What are the solutions to the equation 49x^2+81=0 ?

The solutions to the equation 49x^2+81=0 are x= 9/7 i,x=-9/7 i

Find the zeros of 49x^2+81=0

The zeros of 49x^2+81=0 are x= 9/7 i,x=-9/7 i

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49 x^{2} + 81 = 0

x = \frac{9}{7} i, x = - \frac{9}{7} i

Solution steps

49 x^{2} + 81 = 0

Move

81

to the right side

49 x^{2} = - 81

Divide both sides by

49

x^{2} = - \frac{81}{49}

For

x^{2} = f (a)

the solutions are

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

x = - \frac{81}{49}, x = - - \frac{81}{49}

- \frac{81}{49} = \frac{9}{7} i

- - \frac{81}{49} = - \frac{9}{7} i

x = \frac{9}{7} i, x = - \frac{9}{7} i

4 x^{3} y^{4} z^{3} + 2 x^{2} y^{3} z - 6 x^{2} y^{3} z^{2}

x^{2} - 8 = - 7 x

\frac{2}{6}

i + i

3 - 8

What are the solutions to the equation 49x^2+81=0 ?
The solutions to the equation 49x^2+81=0 are x= 9/7 i,x=-9/7 i
Find the zeros of 49x^2+81=0
The zeros of 49x^2+81=0 are x= 9/7 i,x=-9/7 i