Solve sqrt{(2x-5)}-sqrt{(x-1)}=1

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

Solve for x

$$x=2\sqrt{5}+7\approx 11.472135955$$

Solution Steps

Subtract -\sqrt{x-1} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

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

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

Subtract 1 from 1 to get 0.

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

Subtract x from both sides of the equation.

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

Combine 2x and -x to get x.

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

Square both sides of the equation.

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

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

$$x^{2}-10x+25=\left(2\sqrt{x-1}\right)^{2}$$

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

$$x^{2}-10x+25=2^{2}\left(\sqrt{x-1}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}-10x+25=4\left(\sqrt{x-1}\right)^{2}$$

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

$$x^{2}-10x+25=4\left(x-1\right)$$

Use the distributive property to multiply 4 by x-1.

$$x^{2}-10x+25=4x-4$$

Subtract 4x from both sides.

$$x^{2}-10x+25-4x=-4$$

Combine -10x and -4x to get -14x.

$$x^{2}-14x+25=-4$$

Add 4 to both sides.

$$x^{2}-14x+25+4=0$$

Add 25 and 4 to get 29.

$$x^{2}-14x+29=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -14 for b, and 29 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 29}}{2}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 29}}{2}$$

Multiply -4 times 29.

$$x=\frac{-\left(-14\right)±\sqrt{196-116}}{2}$$

Add 196 to -116.

$$x=\frac{-\left(-14\right)±\sqrt{80}}{2}$$

Take the square root of 80.

$$x=\frac{-\left(-14\right)±4\sqrt{5}}{2}$$

The opposite of -14 is 14.

$$x=\frac{14±4\sqrt{5}}{2}$$

Now solve the equation x=\frac{14±4\sqrt{5}}{2} when ± is plus. Add 14 to 4\sqrt{5}.

$$x=\frac{4\sqrt{5}+14}{2}$$

Divide 14+4\sqrt{5} by 2.

$$x=2\sqrt{5}+7$$

Now solve the equation x=\frac{14±4\sqrt{5}}{2} when ± is minus. Subtract 4\sqrt{5} from 14.

$$x=\frac{14-4\sqrt{5}}{2}$$

Divide 14-4\sqrt{5} by 2.

$$x=7-2\sqrt{5}$$

The equation is now solved.

$$x=2\sqrt{5}+7 x=7-2\sqrt{5}$$

Substitute 2\sqrt{5}+7 for x in the equation \sqrt{2x-5}-\sqrt{x-1}=1.

$$\sqrt{2\left(2\sqrt{5}+7\right)-5}-\sqrt{2\sqrt{5}+7-1}=1$$

Simplify. The value x=2\sqrt{5}+7 satisfies the equation.

$$1=1$$

Substitute 7-2\sqrt{5} for x in the equation \sqrt{2x-5}-\sqrt{x-1}=1.

$$\sqrt{2\left(7-2\sqrt{5}\right)-5}-\sqrt{7-2\sqrt{5}-1}=1$$

Simplify. The value x=7-2\sqrt{5} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-1=1$$

Substitute 2\sqrt{5}+7 for x in the equation \sqrt{2x-5}-\sqrt{x-1}=1.

$$\sqrt{2\left(2\sqrt{5}+7\right)-5}-\sqrt{2\sqrt{5}+7-1}=1$$

Simplify. The value x=2\sqrt{5}+7 satisfies the equation.

$$1=1$$

Equation \sqrt{2x-5}=\sqrt{x-1}+1 has a unique solution.

$$x=2\sqrt{5}+7$$