Solve left(-9x-4right)^2-9=0

$${ \left(-9x-4 \right) }^{ 2 } -9=0$$

Solve for x

$$x=-\frac{7}{9}\approx -0.777777778$$

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

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-9x-4\right)^{2}.

$$81x^{2}+72x+16-9=0$$

Subtract 9 from 16 to get 7.

$$81x^{2}+72x+7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 81x^{2}+ax+bx+7. To find a and b, set up a system to be solved.

$$a+b=72 ab=81\times 7=567$$

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 567.

$$1,567 3,189 7,81 9,63 21,27$$

Calculate the sum for each pair.

$$1+567=568 3+189=192 7+81=88 9+63=72 21+27=48$$

The solution is the pair that gives sum 72.

$$a=9 b=63$$

Rewrite 81x^{2}+72x+7 as \left(81x^{2}+9x\right)+\left(63x+7\right).

$$\left(81x^{2}+9x\right)+\left(63x+7\right)$$

Factor out 9x in the first and 7 in the second group.

$$9x\left(9x+1\right)+7\left(9x+1\right)$$

Factor out common term 9x+1 by using distributive property.

$$\left(9x+1\right)\left(9x+7\right)$$

To find equation solutions, solve 9x+1=0 and 9x+7=0.

$$x=-\frac{1}{9} x=-\frac{7}{9}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-9x-4\right)^{2}.

$$81x^{2}+72x+16-9=0$$

Subtract 9 from 16 to get 7.

$$81x^{2}+72x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 81 for a, 72 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-72±\sqrt{72^{2}-4\times 81\times 7}}{2\times 81}$$

Square 72.

$$x=\frac{-72±\sqrt{5184-4\times 81\times 7}}{2\times 81}$$

Multiply -4 times 81.

$$x=\frac{-72±\sqrt{5184-324\times 7}}{2\times 81}$$

Multiply -324 times 7.

$$x=\frac{-72±\sqrt{5184-2268}}{2\times 81}$$

Add 5184 to -2268.

$$x=\frac{-72±\sqrt{2916}}{2\times 81}$$

Take the square root of 2916.

$$x=\frac{-72±54}{2\times 81}$$

Multiply 2 times 81.

$$x=\frac{-72±54}{162}$$

Now solve the equation x=\frac{-72±54}{162} when ± is plus. Add -72 to 54.

$$x=-\frac{18}{162}$$

Reduce the fraction \frac{-18}{162} to lowest terms by extracting and canceling out 18.

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

Now solve the equation x=\frac{-72±54}{162} when ± is minus. Subtract 54 from -72.

$$x=-\frac{126}{162}$$

Reduce the fraction \frac{-126}{162} to lowest terms by extracting and canceling out 18.

$$x=-\frac{7}{9}$$

The equation is now solved.

$$x=-\frac{1}{9} x=-\frac{7}{9}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-9x-4\right)^{2}.

$$81x^{2}+72x+16-9=0$$

Subtract 9 from 16 to get 7.

$$81x^{2}+72x+7=0$$

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$81x^{2}+72x=-7$$

Divide both sides by 81.

$$\frac{81x^{2}+72x}{81}=-\frac{7}{81}$$

Dividing by 81 undoes the multiplication by 81.

$$x^{2}+\frac{72}{81}x=-\frac{7}{81}$$

Reduce the fraction \frac{72}{81} to lowest terms by extracting and canceling out 9.

$$x^{2}+\frac{8}{9}x=-\frac{7}{81}$$

Divide \frac{8}{9}, the coefficient of the x term, by 2 to get \frac{4}{9}. Then add the square of \frac{4}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{9}x+\left(\frac{4}{9}\right)^{2}=-\frac{7}{81}+\left(\frac{4}{9}\right)^{2}$$

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

$$x^{2}+\frac{8}{9}x+\frac{16}{81}=\frac{-7+16}{81}$$

Add -\frac{7}{81} to \frac{16}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{9}x+\frac{16}{81}=\frac{1}{9}$$

Factor x^{2}+\frac{8}{9}x+\frac{16}{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}.

$$\left(x+\frac{4}{9}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{9}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

$$x+\frac{4}{9}=\frac{1}{3} x+\frac{4}{9}=-\frac{1}{3}$$

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

$$x=-\frac{1}{9} x=-\frac{7}{9}$$