Solve (0)9(x+1)^2-(3x-1)^2=-4

$$( 0 ) 9 ( x + 1 ) ^ { 2 } - ( 3 x - 1 ) ^ { 2 } = - 4$$

Solve for x

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

$$x=1$$

Steps Using Factoring By Grouping

Multiply 0 and 9 to get 0.

$$0\left(x+1\right)^{2}-\left(3x-1\right)^{2}=-4$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$0\left(x^{2}+2x+1\right)-\left(3x-1\right)^{2}=-4$$

Anything times zero gives zero.

$$0-\left(3x-1\right)^{2}=-4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$0-\left(9x^{2}-6x+1\right)=-4$$

To find the opposite of 9x^{2}-6x+1, find the opposite of each term.

$$0-9x^{2}+6x-1=-4$$

Subtract 1 from 0 to get -1.

$$-1-9x^{2}+6x=-4$$

Add 4 to both sides.

$$-1-9x^{2}+6x+4=0$$

Add -1 and 4 to get 3.

$$3-9x^{2}+6x=0$$

Divide both sides by 3.

$$1-3x^{2}+2x=0$$

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

$$-3x^{2}+2x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=2 ab=-3=-3$$

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. The only such pair is the system solution.

$$a=3 b=-1$$

Rewrite -3x^{2}+2x+1 as \left(-3x^{2}+3x\right)+\left(-x+1\right).

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

Factor out 3x in -3x^{2}+3x.

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

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

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

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

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

Steps Using the Quadratic Formula

Multiply 0 and 9 to get 0.

$$0\left(x+1\right)^{2}-\left(3x-1\right)^{2}=-4$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$0\left(x^{2}+2x+1\right)-\left(3x-1\right)^{2}=-4$$

Anything times zero gives zero.

$$0-\left(3x-1\right)^{2}=-4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$0-\left(9x^{2}-6x+1\right)=-4$$

To find the opposite of 9x^{2}-6x+1, find the opposite of each term.

$$0-9x^{2}+6x-1=-4$$

Subtract 1 from 0 to get -1.

$$-1-9x^{2}+6x=-4$$

Add 4 to both sides.

$$-1-9x^{2}+6x+4=0$$

Add -1 and 4 to get 3.

$$3-9x^{2}+6x=0$$

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.

$$-9x^{2}+6x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 6 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-9\right)\times 3}}{2\left(-9\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-9\right)\times 3}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-6±\sqrt{36+36\times 3}}{2\left(-9\right)}$$

Multiply 36 times 3.

$$x=\frac{-6±\sqrt{36+108}}{2\left(-9\right)}$$

Add 36 to 108.

$$x=\frac{-6±\sqrt{144}}{2\left(-9\right)}$$

Take the square root of 144.

$$x=\frac{-6±12}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-6±12}{-18}$$

Now solve the equation x=\frac{-6±12}{-18} when ± is plus. Add -6 to 12.

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

Reduce the fraction \frac{6}{-18} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{-6±12}{-18} when ± is minus. Subtract 12 from -6.

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

Divide -18 by -18.

$$x=1$$

The equation is now solved.

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

Steps for Completing the Square

Multiply 0 and 9 to get 0.

$$0\left(x+1\right)^{2}-\left(3x-1\right)^{2}=-4$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$0\left(x^{2}+2x+1\right)-\left(3x-1\right)^{2}=-4$$

Anything times zero gives zero.

$$0-\left(3x-1\right)^{2}=-4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$0-\left(9x^{2}-6x+1\right)=-4$$

To find the opposite of 9x^{2}-6x+1, find the opposite of each term.

$$0-9x^{2}+6x-1=-4$$

Subtract 1 from 0 to get -1.

$$-1-9x^{2}+6x=-4$$

Add 1 to both sides.

$$-9x^{2}+6x=-4+1$$

Add -4 and 1 to get -3.

$$-9x^{2}+6x=-3$$

Divide both sides by -9.

$$\frac{-9x^{2}+6x}{-9}=-\frac{3}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\frac{6}{-9}x=-\frac{3}{-9}$$

Reduce the fraction \frac{6}{-9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{2}{3}x=-\frac{3}{-9}$$

Reduce the fraction \frac{-3}{-9} to lowest terms by extracting and canceling out 3.

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

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

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

Add \frac{1}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{4}{9}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{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}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

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

Add \frac{1}{3} to both sides of the equation.

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