Solve (x+1)^2=4(x-1)^2

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

Solve for x

$$x=3$$

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

Steps Using Factoring By Grouping

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

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

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

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

Use the distributive property to multiply 4 by x^{2}-2x+1.

$$x^{2}+2x+1=4x^{2}-8x+4$$

Subtract 4x^{2} from both sides.

$$x^{2}+2x+1-4x^{2}=-8x+4$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+2x+1=-8x+4$$

Add 8x to both sides.

$$-3x^{2}+2x+1+8x=4$$

Combine 2x and 8x to get 10x.

$$-3x^{2}+10x+1=4$$

Subtract 4 from both sides.

$$-3x^{2}+10x+1-4=0$$

Subtract 4 from 1 to get -3.

$$-3x^{2}+10x-3=0$$

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

$$a+b=10 ab=-3\left(-3\right)=9$$

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 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 10.

$$a=9 b=1$$

Rewrite -3x^{2}+10x-3 as \left(-3x^{2}+9x\right)+\left(x-3\right).

$$\left(-3x^{2}+9x\right)+\left(x-3\right)$$

Factor out 3x in the first and -1 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

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

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

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

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

Use the distributive property to multiply 4 by x^{2}-2x+1.

$$x^{2}+2x+1=4x^{2}-8x+4$$

Subtract 4x^{2} from both sides.

$$x^{2}+2x+1-4x^{2}=-8x+4$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+2x+1=-8x+4$$

Add 8x to both sides.

$$-3x^{2}+2x+1+8x=4$$

Combine 2x and 8x to get 10x.

$$-3x^{2}+10x+1=4$$

Subtract 4 from both sides.

$$-3x^{2}+10x+1-4=0$$

Subtract 4 from 1 to get -3.

$$-3x^{2}+10x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 10 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-3\right)\left(-3\right)}}{2\left(-3\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-3\right)\left(-3\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-10±\sqrt{100+12\left(-3\right)}}{2\left(-3\right)}$$

Multiply 12 times -3.

$$x=\frac{-10±\sqrt{100-36}}{2\left(-3\right)}$$

Add 100 to -36.

$$x=\frac{-10±\sqrt{64}}{2\left(-3\right)}$$

Take the square root of 64.

$$x=\frac{-10±8}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-10±8}{-6}$$

Now solve the equation x=\frac{-10±8}{-6} when ± is plus. Add -10 to 8.

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

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

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

Now solve the equation x=\frac{-10±8}{-6} when ± is minus. Subtract 8 from -10.

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

Divide -18 by -6.

$$x=3$$

The equation is now solved.

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

Steps for Completing the Square

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

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

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

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

Use the distributive property to multiply 4 by x^{2}-2x+1.

$$x^{2}+2x+1=4x^{2}-8x+4$$

Subtract 4x^{2} from both sides.

$$x^{2}+2x+1-4x^{2}=-8x+4$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+2x+1=-8x+4$$

Add 8x to both sides.

$$-3x^{2}+2x+1+8x=4$$

Combine 2x and 8x to get 10x.

$$-3x^{2}+10x+1=4$$

Subtract 1 from both sides.

$$-3x^{2}+10x=4-1$$

Subtract 1 from 4 to get 3.

$$-3x^{2}+10x=3$$

Divide both sides by -3.

$$\frac{-3x^{2}+10x}{-3}=\frac{3}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{10}{-3}x=\frac{3}{-3}$$

Divide 10 by -3.

$$x^{2}-\frac{10}{3}x=\frac{3}{-3}$$

Divide 3 by -3.

$$x^{2}-\frac{10}{3}x=-1$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=-1+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=-1+\frac{25}{9}$$

Add -1 to \frac{25}{9}.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{16}{9}$$

Factor x^{2}-\frac{10}{3}x+\frac{25}{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}.

$$\left(x-\frac{5}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$x-\frac{5}{3}=\frac{4}{3} x-\frac{5}{3}=-\frac{4}{3}$$

Add \frac{5}{3} to both sides of the equation.

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