Solve x+1/4/x=17/8

$$x+ \frac{ \frac{ 1 }{ 4 } }{ x } = \frac{ 17 }{ 8 }$$

Solve for x

$$x=\frac{1}{8}=0.125$$

$$x=2$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 8x, the least common multiple of x,8.

$$8xx+8\times \frac{1}{4}=17x$$

Multiply x and x to get x^{2}.

$$8x^{2}+8\times \frac{1}{4}=17x$$

Multiply 8 and \frac{1}{4} to get \frac{8}{4}.

$$8x^{2}+\frac{8}{4}=17x$$

Divide 8 by 4 to get 2.

$$8x^{2}+2=17x$$

Subtract 17x from both sides.

$$8x^{2}+2-17x=0$$

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

$$8x^{2}-17x+2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=-17 ab=8\times 2=16$$

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 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -17.

$$a=-16 b=-1$$

Rewrite 8x^{2}-17x+2 as \left(8x^{2}-16x\right)+\left(-x+2\right).

$$\left(8x^{2}-16x\right)+\left(-x+2\right)$$

Factor out 8x in the first and -1 in the second group.

$$8x\left(x-2\right)-\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(8x-1\right)$$

To find equation solutions, solve x-2=0 and 8x-1=0.

$$x=2 x=\frac{1}{8}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 8x, the least common multiple of x,8.

$$8xx+8\times \frac{1}{4}=17x$$

Multiply x and x to get x^{2}.

$$8x^{2}+8\times \frac{1}{4}=17x$$

Multiply 8 and \frac{1}{4} to get \frac{8}{4}.

$$8x^{2}+\frac{8}{4}=17x$$

Divide 8 by 4 to get 2.

$$8x^{2}+2=17x$$

Subtract 17x from both sides.

$$8x^{2}+2-17x=0$$

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.

$$8x^{2}-17x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -17 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 8\times 2}}{2\times 8}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 8\times 2}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-17\right)±\sqrt{289-32\times 2}}{2\times 8}$$

Multiply -32 times 2.

$$x=\frac{-\left(-17\right)±\sqrt{289-64}}{2\times 8}$$

Add 289 to -64.

$$x=\frac{-\left(-17\right)±\sqrt{225}}{2\times 8}$$

Take the square root of 225.

$$x=\frac{-\left(-17\right)±15}{2\times 8}$$

The opposite of -17 is 17.

$$x=\frac{17±15}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{17±15}{16}$$

Now solve the equation x=\frac{17±15}{16} when ± is plus. Add 17 to 15.

$$x=\frac{32}{16}$$

Divide 32 by 16.

$$x=2$$

Now solve the equation x=\frac{17±15}{16} when ± is minus. Subtract 15 from 17.

$$x=\frac{2}{16}$$

Reduce the fraction \frac{2}{16} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{8}$$

The equation is now solved.

$$x=2 x=\frac{1}{8}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 8x, the least common multiple of x,8.

$$8xx+8\times \frac{1}{4}=17x$$

Multiply x and x to get x^{2}.

$$8x^{2}+8\times \frac{1}{4}=17x$$

Multiply 8 and \frac{1}{4} to get \frac{8}{4}.

$$8x^{2}+\frac{8}{4}=17x$$

Divide 8 by 4 to get 2.

$$8x^{2}+2=17x$$

Subtract 17x from both sides.

$$8x^{2}+2-17x=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$8x^{2}-17x=-2$$

Divide both sides by 8.

$$\frac{8x^{2}-17x}{8}=-\frac{2}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}-\frac{17}{8}x=-\frac{2}{8}$$

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

$$x^{2}-\frac{17}{8}x=-\frac{1}{4}$$

Divide -\frac{17}{8}, the coefficient of the x term, by 2 to get -\frac{17}{16}. Then add the square of -\frac{17}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{17}{8}x+\left(-\frac{17}{16}\right)^{2}=-\frac{1}{4}+\left(-\frac{17}{16}\right)^{2}$$

Square -\frac{17}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{17}{8}x+\frac{289}{256}=-\frac{1}{4}+\frac{289}{256}$$

Add -\frac{1}{4} to \frac{289}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{17}{8}x+\frac{289}{256}=\frac{225}{256}$$

Factor x^{2}-\frac{17}{8}x+\frac{289}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{17}{16}\right)^{2}=\frac{225}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{16}\right)^{2}}=\sqrt{\frac{225}{256}}$$

Simplify.

$$x-\frac{17}{16}=\frac{15}{16} x-\frac{17}{16}=-\frac{15}{16}$$

Add \frac{17}{16} to both sides of the equation.

$$x=2 x=\frac{1}{8}$$