Solve sqrt{3z^2+16}=sqrt{2z^2-8z}

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

Solve for z

$$z=-4$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3z^{2}+16}\right)^{2}=\left(\sqrt{2z^{2}-8z}\right)^{2}$$

Calculate \sqrt{3z^{2}+16} to the power of 2 and get 3z^{2}+16.

$$3z^{2}+16=\left(\sqrt{2z^{2}-8z}\right)^{2}$$

Calculate \sqrt{2z^{2}-8z} to the power of 2 and get 2z^{2}-8z.

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

Subtract 2z^{2} from both sides.

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

Combine 3z^{2} and -2z^{2} to get z^{2}.

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

Add 8z to both sides.

$$z^{2}+16+8z=0$$

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

$$z^{2}+8z+16=0$$

To solve the equation, factor z^{2}+8z+16 using formula z^{2}+\left(a+b\right)z+ab=\left(z+a\right)\left(z+b\right). To find a and b, set up a system to be solved.

$$a+b=8 ab=16$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 16.

$$1,16 2,8 4,4$$

Calculate the sum for each pair.

$$1+16=17 2+8=10 4+4=8$$

The solution is the pair that gives sum 8.

$$a=4 b=4$$

Rewrite factored expression \left(z+a\right)\left(z+b\right) using the obtained values.

$$\left(z+4\right)\left(z+4\right)$$

Rewrite as a binomial square.

$$\left(z+4\right)^{2}$$

To find equation solution, solve z+4=0.

$$z=-4$$

Substitute -4 for z in the equation \sqrt{3z^{2}+16}=\sqrt{2z^{2}-8z}.

$$\sqrt{3\left(-4\right)^{2}+16}=\sqrt{2\left(-4\right)^{2}-8\left(-4\right)}$$

Simplify. The value z=-4 satisfies the equation.

$$8=8$$

Equation \sqrt{3z^{2}+16}=\sqrt{2z^{2}-8z} has a unique solution.

$$z=-4$$