Solve 7(x-1)(x+1)=(x+1)^2

$$7 ( x - 1 ) ( x + 1 ) = ( x + 1 ) ^ { 2 }$$

Solve for x

$$x=-1$$

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 7 by x-1.

$$\left(7x-7\right)\left(x+1\right)=\left(x+1\right)^{2}$$

Use the distributive property to multiply 7x-7 by x+1 and combine like terms.

$$7x^{2}-7=\left(x+1\right)^{2}$$

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

$$7x^{2}-7=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$7x^{2}-7-x^{2}=2x+1$$

Combine 7x^{2} and -x^{2} to get 6x^{2}.

$$6x^{2}-7=2x+1$$

Subtract 2x from both sides.

$$6x^{2}-7-2x=1$$

Subtract 1 from both sides.

$$6x^{2}-7-2x-1=0$$

Subtract 1 from -7 to get -8.

$$6x^{2}-8-2x=0$$

Divide both sides by 2.

$$3x^{2}-4-x=0$$

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

$$3x^{2}-x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-1 ab=3\left(-4\right)=-12$$

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 -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -1.

$$a=-4 b=3$$

Rewrite 3x^{2}-x-4 as \left(3x^{2}-4x\right)+\left(3x-4\right).

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

Factor out x in 3x^{2}-4x.

$$x\left(3x-4\right)+3x-4$$

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(x+1\right)$$

To find equation solutions, solve 3x-4=0 and x+1=0.

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

Steps Using the Quadratic Formula

Use the distributive property to multiply 7 by x-1.

$$\left(7x-7\right)\left(x+1\right)=\left(x+1\right)^{2}$$

Use the distributive property to multiply 7x-7 by x+1 and combine like terms.

$$7x^{2}-7=\left(x+1\right)^{2}$$

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

$$7x^{2}-7=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$7x^{2}-7-x^{2}=2x+1$$

Combine 7x^{2} and -x^{2} to get 6x^{2}.

$$6x^{2}-7=2x+1$$

Subtract 2x from both sides.

$$6x^{2}-7-2x=1$$

Subtract 1 from both sides.

$$6x^{2}-7-2x-1=0$$

Subtract 1 from -7 to get -8.

$$6x^{2}-8-2x=0$$

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.

$$6x^{2}-2x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -2 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 6\left(-8\right)}}{2\times 6}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 6\left(-8\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-2\right)±\sqrt{4-24\left(-8\right)}}{2\times 6}$$

Multiply -24 times -8.

$$x=\frac{-\left(-2\right)±\sqrt{4+192}}{2\times 6}$$

Add 4 to 192.

$$x=\frac{-\left(-2\right)±\sqrt{196}}{2\times 6}$$

Take the square root of 196.

$$x=\frac{-\left(-2\right)±14}{2\times 6}$$

The opposite of -2 is 2.

$$x=\frac{2±14}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{2±14}{12}$$

Now solve the equation x=\frac{2±14}{12} when ± is plus. Add 2 to 14.

$$x=\frac{16}{12}$$

Reduce the fraction \frac{16}{12} to lowest terms by extracting and canceling out 4.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{2±14}{12} when ± is minus. Subtract 14 from 2.

$$x=-\frac{12}{12}$$

Divide -12 by 12.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply 7 by x-1.

$$\left(7x-7\right)\left(x+1\right)=\left(x+1\right)^{2}$$

Use the distributive property to multiply 7x-7 by x+1 and combine like terms.

$$7x^{2}-7=\left(x+1\right)^{2}$$

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

$$7x^{2}-7=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$7x^{2}-7-x^{2}=2x+1$$

Combine 7x^{2} and -x^{2} to get 6x^{2}.

$$6x^{2}-7=2x+1$$

Subtract 2x from both sides.

$$6x^{2}-7-2x=1$$

Add 7 to both sides.

$$6x^{2}-2x=1+7$$

Add 1 and 7 to get 8.

$$6x^{2}-2x=8$$

Divide both sides by 6.

$$\frac{6x^{2}-2x}{6}=\frac{8}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{2}{6}\right)x=\frac{8}{6}$$

Reduce the fraction \frac{-2}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{3}x=\frac{8}{6}$$

Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

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

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{4}{3}+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{4}{3}+\frac{1}{36}$$

Add \frac{4}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{49}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{7}{6} x-\frac{1}{6}=-\frac{7}{6}$$

Add \frac{1}{6} to both sides of the equation.

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