Solve (x+6)-(3x+1)^2=1

$$( x + 6 ) - ( 3 x + 1 ) ^ { 2 } = 1$$

Solve for x

$$x=-1$$

$$x=\frac{4}{9}\approx 0.444444444$$

Steps Using Factoring By Grouping

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

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

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

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

Combine x and -6x to get -5x.

$$-5x+6-9x^{2}-1=1$$

Subtract 1 from 6 to get 5.

$$-5x+5-9x^{2}=1$$

Subtract 1 from both sides.

$$-5x+5-9x^{2}-1=0$$

Subtract 1 from 5 to get 4.

$$-5x+4-9x^{2}=0$$

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

$$-9x^{2}-5x+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -9x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-9\times 4=-36$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -5.

$$a=4 b=-9$$

Rewrite -9x^{2}-5x+4 as \left(-9x^{2}+4x\right)+\left(-9x+4\right).

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

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

$$-x\left(9x-4\right)-\left(9x-4\right)$$

Factor out common term 9x-4 by using distributive property.

$$\left(9x-4\right)\left(-x-1\right)$$

To find equation solutions, solve 9x-4=0 and -x-1=0.

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

Steps Using the Quadratic Formula

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

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

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

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

Combine x and -6x to get -5x.

$$-5x+6-9x^{2}-1=1$$

Subtract 1 from 6 to get 5.

$$-5x+5-9x^{2}=1$$

Subtract 1 from both sides.

$$-5x+5-9x^{2}-1=0$$

Subtract 1 from 5 to get 4.

$$-5x+4-9x^{2}=0$$

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.

$$-9x^{2}-5x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, -5 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-9\right)\times 4}}{2\left(-9\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-9\right)\times 4}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-\left(-5\right)±\sqrt{25+36\times 4}}{2\left(-9\right)}$$

Multiply 36 times 4.

$$x=\frac{-\left(-5\right)±\sqrt{25+144}}{2\left(-9\right)}$$

Add 25 to 144.

$$x=\frac{-\left(-5\right)±\sqrt{169}}{2\left(-9\right)}$$

Take the square root of 169.

$$x=\frac{-\left(-5\right)±13}{2\left(-9\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±13}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{5±13}{-18}$$

Now solve the equation x=\frac{5±13}{-18} when ± is plus. Add 5 to 13.

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

Divide 18 by -18.

$$x=-1$$

Now solve the equation x=\frac{5±13}{-18} when ± is minus. Subtract 13 from 5.

$$x=-\frac{8}{-18}$$

Reduce the fraction \frac{-8}{-18} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

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

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

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

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

Combine x and -6x to get -5x.

$$-5x+6-9x^{2}-1=1$$

Subtract 1 from 6 to get 5.

$$-5x+5-9x^{2}=1$$

Subtract 5 from both sides.

$$-5x-9x^{2}=1-5$$

Subtract 5 from 1 to get -4.

$$-5x-9x^{2}=-4$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-9x^{2}-5x=-4$$

Divide both sides by -9.

$$\frac{-9x^{2}-5x}{-9}=-\frac{4}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\left(-\frac{5}{-9}\right)x=-\frac{4}{-9}$$

Divide -5 by -9.

$$x^{2}+\frac{5}{9}x=-\frac{4}{-9}$$

Divide -4 by -9.

$$x^{2}+\frac{5}{9}x=\frac{4}{9}$$

Divide \frac{5}{9}, the coefficient of the x term, by 2 to get \frac{5}{18}. Then add the square of \frac{5}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{9}x+\left(\frac{5}{18}\right)^{2}=\frac{4}{9}+\left(\frac{5}{18}\right)^{2}$$

Square \frac{5}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{9}x+\frac{25}{324}=\frac{4}{9}+\frac{25}{324}$$

Add \frac{4}{9} to \frac{25}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{9}x+\frac{25}{324}=\frac{169}{324}$$

Factor x^{2}+\frac{5}{9}x+\frac{25}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{18}\right)^{2}=\frac{169}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{18}\right)^{2}}=\sqrt{\frac{169}{324}}$$

Simplify.

$$x+\frac{5}{18}=\frac{13}{18} x+\frac{5}{18}=-\frac{13}{18}$$

Subtract \frac{5}{18} from both sides of the equation.

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