Solve 10x-1/x=3.xneq0

$$10 x - \frac { 1 } { x } = 3 . x \neq 0$$

Solve for x

$$x=-\frac{1}{5}=-0.2$$

$$x=\frac{1}{2}=0.5$$

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 x.

$$10xx-1=3x$$

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

$$10x^{2}-1=3x$$

Subtract 3x from both sides.

$$10x^{2}-1-3x=0$$

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

$$10x^{2}-3x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 10x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=-3 ab=10\left(-1\right)=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

Rewrite 10x^{2}-3x-1 as \left(10x^{2}-5x\right)+\left(2x-1\right).

$$\left(10x^{2}-5x\right)+\left(2x-1\right)$$

Factor out 5x in 10x^{2}-5x.

$$5x\left(2x-1\right)+2x-1$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(5x+1\right)$$

To find equation solutions, solve 2x-1=0 and 5x+1=0.

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

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 x.

$$10xx-1=3x$$

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

$$10x^{2}-1=3x$$

Subtract 3x from both sides.

$$10x^{2}-1-3x=0$$

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.

$$10x^{2}-3x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -3 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 10\left(-1\right)}}{2\times 10}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 10\left(-1\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-3\right)±\sqrt{9-40\left(-1\right)}}{2\times 10}$$

Multiply -40 times -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+40}}{2\times 10}$$

Add 9 to 40.

$$x=\frac{-\left(-3\right)±\sqrt{49}}{2\times 10}$$

Take the square root of 49.

$$x=\frac{-\left(-3\right)±7}{2\times 10}$$

The opposite of -3 is 3.

$$x=\frac{3±7}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{3±7}{20}$$

Now solve the equation x=\frac{3±7}{20} when ± is plus. Add 3 to 7.

$$x=\frac{10}{20}$$

Reduce the fraction \frac{10}{20} to lowest terms by extracting and canceling out 10.

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

Now solve the equation x=\frac{3±7}{20} when ± is minus. Subtract 7 from 3.

$$x=-\frac{4}{20}$$

Reduce the fraction \frac{-4}{20} to lowest terms by extracting and canceling out 4.

$$x=-\frac{1}{5}$$

The equation is now solved.

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

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 x.

$$10xx-1=3x$$

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

$$10x^{2}-1=3x$$

Subtract 3x from both sides.

$$10x^{2}-1-3x=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$10x^{2}-3x=1$$

Divide both sides by 10.

$$\frac{10x^{2}-3x}{10}=\frac{1}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}-\frac{3}{10}x=\frac{1}{10}$$

Divide -\frac{3}{10}, the coefficient of the x term, by 2 to get -\frac{3}{20}. Then add the square of -\frac{3}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{10}x+\left(-\frac{3}{20}\right)^{2}=\frac{1}{10}+\left(-\frac{3}{20}\right)^{2}$$

Square -\frac{3}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{10}x+\frac{9}{400}=\frac{1}{10}+\frac{9}{400}$$

Add \frac{1}{10} to \frac{9}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{10}x+\frac{9}{400}=\frac{49}{400}$$

Factor x^{2}-\frac{3}{10}x+\frac{9}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{20}\right)^{2}=\frac{49}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{20}\right)^{2}}=\sqrt{\frac{49}{400}}$$

Simplify.

$$x-\frac{3}{20}=\frac{7}{20} x-\frac{3}{20}=-\frac{7}{20}$$

Add \frac{3}{20} to both sides of the equation.

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