Solve sqrt{2x-1}-2=sqrt{x-4}

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

Solve for x

$$x=13$$

$$x=5$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Add -1 and 4 to get 3.

$$2x+3-4\sqrt{2x-1}=\left(\sqrt{x-4}\right)^{2}$$

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

$$2x+3-4\sqrt{2x-1}=x-4$$

Subtract 2x+3 from both sides of the equation.

$$-4\sqrt{2x-1}=x-4-\left(2x+3\right)$$

To find the opposite of 2x+3, find the opposite of each term.

$$-4\sqrt{2x-1}=x-4-2x-3$$

Combine x and -2x to get -x.

$$-4\sqrt{2x-1}=-x-4-3$$

Subtract 3 from -4 to get -7.

$$-4\sqrt{2x-1}=-x-7$$

Square both sides of the equation.

$$\left(-4\sqrt{2x-1}\right)^{2}=\left(-x-7\right)^{2}$$

Expand \left(-4\sqrt{2x-1}\right)^{2}.

$$\left(-4\right)^{2}\left(\sqrt{2x-1}\right)^{2}=\left(-x-7\right)^{2}$$

Calculate -4 to the power of 2 and get 16.

$$16\left(\sqrt{2x-1}\right)^{2}=\left(-x-7\right)^{2}$$

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

$$16\left(2x-1\right)=\left(-x-7\right)^{2}$$

Use the distributive property to multiply 16 by 2x-1.

$$32x-16=\left(-x-7\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-x-7\right)^{2}.

$$32x-16=x^{2}+14x+49$$

Subtract x^{2} from both sides.

$$32x-16-x^{2}=14x+49$$

Subtract 14x from both sides.

$$32x-16-x^{2}-14x=49$$

Combine 32x and -14x to get 18x.

$$18x-16-x^{2}=49$$

Subtract 49 from both sides.

$$18x-16-x^{2}-49=0$$

Subtract 49 from -16 to get -65.

$$18x-65-x^{2}=0$$

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-65. To find a and b, set up a system to be solved.

$$a+b=18 ab=-\left(-65\right)=65$$

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 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 -x^{2}+18x-65 as \left(-x^{2}+13x\right)+\left(5x-65\right).

$$\left(-x^{2}+13x\right)+\left(5x-65\right)$$

Factor out -x in the first and 5 in the second group.

$$-x\left(x-13\right)+5\left(x-13\right)$$

Factor out common term x-13 by using distributive property.

$$\left(x-13\right)\left(-x+5\right)$$

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

$$x=13 x=5$$

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

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

Simplify. The value x=13 satisfies the equation.

$$3=3$$

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

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

Simplify. The value x=5 satisfies the equation.

$$1=1$$

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

$$x=13 x=5$$