Solve sqrt{x-1}-sqrt{2x-10}=2

$$\sqrt { x - 1 } - \sqrt { 2 x - 10 } = 2$$

Solve for x

$$x=5$$

Solution Steps

Subtract -\sqrt{2x-10} from both sides of the equation.

$$\sqrt{x-1}=2+\sqrt{2x-10}$$

Square both sides of the equation.

$$\left(\sqrt{x-1}\right)^{2}=\left(2+\sqrt{2x-10}\right)^{2}$$

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

$$x-1=\left(2+\sqrt{2x-10}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{2x-10}\right)^{2}.

$$x-1=4+4\sqrt{2x-10}+\left(\sqrt{2x-10}\right)^{2}$$

Calculate \sqrt{2x-10} to the power of 2 and get 2x-10.

$$x-1=4+4\sqrt{2x-10}+2x-10$$

Subtract 10 from 4 to get -6.

$$x-1=-6+4\sqrt{2x-10}+2x$$

Subtract -6+2x from both sides of the equation.

$$x-1-\left(-6+2x\right)=4\sqrt{2x-10}$$

To find the opposite of -6+2x, find the opposite of each term.

$$x-1+6-2x=4\sqrt{2x-10}$$

Add -1 and 6 to get 5.

$$x+5-2x=4\sqrt{2x-10}$$

Combine x and -2x to get -x.

$$-x+5=4\sqrt{2x-10}$$

Square both sides of the equation.

$$\left(-x+5\right)^{2}=\left(4\sqrt{2x-10}\right)^{2}$$

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

$$x^{2}-10x+25=\left(4\sqrt{2x-10}\right)^{2}$$

Expand \left(4\sqrt{2x-10}\right)^{2}.

$$x^{2}-10x+25=4^{2}\left(\sqrt{2x-10}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$x^{2}-10x+25=16\left(\sqrt{2x-10}\right)^{2}$$

Calculate \sqrt{2x-10} to the power of 2 and get 2x-10.

$$x^{2}-10x+25=16\left(2x-10\right)$$

Use the distributive property to multiply 16 by 2x-10.

$$x^{2}-10x+25=32x-160$$

Subtract 32x from both sides.

$$x^{2}-10x+25-32x=-160$$

Combine -10x and -32x to get -42x.

$$x^{2}-42x+25=-160$$

Add 160 to both sides.

$$x^{2}-42x+25+160=0$$

Add 25 and 160 to get 185.

$$x^{2}-42x+185=0$$

To solve the equation, factor x^{2}-42x+185 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-42 ab=185$$

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 185.

$$-1,-185 -5,-37$$

Calculate the sum for each pair.

$$-1-185=-186 -5-37=-42$$

The solution is the pair that gives sum -42.

$$a=-37 b=-5$$

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

$$\left(x-37\right)\left(x-5\right)$$

To find equation solutions, solve x-37=0 and x-5=0.

$$x=37 x=5$$

Substitute 37 for x in the equation \sqrt{x-1}-\sqrt{2x-10}=2.

$$\sqrt{37-1}-\sqrt{2\times 37-10}=2$$

Simplify. The value x=37 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-2=2$$

Substitute 5 for x in the equation \sqrt{x-1}-\sqrt{2x-10}=2.

$$\sqrt{5-1}-\sqrt{2\times 5-10}=2$$

Simplify. The value x=5 satisfies the equation.

$$2=2$$

Equation \sqrt{x-1}=\sqrt{2x-10}+2 has a unique solution.

$$x=5$$