Solve 2x^2/27+x/3-3=0

$$\frac { 2 x ^ { 2 } } { 27 } + \frac { x } { 3 } - 3 = 0$$

Solve for x

$$x=-9$$

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 27, the least common multiple of 27,3.

$$2x^{2}+9x-81=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-81. To find a and b, set up a system to be solved.

$$a+b=9 ab=2\left(-81\right)=-162$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -162.

$$-1,162 -2,81 -3,54 -6,27 -9,18$$

Calculate the sum for each pair.

$$-1+162=161 -2+81=79 -3+54=51 -6+27=21 -9+18=9$$

The solution is the pair that gives sum 9.

$$a=-9 b=18$$

Rewrite 2x^{2}+9x-81 as \left(2x^{2}-9x\right)+\left(18x-81\right).

$$\left(2x^{2}-9x\right)+\left(18x-81\right)$$

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

$$x\left(2x-9\right)+9\left(2x-9\right)$$

Factor out common term 2x-9 by using distributive property.

$$\left(2x-9\right)\left(x+9\right)$$

To find equation solutions, solve 2x-9=0 and x+9=0.

$$x=\frac{9}{2} x=-9$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 27, the least common multiple of 27,3.

$$2x^{2}+9x-81=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 9 for b, and -81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\times 2\left(-81\right)}}{2\times 2}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 2\left(-81\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-9±\sqrt{81-8\left(-81\right)}}{2\times 2}$$

Multiply -8 times -81.

$$x=\frac{-9±\sqrt{81+648}}{2\times 2}$$

Add 81 to 648.

$$x=\frac{-9±\sqrt{729}}{2\times 2}$$

Take the square root of 729.

$$x=\frac{-9±27}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-9±27}{4}$$

Now solve the equation x=\frac{-9±27}{4} when ± is plus. Add -9 to 27.

$$x=\frac{18}{4}$$

Reduce the fraction \frac{18}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{9}{2}$$

Now solve the equation x=\frac{-9±27}{4} when ± is minus. Subtract 27 from -9.

$$x=-\frac{36}{4}$$

Divide -36 by 4.

$$x=-9$$

The equation is now solved.

$$x=\frac{9}{2} x=-9$$

Steps for Completing the Square

Multiply both sides of the equation by 27, the least common multiple of 27,3.

$$2x^{2}+9x-81=0$$

Add 81 to both sides. Anything plus zero gives itself.

$$2x^{2}+9x=81$$

Divide both sides by 2.

$$\frac{2x^{2}+9x}{2}=\frac{81}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{9}{2}x=\frac{81}{2}$$

Divide \frac{9}{2}, the coefficient of the x term, by 2 to get \frac{9}{4}. Then add the square of \frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{9}{2}x+\left(\frac{9}{4}\right)^{2}=\frac{81}{2}+\left(\frac{9}{4}\right)^{2}$$

Square \frac{9}{4} by squaring both the numerator and the denominator of the fraction.

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

Add \frac{81}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=\frac{729}{16}$$

Factor x^{2}+\frac{9}{2}x+\frac{81}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{9}{4}\right)^{2}=\frac{729}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{4}\right)^{2}}=\sqrt{\frac{729}{16}}$$

Simplify.

$$x+\frac{9}{4}=\frac{27}{4} x+\frac{9}{4}=-\frac{27}{4}$$

Subtract \frac{9}{4} from both sides of the equation.

$$x=\frac{9}{2} x=-9$$