Solve (x^2+1)=2x

$$( x ^ { 2 } + 1 ) = 2 x$$

Solve for x

$$x=1$$

Steps Using Factoring

Subtract 2x from both sides.

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

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 x^{2}-2x+1 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.

$$a+b=-2 ab=1$$

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

$$a=-1 b=-1$$

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

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

Rewrite as a binomial square.

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

To find equation solution, solve x-1=0.

$$x=1$$

Steps Using Factoring By Grouping

Subtract 2x from both sides.

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

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.

$$a+b=-2 ab=1\times 1=1$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. 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 the first and -1 in the second group.

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

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

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

Rewrite as a binomial square.

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

To find equation solution, solve x-1=0.

$$x=1$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Square -2.

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

Add 4 to -4.

$$x=\frac{-\left(-2\right)±\sqrt{0}}{2}$$

Take the square root of 0.

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

The opposite of -2 is 2.

$$x=\frac{2}{2}$$

Divide 2 by 2.

$$x=1$$

Steps for Completing the Square

Subtract 2x from both sides.

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

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Add -1 to 1.

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

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-1=0 x-1=0$$

Add 1 to both sides of the equation.

$$x=1 x=1$$

The equation is now solved. Solutions are the same.

$$x=1$$