Solve 0=1/2sqrt{x_{0-1}}*(0-x_{0})+sqrt{x_{0}-1}

$$0 = \frac { 1 } { 2 \sqrt { x _ { 0 } - 1 } } \cdot ( 0 - x _ { 0 } ) + \sqrt { x _ { 0 } - 1 }$$

Solve for x_0

$$x_{0}=2$$

Solution Steps

Subtract \sqrt{x_{0}-1} from both sides of the equation.

$$-\sqrt{x_{0}-1}=\frac{1}{2\sqrt{x_{0}-1}}\left(-1\right)x_{0}$$

Express \frac{1}{2\sqrt{x_{0}-1}}x_{0} as a single fraction.

$$-\sqrt{x_{0}-1}=\frac{x_{0}}{2\sqrt{x_{0}-1}}\left(-1\right)$$

Cancel out -1 on both sides.

$$\sqrt{x_{0}-1}=\frac{x_{0}}{2\sqrt{x_{0}-1}}$$

Square both sides of the equation.

$$\left(\sqrt{x_{0}-1}\right)^{2}=\left(\frac{x_{0}}{2\sqrt{x_{0}-1}}\right)^{2}$$

Calculate \sqrt{x_{0}-1} to the power of 2 and get x_{0}-1.

$$x_{0}-1=\left(\frac{x_{0}}{2\sqrt{x_{0}-1}}\right)^{2}$$

To raise \frac{x_{0}}{2\sqrt{x_{0}-1}} to a power, raise both numerator and denominator to the power and then divide.

$$x_{0}-1=\frac{x_{0}^{2}}{\left(2\sqrt{x_{0}-1}\right)^{2}}$$

Expand \left(2\sqrt{x_{0}-1}\right)^{2}.

$$x_{0}-1=\frac{x_{0}^{2}}{2^{2}\left(\sqrt{x_{0}-1}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$x_{0}-1=\frac{x_{0}^{2}}{4\left(\sqrt{x_{0}-1}\right)^{2}}$$

Calculate \sqrt{x_{0}-1} to the power of 2 and get x_{0}-1.

$$x_{0}-1=\frac{x_{0}^{2}}{4\left(x_{0}-1\right)}$$

Multiply both sides of the equation by 4\left(x_{0}-1\right).

$$4\left(x_{0}-1\right)x_{0}+4\left(x_{0}-1\right)\left(-1\right)=x_{0}^{2}$$

Reorder the terms.

$$4x_{0}\left(x_{0}-1\right)-4\left(x_{0}-1\right)=x_{0}^{2}$$

Use the distributive property to multiply 4x_{0} by x_{0}-1.

$$4x_{0}^{2}-4x_{0}-4\left(x_{0}-1\right)=x_{0}^{2}$$

Use the distributive property to multiply -4 by x_{0}-1.

$$4x_{0}^{2}-4x_{0}-4x_{0}+4=x_{0}^{2}$$

Combine -4x_{0} and -4x_{0} to get -8x_{0}.

$$4x_{0}^{2}-8x_{0}+4=x_{0}^{2}$$

Subtract x_{0}^{2} from both sides.

$$4x_{0}^{2}-8x_{0}+4-x_{0}^{2}=0$$

Combine 4x_{0}^{2} and -x_{0}^{2} to get 3x_{0}^{2}.

$$3x_{0}^{2}-8x_{0}+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x_{0}^{2}+ax_{0}+bx_{0}+4. To find a and b, set up a system to be solved.

$$a+b=-8 ab=3\times 4=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -8.

$$a=-6 b=-2$$

Rewrite 3x_{0}^{2}-8x_{0}+4 as \left(3x_{0}^{2}-6x_{0}\right)+\left(-2x_{0}+4\right).

$$\left(3x_{0}^{2}-6x_{0}\right)+\left(-2x_{0}+4\right)$$

Factor out 3x_{0} in the first and -2 in the second group.

$$3x_{0}\left(x_{0}-2\right)-2\left(x_{0}-2\right)$$

Factor out common term x_{0}-2 by using distributive property.

$$\left(x_{0}-2\right)\left(3x_{0}-2\right)$$

To find equation solutions, solve x_{0}-2=0 and 3x_{0}-2=0.

$$x_{0}=2 x_{0}=\frac{2}{3}$$

Substitute 2 for x_{0} in the equation 0=\frac{1}{2\sqrt{x_{0}-1}}\left(0-x_{0}\right)+\sqrt{x_{0}-1}.

$$0=\frac{1}{2\sqrt{2-1}}\left(0-2\right)+\sqrt{2-1}$$

Simplify. The value x_{0}=2 satisfies the equation.

$$0=0$$

Substitute \frac{2}{3} for x_{0} in the equation 0=\frac{1}{2\sqrt{x_{0}-1}}\left(0-x_{0}\right)+\sqrt{x_{0}-1}. The expression \sqrt{\frac{2}{3}-1} is undefined because the radicand cannot be negative.

$$0=\frac{1}{2\sqrt{\frac{2}{3}-1}}\left(0-\frac{2}{3}\right)+\sqrt{\frac{2}{3}-1}$$

Equation \sqrt{x_{0}-1}=\frac{x_{0}}{2\sqrt{x_{0}-1}} has a unique solution.

$$x_{0}=2$$