x = 1

\frac{1}{8} x = \frac{1}{2}

\frac{- 1}{2} \leq \frac{4 - 3 x}{5} \leq \frac{1}{4}

What are the solutions to the equation 5^2=-x^2+4^2 ?
The solutions to the equation 5^2=-x^2+4^2 are x=3i,x=-3i
Find the zeros of 5^2=-x^2+4^2
The zeros of 5^2=-x^2+4^2 are x=3i,x=-3i