Solve sqrt{2z+3}=-z

$$\sqrt { 2 z + 3 } = - z$$

Solve for z

$$z=-1$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2z+3}\right)^{2}=\left(-z\right)^{2}$$

Calculate \sqrt{2z+3} to the power of 2 and get 2z+3.

$$2z+3=\left(-z\right)^{2}$$

Calculate -z to the power of 2 and get z^{2}.

$$2z+3=z^{2}$$

Subtract z^{2} from both sides.

$$2z+3-z^{2}=0$$

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

$$-z^{2}+2z+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -z^{2}+az+bz+3. To find a and b, set up a system to be solved.

$$a+b=2 ab=-3=-3$$

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.

$$a=3 b=-1$$

Rewrite -z^{2}+2z+3 as \left(-z^{2}+3z\right)+\left(-z+3\right).

$$\left(-z^{2}+3z\right)+\left(-z+3\right)$$

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

$$-z\left(z-3\right)-\left(z-3\right)$$

Factor out common term z-3 by using distributive property.

$$\left(z-3\right)\left(-z-1\right)$$

To find equation solutions, solve z-3=0 and -z-1=0.

$$z=3 z=-1$$

Substitute 3 for z in the equation \sqrt{2z+3}=-z.

$$\sqrt{2\times 3+3}=-3$$

Simplify. The value z=3 does not satisfy the equation because the left and the right hand side have opposite signs.

$$3=-3$$

Substitute -1 for z in the equation \sqrt{2z+3}=-z.

$$\sqrt{2\left(-1\right)+3}=-\left(-1\right)$$

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

$$1=1$$

Equation \sqrt{2z+3}=-z has a unique solution.

$$z=-1$$