Solve sqrt{5z+6}=z+2

$$\sqrt { 5 z + 6 } = z + 2$$

Solve for z

$$z=2$$

$$z=-1$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{5z+6}\right)^{2}=\left(z+2\right)^{2}$$

Calculate \sqrt{5z+6} to the power of 2 and get 5z+6.

$$5z+6=\left(z+2\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(z+2\right)^{2}.

$$5z+6=z^{2}+4z+4$$

Subtract z^{2} from both sides.

$$5z+6-z^{2}=4z+4$$

Subtract 4z from both sides.

$$5z+6-z^{2}-4z=4$$

Combine 5z and -4z to get z.

$$z+6-z^{2}=4$$

Subtract 4 from both sides.

$$z+6-z^{2}-4=0$$

Subtract 4 from 6 to get 2.

$$z+2-z^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-z^{2}+z+2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -z^{2}+az+bz+2. To find a and b, set up a system to be solved.

$$a+b=1 ab=-2=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=2 b=-1$$

Rewrite -z^{2}+z+2 as \left(-z^{2}+2z\right)+\left(-z+2\right).

$$\left(-z^{2}+2z\right)+\left(-z+2\right)$$

Factor out -z in the first and -1 in the second group.

$$-z\left(z-2\right)-\left(z-2\right)$$

Factor out common term z-2 by using distributive property.

$$\left(z-2\right)\left(-z-1\right)$$

To find equation solutions, solve z-2=0 and -z-1=0.

$$z=2 z=-1$$

Substitute 2 for z in the equation \sqrt{5z+6}=z+2.

$$\sqrt{5\times 2+6}=2+2$$

Simplify. The value z=2 satisfies the equation.

$$4=4$$

Substitute -1 for z in the equation \sqrt{5z+6}=z+2.

$$\sqrt{5\left(-1\right)+6}=-1+2$$

Simplify. The value z=-1 satisfies the equation.

$$1=1$$

List all solutions of \sqrt{5z+6}=z+2.

$$z=2 z=-1$$