Solve z^2(1-lambda/2a)=k

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

Solve for k

$$k=-\frac{\lambda z^{2}}{2a}+z^{2}$$

$$a\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2a.

$$z^{2}\left(1-\frac{\lambda }{2a}\right)\times 2a=k\times 2a$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2a}{2a}.

$$z^{2}\left(\frac{2a}{2a}-\frac{\lambda }{2a}\right)\times 2a=k\times 2a$$

Since \frac{2a}{2a} and \frac{\lambda }{2a} have the same denominator, subtract them by subtracting their numerators.

$$z^{2}\times \frac{2a-\lambda }{2a}\times 2a=k\times 2a$$

Express z^{2}\times \frac{2a-\lambda }{2a} as a single fraction.

$$\frac{z^{2}\left(2a-\lambda \right)}{2a}\times 2a=k\times 2a$$

Express \frac{z^{2}\left(2a-\lambda \right)}{2a}\times 2 as a single fraction.

$$\frac{z^{2}\left(2a-\lambda \right)\times 2}{2a}a=k\times 2a$$

Cancel out 2 in both numerator and denominator.

$$\frac{\left(-\lambda +2a\right)z^{2}}{a}a=k\times 2a$$

Express \frac{\left(-\lambda +2a\right)z^{2}}{a}a as a single fraction.

$$\frac{\left(-\lambda +2a\right)z^{2}a}{a}=k\times 2a$$

Cancel out a in both numerator and denominator.

$$\left(-\lambda +2a\right)z^{2}=k\times 2a$$

Use the distributive property to multiply -\lambda +2a by z^{2}.

$$-\lambda z^{2}+2az^{2}=k\times 2a$$

Swap sides so that all variable terms are on the left hand side.

$$k\times 2a=-\lambda z^{2}+2az^{2}$$

The equation is in standard form.

$$2ak=2az^{2}-\lambda z^{2}$$

Divide both sides by 2a.

$$\frac{2ak}{2a}=\frac{\left(2a-\lambda \right)z^{2}}{2a}$$

Dividing by 2a undoes the multiplication by 2a.

$$k=\frac{\left(2a-\lambda \right)z^{2}}{2a}$$

Divide \left(2a-\lambda \right)z^{2} by 2a.

$$k=-\frac{\lambda z^{2}}{2a}+z^{2}$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=-\frac{\lambda z^{2}}{2\left(k-z^{2}\right)}\text{, }&z\neq 0\text{ and }\lambda \neq 0\text{ and }z\neq -\sqrt{k}\text{ and }z\neq \sqrt{k}\text{ and }k\neq z^{2}\\a\neq 0\text{, }&\left(k=0\text{ and }z=0\right)\text{ or }\left(z\neq 0\text{ and }k=z^{2}\text{ and }\lambda =0\right)\end{matrix}\right.$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{\lambda z^{2}}{2\left(k-z^{2}\right)}\text{, }&z\neq 0\text{ and }\lambda \neq 0\text{ and }\left(|z|\neq \sqrt{k}\text{ or }k<0\right)\text{ and }k\neq z^{2}\\a\neq 0\text{, }&\left(k=0\text{ and }z=0\right)\text{ or }\left(z\neq 0\text{ and }k=z^{2}\text{ and }\lambda =0\right)\end{matrix}\right.$$