Solve 2x^2=-27x-81

$$2 x ^ { 2 } = - 27 x - 81$$

Solve for x

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

$$x=-9$$

Steps Using Factoring By Grouping

Add 27x to both sides.

$$2x^{2}+27x=-81$$

Add 81 to both sides.

$$2x^{2}+27x+81=0$$

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.

$$a+b=27 ab=2\times 81=162$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. 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.

$$1+162=163 2+81=83 3+54=57 6+27=33 9+18=27$$

The solution is the pair that gives sum 27.

$$a=9 b=18$$

Rewrite 2x^{2}+27x+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

Add 27x to both sides.

$$2x^{2}+27x=-81$$

Add 81 to both sides.

$$2x^{2}+27x+81=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 27 for b, and 81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-27±\sqrt{27^{2}-4\times 2\times 81}}{2\times 2}$$

Square 27.

$$x=\frac{-27±\sqrt{729-4\times 2\times 81}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-27±\sqrt{729-8\times 81}}{2\times 2}$$

Multiply -8 times 81.

$$x=\frac{-27±\sqrt{729-648}}{2\times 2}$$

Add 729 to -648.

$$x=\frac{-27±\sqrt{81}}{2\times 2}$$

Take the square root of 81.

$$x=\frac{-27±9}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-27±9}{4}$$

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

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

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{-27±9}{4} when ± is minus. Subtract 9 from -27.

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

Divide -36 by 4.

$$x=-9$$

The equation is now solved.

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

Steps for Completing the Square

Add 27x to both sides.

$$2x^{2}+27x=-81$$

Divide both sides by 2.

$$\frac{2x^{2}+27x}{2}=-\frac{81}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{27}{2}x=-\frac{81}{2}$$

Divide \frac{27}{2}, the coefficient of the x term, by 2 to get \frac{27}{4}. Then add the square of \frac{27}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{27}{2}x+\left(\frac{27}{4}\right)^{2}=-\frac{81}{2}+\left(\frac{27}{4}\right)^{2}$$

Square \frac{27}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{27}{2}x+\frac{729}{16}=-\frac{81}{2}+\frac{729}{16}$$

Add -\frac{81}{2} to \frac{729}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{27}{2}x+\frac{729}{16}=\frac{81}{16}$$

Factor x^{2}+\frac{27}{2}x+\frac{729}{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}.

$$\left(x+\frac{27}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{27}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

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

Subtract \frac{27}{4} from both sides of the equation.

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