Solve 54x+6/x=60

$$54 x + \frac { 6 } { x } = 60$$

Solve for x

$$x=\frac{1}{9}\approx 0.111111111$$

$$x=1$$

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.

$$54xx+6=60x$$

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

$$54x^{2}+6=60x$$

Subtract 60x from both sides.

$$54x^{2}+6-60x=0$$

Divide both sides by 6.

$$9x^{2}+1-10x=0$$

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

$$9x^{2}-10x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-10 ab=9\times 1=9$$

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 9x^{2}-10x+1 as \left(9x^{2}-9x\right)+\left(-x+1\right).

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

Factor out 9x in the first and -1 in the second group.

$$9x\left(x-1\right)-\left(x-1\right)$$

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

$$\left(x-1\right)\left(9x-1\right)$$

To find equation solutions, solve x-1=0 and 9x-1=0.

$$x=1 x=\frac{1}{9}$$

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.

$$54xx+6=60x$$

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

$$54x^{2}+6=60x$$

Subtract 60x from both sides.

$$54x^{2}+6-60x=0$$

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.

$$54x^{2}-60x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 54 for a, -60 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-60\right)±\sqrt{\left(-60\right)^{2}-4\times 54\times 6}}{2\times 54}$$

Square -60.

$$x=\frac{-\left(-60\right)±\sqrt{3600-4\times 54\times 6}}{2\times 54}$$

Multiply -4 times 54.

$$x=\frac{-\left(-60\right)±\sqrt{3600-216\times 6}}{2\times 54}$$

Multiply -216 times 6.

$$x=\frac{-\left(-60\right)±\sqrt{3600-1296}}{2\times 54}$$

Add 3600 to -1296.

$$x=\frac{-\left(-60\right)±\sqrt{2304}}{2\times 54}$$

Take the square root of 2304.

$$x=\frac{-\left(-60\right)±48}{2\times 54}$$

The opposite of -60 is 60.

$$x=\frac{60±48}{2\times 54}$$

Multiply 2 times 54.

$$x=\frac{60±48}{108}$$

Now solve the equation x=\frac{60±48}{108} when ± is plus. Add 60 to 48.

$$x=\frac{108}{108}$$

Divide 108 by 108.

$$x=1$$

Now solve the equation x=\frac{60±48}{108} when ± is minus. Subtract 48 from 60.

$$x=\frac{12}{108}$$

Reduce the fraction \frac{12}{108} to lowest terms by extracting and canceling out 12.

$$x=\frac{1}{9}$$

The equation is now solved.

$$x=1 x=\frac{1}{9}$$

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.

$$54xx+6=60x$$

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

$$54x^{2}+6=60x$$

Subtract 60x from both sides.

$$54x^{2}+6-60x=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$54x^{2}-60x=-6$$

Divide both sides by 54.

$$\frac{54x^{2}-60x}{54}=-\frac{6}{54}$$

Dividing by 54 undoes the multiplication by 54.

$$x^{2}+\left(-\frac{60}{54}\right)x=-\frac{6}{54}$$

Reduce the fraction \frac{-60}{54} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{10}{9}x=-\frac{6}{54}$$

Reduce the fraction \frac{-6}{54} to lowest terms by extracting and canceling out 6.

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

Divide -\frac{10}{9}, the coefficient of the x term, by 2 to get -\frac{5}{9}. Then add the square of -\frac{5}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{9}x+\left(-\frac{5}{9}\right)^{2}=-\frac{1}{9}+\left(-\frac{5}{9}\right)^{2}$$

Square -\frac{5}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{9}x+\frac{25}{81}=-\frac{1}{9}+\frac{25}{81}$$

Add -\frac{1}{9} to \frac{25}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{10}{9}x+\frac{25}{81}=\frac{16}{81}$$

Factor x^{2}-\frac{10}{9}x+\frac{25}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{9}\right)^{2}=\frac{16}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{9}\right)^{2}}=\sqrt{\frac{16}{81}}$$

Simplify.

$$x-\frac{5}{9}=\frac{4}{9} x-\frac{5}{9}=-\frac{4}{9}$$

Add \frac{5}{9} to both sides of the equation.

$$x=1 x=\frac{1}{9}$$