Solve x^2+1/x=31/3

$$\frac { x ^ { 2 } + 1 } { x } = 3 \frac { 1 } { 3 }$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=3$$

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 3x, the least common multiple of x,3.

$$3\left(x^{2}+1\right)=x\left(3\times 3+1\right)$$

Use the distributive property to multiply 3 by x^{2}+1.

$$3x^{2}+3=x\left(3\times 3+1\right)$$

Multiply 3 and 3 to get 9.

$$3x^{2}+3=x\left(9+1\right)$$

Add 9 and 1 to get 10.

$$3x^{2}+3=x\times 10$$

Subtract x\times 10 from both sides.

$$3x^{2}+3-x\times 10=0$$

Multiply -1 and 10 to get -10.

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

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

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

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

$$a+b=-10 ab=3\times 3=9$$

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 9.

$$-1,-9 -3,-3$$

Calculate the sum for each pair.

$$-1-9=-10 -3-3=-6$$

The solution is the pair that gives sum -10.

$$a=-9 b=-1$$

Rewrite 3x^{2}-10x+3 as \left(3x^{2}-9x\right)+\left(-x+3\right).

$$\left(3x^{2}-9x\right)+\left(-x+3\right)$$

Factor out 3x in the first and -1 in the second group.

$$3x\left(x-3\right)-\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(3x-1\right)$$

To find equation solutions, solve x-3=0 and 3x-1=0.

$$x=3 x=\frac{1}{3}$$

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 3x, the least common multiple of x,3.

$$3\left(x^{2}+1\right)=x\left(3\times 3+1\right)$$

Use the distributive property to multiply 3 by x^{2}+1.

$$3x^{2}+3=x\left(3\times 3+1\right)$$

Multiply 3 and 3 to get 9.

$$3x^{2}+3=x\left(9+1\right)$$

Add 9 and 1 to get 10.

$$3x^{2}+3=x\times 10$$

Subtract x\times 10 from both sides.

$$3x^{2}+3-x\times 10=0$$

Multiply -1 and 10 to get -10.

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

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.

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

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

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

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 3\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-10\right)±\sqrt{100-12\times 3}}{2\times 3}$$

Multiply -12 times 3.

$$x=\frac{-\left(-10\right)±\sqrt{100-36}}{2\times 3}$$

Add 100 to -36.

$$x=\frac{-\left(-10\right)±\sqrt{64}}{2\times 3}$$

Take the square root of 64.

$$x=\frac{-\left(-10\right)±8}{2\times 3}$$

The opposite of -10 is 10.

$$x=\frac{10±8}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{10±8}{6}$$

Now solve the equation x=\frac{10±8}{6} when ± is plus. Add 10 to 8.

$$x=\frac{18}{6}$$

Divide 18 by 6.

$$x=3$$

Now solve the equation x=\frac{10±8}{6} when ± is minus. Subtract 8 from 10.

$$x=\frac{2}{6}$$

Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{3}$$

The equation is now solved.

$$x=3 x=\frac{1}{3}$$

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 3x, the least common multiple of x,3.

$$3\left(x^{2}+1\right)=x\left(3\times 3+1\right)$$

Use the distributive property to multiply 3 by x^{2}+1.

$$3x^{2}+3=x\left(3\times 3+1\right)$$

Multiply 3 and 3 to get 9.

$$3x^{2}+3=x\left(9+1\right)$$

Add 9 and 1 to get 10.

$$3x^{2}+3=x\times 10$$

Subtract x\times 10 from both sides.

$$3x^{2}+3-x\times 10=0$$

Multiply -1 and 10 to get -10.

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

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide -3 by 3.

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

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=-1+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=-1+\frac{25}{9}$$

Add -1 to \frac{25}{9}.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{16}{9}$$

Factor x^{2}-\frac{10}{3}x+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$x-\frac{5}{3}=\frac{4}{3} x-\frac{5}{3}=-\frac{4}{3}$$

Add \frac{5}{3} to both sides of the equation.

$$x=3 x=\frac{1}{3}$$