Solve frac{sqrt{x-1}}{x}=-1/xsqrt{x^2-x}

$$\frac { \sqrt { x - 1 } } { x } = - \frac { 1 } { x } \sqrt { x ^ { 2 } - x }$$

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

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

To raise \frac{\sqrt{x-1}}{x} to a power, raise both numerator and denominator to the power and then divide.

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

Express \left(-\frac{1}{x}\right)\sqrt{x^{2}-x} as a single fraction.

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

To raise \frac{-\sqrt{x^{2}-x}}{x} to a power, raise both numerator and denominator to the power and then divide.

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

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

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

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

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

Calculate -1 to the power of 2 and get 1.

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

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

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

Factor the expressions that are not already factored in \frac{1\left(x^{2}-x\right)}{x^{2}}.

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

Cancel out x in both numerator and denominator.

$$\frac{x-1}{x^{2}}=\frac{x-1}{x}$$

Multiply both sides of the equation by x^{2}, the least common multiple of x^{2},x.

$$x-1=x\left(x-1\right)$$

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

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

Subtract x^{2} from both sides.

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

Add x to both sides.

$$x-1-x^{2}+x=0$$

Combine x and x to get 2x.

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

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

$$-x^{2}+2x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=2 ab=-\left(-1\right)=1$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=1$$

Rewrite -x^{2}+2x-1 as \left(-x^{2}+x\right)+\left(x-1\right).

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

Factor out -x in -x^{2}+x.

$$-x\left(x-1\right)+x-1$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(-x+1\right)$$

To find equation solutions, solve x-1=0 and -x+1=0.

$$x=1 x=1$$

Substitute 1 for x in the equation \frac{\sqrt{x-1}}{x}=\left(-\frac{1}{x}\right)\sqrt{x^{2}-x}.

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

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Substitute 1 for x in the equation \frac{\sqrt{x-1}}{x}=\left(-\frac{1}{x}\right)\sqrt{x^{2}-x}.

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

Simplify. The value x=1 satisfies the equation.

$$0=0$$

List all solutions of \frac{\sqrt{x-1}}{x}=\left(-\frac{1}{x}\right)\sqrt{x^{2}-x}.

$$x=1 x=1$$