What are the solutions to the equation 42+12x^2=0 ?

The solutions to the equation 42+12x^2=0 are x=sqrt(7/2)i,x=-sqrt(7/2)i

Find the zeros of 42+12x^2=0

The zeros of 42+12x^2=0 are x=sqrt(7/2)i,x=-sqrt(7/2)i

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42 + 12 x^{2} = 0

x = \frac{7}{2} i, x = - \frac{7}{2} i

Solution steps

42 + 12 x^{2} = 0

Move

42

to the right side

12 x^{2} = - 42

Divide both sides by

12

x^{2} = - \frac{7}{2}

For

x^{2} = f (a)

the solutions are

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

x = - \frac{7}{2}, x = - - \frac{7}{2}

- \frac{7}{2} = \frac{7}{2} i

- - \frac{7}{2} = - \frac{7}{2} i

x = \frac{7}{2} i, x = - \frac{7}{2} i

x^{2} + 12 x - 14 = 0

x^{2} = - 8 x - 18

3 x^{2} - 1 = 8

- 15 x^{2} + 44 = 2

\frac{3 x ^{2}}{2} + 43 = 139

What are the solutions to the equation 42+12x^2=0 ?
The solutions to the equation 42+12x^2=0 are x=sqrt(7/2)i,x=-sqrt(7/2)i
Find the zeros of 42+12x^2=0
The zeros of 42+12x^2=0 are x=sqrt(7/2)i,x=-sqrt(7/2)i