Solve sqrt{2x-5}-2=sqrt{x-2}

$$\sqrt { 2 x - 5 } - 2 = \sqrt { x - 2 }$$

Solve for x

$$x=27$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2x-5}-2\right)^{2}=\left(\sqrt{x-2}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{2x-5}-2\right)^{2}.

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

Calculate \sqrt{2x-5} to the power of 2 and get 2x-5.

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

Add -5 and 4 to get -1.

$$2x-1-4\sqrt{2x-5}=\left(\sqrt{x-2}\right)^{2}$$

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

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

Subtract 2x-1 from both sides of the equation.

$$-4\sqrt{2x-5}=x-2-\left(2x-1\right)$$

To find the opposite of 2x-1, find the opposite of each term.

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

Combine x and -2x to get -x.

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

Add -2 and 1 to get -1.

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

Square both sides of the equation.

$$\left(-4\sqrt{2x-5}\right)^{2}=\left(-x-1\right)^{2}$$

Expand \left(-4\sqrt{2x-5}\right)^{2}.

$$\left(-4\right)^{2}\left(\sqrt{2x-5}\right)^{2}=\left(-x-1\right)^{2}$$

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

$$16\left(\sqrt{2x-5}\right)^{2}=\left(-x-1\right)^{2}$$

Calculate \sqrt{2x-5} to the power of 2 and get 2x-5.

$$16\left(2x-5\right)=\left(-x-1\right)^{2}$$

Use the distributive property to multiply 16 by 2x-5.

$$32x-80=\left(-x-1\right)^{2}$$

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

$$32x-80=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$32x-80-x^{2}=2x+1$$

Subtract 2x from both sides.

$$32x-80-x^{2}-2x=1$$

Combine 32x and -2x to get 30x.

$$30x-80-x^{2}=1$$

Subtract 1 from both sides.

$$30x-80-x^{2}-1=0$$

Subtract 1 from -80 to get -81.

$$30x-81-x^{2}=0$$

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

$$-x^{2}+30x-81=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-81. To find a and b, set up a system to be solved.

$$a+b=30 ab=-\left(-81\right)=81$$

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 81.

$$1,81 3,27 9,9$$

Calculate the sum for each pair.

$$1+81=82 3+27=30 9+9=18$$

The solution is the pair that gives sum 30.

$$a=27 b=3$$

Rewrite -x^{2}+30x-81 as \left(-x^{2}+27x\right)+\left(3x-81\right).

$$\left(-x^{2}+27x\right)+\left(3x-81\right)$$

Factor out -x in the first and 3 in the second group.

$$-x\left(x-27\right)+3\left(x-27\right)$$

Factor out common term x-27 by using distributive property.

$$\left(x-27\right)\left(-x+3\right)$$

To find equation solutions, solve x-27=0 and -x+3=0.

$$x=27 x=3$$

Substitute 27 for x in the equation \sqrt{2x-5}-2=\sqrt{x-2}.

$$\sqrt{2\times 27-5}-2=\sqrt{27-2}$$

Simplify. The value x=27 satisfies the equation.

$$5=5$$

Substitute 3 for x in the equation \sqrt{2x-5}-2=\sqrt{x-2}.

$$\sqrt{2\times 3-5}-2=\sqrt{3-2}$$

Simplify. The value x=3 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-1=1$$

Substitute 27 for x in the equation \sqrt{2x-5}-2=\sqrt{x-2}.

$$\sqrt{2\times 27-5}-2=\sqrt{27-2}$$

Simplify. The value x=27 satisfies the equation.

$$5=5$$

Equation \sqrt{2x-5}-2=\sqrt{x-2} has a unique solution.

$$x=27$$