Solve sqrt{2m-1}=2+sqrt{m-4}

$$\sqrt { 2 m - 1 } = 2 + \sqrt { m - 4 }$$

Solve for m

$$m=13$$

$$m=5$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2m-1}\right)^{2}=\left(2+\sqrt{m-4}\right)^{2}$$

Calculate \sqrt{2m-1} to the power of 2 and get 2m-1.

$$2m-1=\left(2+\sqrt{m-4}\right)^{2}$$

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

$$2m-1=4+4\sqrt{m-4}+\left(\sqrt{m-4}\right)^{2}$$

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

$$2m-1=4+4\sqrt{m-4}+m-4$$

Subtract 4 from 4 to get 0.

$$2m-1=4\sqrt{m-4}+m$$

Subtract m from both sides of the equation.

$$2m-1-m=4\sqrt{m-4}$$

Combine 2m and -m to get m.

$$m-1=4\sqrt{m-4}$$

Square both sides of the equation.

$$\left(m-1\right)^{2}=\left(4\sqrt{m-4}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(m-1\right)^{2}.

$$m^{2}-2m+1=\left(4\sqrt{m-4}\right)^{2}$$

Expand \left(4\sqrt{m-4}\right)^{2}.

$$m^{2}-2m+1=4^{2}\left(\sqrt{m-4}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$m^{2}-2m+1=16\left(\sqrt{m-4}\right)^{2}$$

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

$$m^{2}-2m+1=16\left(m-4\right)$$

Use the distributive property to multiply 16 by m-4.

$$m^{2}-2m+1=16m-64$$

Subtract 16m from both sides.

$$m^{2}-2m+1-16m=-64$$

Combine -2m and -16m to get -18m.

$$m^{2}-18m+1=-64$$

Add 64 to both sides.

$$m^{2}-18m+1+64=0$$

Add 1 and 64 to get 65.

$$m^{2}-18m+65=0$$

To solve the equation, factor m^{2}-18m+65 using formula m^{2}+\left(a+b\right)m+ab=\left(m+a\right)\left(m+b\right). To find a and b, set up a system to be solved.

$$a+b=-18 ab=65$$

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 65.

$$-1,-65 -5,-13$$

Calculate the sum for each pair.

$$-1-65=-66 -5-13=-18$$

The solution is the pair that gives sum -18.

$$a=-13 b=-5$$

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

$$\left(m-13\right)\left(m-5\right)$$

To find equation solutions, solve m-13=0 and m-5=0.

$$m=13 m=5$$

Substitute 13 for m in the equation \sqrt{2m-1}=2+\sqrt{m-4}.

$$\sqrt{2\times 13-1}=2+\sqrt{13-4}$$

Simplify. The value m=13 satisfies the equation.

$$5=5$$

Substitute 5 for m in the equation \sqrt{2m-1}=2+\sqrt{m-4}.

$$\sqrt{2\times 5-1}=2+\sqrt{5-4}$$

Simplify. The value m=5 satisfies the equation.

$$3=3$$

List all solutions of \sqrt{2m-1}=\sqrt{m-4}+2.

$$m=13 m=5$$