What is the answer to (6q^2+1)/2 = 7/(2q) ?

The answer to (6q^2+1)/2 = 7/(2q) is q=1,q=-1/2+i(sqrt(33))/6 ,q=-1/2-i(sqrt(33))/6

Solve for q: (6q^2+1)/2 = 7/(2q)

The solution is q=1,q=-1/2+i(sqrt(33))/6 ,q=-1/2-i(sqrt(33))/6

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\frac{6 q ^{2} + 1}{2} = \frac{7}{2 q}

q = 1, q = - \frac{1}{2} + i \frac{33}{6}, q = - \frac{1}{2} - i \frac{33}{6}

Solution steps

\frac{6 q ^{2} + 1}{2} = \frac{7}{2 q}

Cross multiply

(6 q^{2} + 1) \cdot 2 q = 14

Solve

(6 q^{2} + 1) \cdot 2 q = 14 : q = 1, q = - \frac{1}{2} + i \frac{33}{6}, q = - \frac{1}{2} - i \frac{33}{6}

q = 1, q = - \frac{1}{2} + i \frac{33}{6}, q = - \frac{1}{2} - i \frac{33}{6}

Verify Solutions

Find undefined (singularity) points:

q = 0

Combine undefined points with solutions:

q = 1, q = - \frac{1}{2} + i \frac{33}{6}, q = - \frac{1}{2} - i \frac{33}{6}

\frac{\frac{16}{7}}{\frac{4}{9}} = \frac{\frac{9}{14}}{t}

(\frac{x}{( x + 5 )}) - (\frac{11 x}{( 11 x - 8 )}) + (\frac{7}{( 6 - 4 x )}) = 0

x^{- 4} - 6 x^{- 2} + 8 = 0

\frac{( x ) ^{2}}{( ( \frac{4}{5} ) ^{- 4} )} = (\frac{1}{x})^{2}

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

What is the answer to (6q^2+1)/2 = 7/(2q) ?
The answer to (6q^2+1)/2 = 7/(2q) is q=1,q=-1/2+i(sqrt(33))/6 ,q=-1/2-i(sqrt(33))/6
Solve for q: (6q^2+1)/2 = 7/(2q)
The solution is q=1,q=-1/2+i(sqrt(33))/6 ,q=-1/2-i(sqrt(33))/6