Solve x^2-x+1/4=x/4

$${ x }^{ 2 } -x+ \frac{ 1 }{ 4 } = \frac{ x }{ 4 }$$

Solve for x

$$x=\frac{1}{4}=0.25$$

$$x=1$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 4.

$$4x^{2}-4x+1=x$$

Subtract x from both sides.

$$4x^{2}-4x+1-x=0$$

Combine -4x and -x to get -5x.

$$4x^{2}-5x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-5 ab=4\times 1=4$$

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 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$a=-4 b=-1$$

Rewrite 4x^{2}-5x+1 as \left(4x^{2}-4x\right)+\left(-x+1\right).

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

Factor out 4x in the first and -1 in the second group.

$$4x\left(x-1\right)-\left(x-1\right)$$

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

$$\left(x-1\right)\left(4x-1\right)$$

To find equation solutions, solve x-1=0 and 4x-1=0.

$$x=1 x=\frac{1}{4}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$4x^{2}-4x+1=x$$

Subtract x from both sides.

$$4x^{2}-4x+1-x=0$$

Combine -4x and -x to get -5x.

$$4x^{2}-5x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -5 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 4}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-5\right)±\sqrt{25-16}}{2\times 4}$$

Add 25 to -16.

$$x=\frac{-\left(-5\right)±\sqrt{9}}{2\times 4}$$

Take the square root of 9.

$$x=\frac{-\left(-5\right)±3}{2\times 4}$$

The opposite of -5 is 5.

$$x=\frac{5±3}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{5±3}{8}$$

Now solve the equation x=\frac{5±3}{8} when ± is plus. Add 5 to 3.

$$x=\frac{8}{8}$$

Divide 8 by 8.

$$x=1$$

Now solve the equation x=\frac{5±3}{8} when ± is minus. Subtract 3 from 5.

$$x=\frac{2}{8}$$

Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{4}$$

The equation is now solved.

$$x=1 x=\frac{1}{4}$$

Steps for Completing the Square

Multiply both sides of the equation by 4.

$$4x^{2}-4x+1=x$$

Subtract x from both sides.

$$4x^{2}-4x+1-x=0$$

Combine -4x and -x to get -5x.

$$4x^{2}-5x+1=0$$

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

$$4x^{2}-5x=-1$$

Divide both sides by 4.

$$\frac{4x^{2}-5x}{4}=-\frac{1}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{5}{4}x=-\frac{1}{4}$$

Divide -\frac{5}{4}, the coefficient of the x term, by 2 to get -\frac{5}{8}. Then add the square of -\frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{4}x+\left(-\frac{5}{8}\right)^{2}=-\frac{1}{4}+\left(-\frac{5}{8}\right)^{2}$$

Square -\frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{4}x+\frac{25}{64}=-\frac{1}{4}+\frac{25}{64}$$

Add -\frac{1}{4} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{4}x+\frac{25}{64}=\frac{9}{64}$$

Factor x^{2}-\frac{5}{4}x+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{8}\right)^{2}=\frac{9}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{8}\right)^{2}}=\sqrt{\frac{9}{64}}$$

Simplify.

$$x-\frac{5}{8}=\frac{3}{8} x-\frac{5}{8}=-\frac{3}{8}$$

Add \frac{5}{8} to both sides of the equation.

$$x=1 x=\frac{1}{4}$$