Solve v-2=sqrt{2v-4}

$$v - 2 = \sqrt { 2 v - 4 }$$

Solve for v

$$v=4$$

$$v=2$$

Solution Steps

Square both sides of the equation.

$$\left(v-2\right)^{2}=\left(\sqrt{2v-4}\right)^{2}$$

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

$$v^{2}-4v+4=\left(\sqrt{2v-4}\right)^{2}$$

Calculate \sqrt{2v-4} to the power of 2 and get 2v-4.

$$v^{2}-4v+4=2v-4$$

Subtract 2v from both sides.

$$v^{2}-4v+4-2v=-4$$

Combine -4v and -2v to get -6v.

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

Add 4 to both sides.

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

Add 4 and 4 to get 8.

$$v^{2}-6v+8=0$$

To solve the equation, factor v^{2}-6v+8 using formula v^{2}+\left(a+b\right)v+ab=\left(v+a\right)\left(v+b\right). To find a and b, set up a system to be solved.

$$a+b=-6 ab=8$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -6.

$$a=-4 b=-2$$

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

$$\left(v-4\right)\left(v-2\right)$$

To find equation solutions, solve v-4=0 and v-2=0.

$$v=4 v=2$$

Substitute 4 for v in the equation v-2=\sqrt{2v-4}.

$$4-2=\sqrt{2\times 4-4}$$

Simplify. The value v=4 satisfies the equation.

$$2=2$$

Substitute 2 for v in the equation v-2=\sqrt{2v-4}.

$$2-2=\sqrt{2\times 2-4}$$

Simplify. The value v=2 satisfies the equation.

$$0=0$$

List all solutions of v-2=\sqrt{2v-4}.

$$v=4 v=2$$