Solve 2v=sqrt{5v^2-6v+5}

$$2 v = \sqrt { 5 v ^ { 2 } - 6 v + 5 }$$

Solve for v

$$v=5$$

$$v=1$$

Solution Steps

Square both sides of the equation.

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

Expand \left(2v\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$4v^{2}=\left(\sqrt{5v^{2}-6v+5}\right)^{2}$$

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

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

Subtract 5v^{2} from both sides.

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

Combine 4v^{2} and -5v^{2} to get -v^{2}.

$$-v^{2}=-6v+5$$

Add 6v to both sides.

$$-v^{2}+6v=5$$

Subtract 5 from both sides.

$$-v^{2}+6v-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -v^{2}+av+bv-5. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-5\right)=5$$

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

$$a=5 b=1$$

Rewrite -v^{2}+6v-5 as \left(-v^{2}+5v\right)+\left(v-5\right).

$$\left(-v^{2}+5v\right)+\left(v-5\right)$$

Factor out -v in -v^{2}+5v.

$$-v\left(v-5\right)+v-5$$

Factor out common term v-5 by using distributive property.

$$\left(v-5\right)\left(-v+1\right)$$

To find equation solutions, solve v-5=0 and -v+1=0.

$$v=5 v=1$$

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

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

Simplify. The value v=5 satisfies the equation.

$$10=10$$

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

$$2\times 1=\sqrt{5\times 1^{2}-6+5}$$

Simplify. The value v=1 satisfies the equation.

$$2=2$$

List all solutions of 2v=\sqrt{5v^{2}-6v+5}.

$$v=5 v=1$$