Solve sqrt{-6z+3}+z=-4

$$\sqrt { - 6 z + 3 } + z = - 4$$

Solve for z

$$z=-13$$

Solution Steps

Subtract z from both sides of the equation.

$$\sqrt{-6z+3}=-4-z$$

Square both sides of the equation.

$$\left(\sqrt{-6z+3}\right)^{2}=\left(-4-z\right)^{2}$$

Calculate \sqrt{-6z+3} to the power of 2 and get -6z+3.

$$-6z+3=\left(-4-z\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4-z\right)^{2}.

$$-6z+3=16+8z+z^{2}$$

Subtract 16 from both sides.

$$-6z+3-16=8z+z^{2}$$

Subtract 16 from 3 to get -13.

$$-6z-13=8z+z^{2}$$

Subtract 8z from both sides.

$$-6z-13-8z=z^{2}$$

Combine -6z and -8z to get -14z.

$$-14z-13=z^{2}$$

Subtract z^{2} from both sides.

$$-14z-13-z^{2}=0$$

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

$$-z^{2}-14z-13=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -z^{2}+az+bz-13. To find a and b, set up a system to be solved.

$$a+b=-14 ab=-\left(-13\right)=13$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-1 b=-13$$

Rewrite -z^{2}-14z-13 as \left(-z^{2}-z\right)+\left(-13z-13\right).

$$\left(-z^{2}-z\right)+\left(-13z-13\right)$$

Factor out z in the first and 13 in the second group.

$$z\left(-z-1\right)+13\left(-z-1\right)$$

Factor out common term -z-1 by using distributive property.

$$\left(-z-1\right)\left(z+13\right)$$

To find equation solutions, solve -z-1=0 and z+13=0.

$$z=-1 z=-13$$

Substitute -1 for z in the equation \sqrt{-6z+3}+z=-4.

$$\sqrt{-6\left(-1\right)+3}-1=-4$$

Simplify. The value z=-1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$2=-4$$

Substitute -13 for z in the equation \sqrt{-6z+3}+z=-4.

$$\sqrt{-6\left(-13\right)+3}-13=-4$$

Simplify. The value z=-13 satisfies the equation.

$$-4=-4$$

Equation \sqrt{3-6z}=-z-4 has a unique solution.

$$z=-13$$