What is the answer to solvefor a,k=(a^2)/((1-a)^2) ?

The answer to solvefor a,k=(a^2)/((1-a)^2) is a=(k+sqrt(k))/(k-1),a=(k-sqrt(k))/(k-1);\quad k\ne 1

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solvefor a,k=(a^2)/((1-a)^2)

solve for

a, k = \frac{a ^{2}}{( 1 - a ) ^{2}}

a = \frac{k + k}{k - 1}, a = \frac{k - k}{k - 1}; k \neq = 1

Solution steps

k = \frac{a ^{2}}{( 1 - a ) ^{2}}

Multiply both sides by

(1 - a)^{2}

k (1 - a)^{2} = a^{2}

Solve

k (1 - a)^{2} = a^{2} : a = \frac{k + k}{k - 1}, a = \frac{k - k}{k - 1}; k \neq = 1

a = \frac{k + k}{k - 1}, a = \frac{k - k}{k - 1}; k \neq = 1

\frac{1}{3 ( x + \frac{1}{2} )} = \frac{7}{30}

\frac{a ^{7}}{8} = \frac{5}{x}

\frac{x ^{2} + 1}{( 25 x ^{2} )} = \frac{83}{5}

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

x^{2} + (\frac{16}{x ^{2}}) = 8

What is the answer to solvefor a,k=(a^2)/((1-a)^2) ?
The answer to solvefor a,k=(a^2)/((1-a)^2) is a=(k+sqrt(k))/(k-1),a=(k-sqrt(k))/(k-1);\quad k\ne 1