Solve x^2-4x/(x-2)^2=-3

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

Solve for x

$$x=1$$

$$x=3$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)^{2}.

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

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

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

Use the distributive property to multiply -3 by x^{2}-4x+4.

$$x^{2}-4x=-3x^{2}+12x-12$$

Add 3x^{2} to both sides.

$$x^{2}-4x+3x^{2}=12x-12$$

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

$$4x^{2}-4x=12x-12$$

Subtract 12x from both sides.

$$4x^{2}-4x-12x=-12$$

Combine -4x and -12x to get -16x.

$$4x^{2}-16x=-12$$

Add 12 to both sides.

$$4x^{2}-16x+12=0$$

Divide both sides by 4.

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

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

$$a+b=-4 ab=1\times 3=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-3 b=-1$$

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

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

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

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

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

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

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

$$x=3 x=1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)^{2}.

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

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

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

Use the distributive property to multiply -3 by x^{2}-4x+4.

$$x^{2}-4x=-3x^{2}+12x-12$$

Add 3x^{2} to both sides.

$$x^{2}-4x+3x^{2}=12x-12$$

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

$$4x^{2}-4x=12x-12$$

Subtract 12x from both sides.

$$4x^{2}-4x-12x=-12$$

Combine -4x and -12x to get -16x.

$$4x^{2}-16x=-12$$

Add 12 to both sides.

$$4x^{2}-16x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -16 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 4\times 12}}{2\times 4}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 4\times 12}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-16\right)±\sqrt{256-16\times 12}}{2\times 4}$$

Multiply -16 times 12.

$$x=\frac{-\left(-16\right)±\sqrt{256-192}}{2\times 4}$$

Add 256 to -192.

$$x=\frac{-\left(-16\right)±\sqrt{64}}{2\times 4}$$

Take the square root of 64.

$$x=\frac{-\left(-16\right)±8}{2\times 4}$$

The opposite of -16 is 16.

$$x=\frac{16±8}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{16±8}{8}$$

Now solve the equation x=\frac{16±8}{8} when ± is plus. Add 16 to 8.

$$x=\frac{24}{8}$$

Divide 24 by 8.

$$x=3$$

Now solve the equation x=\frac{16±8}{8} when ± is minus. Subtract 8 from 16.

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

Divide 8 by 8.

$$x=1$$

The equation is now solved.

$$x=3 x=1$$

Steps for Completing the Square

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)^{2}.

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

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

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

Use the distributive property to multiply -3 by x^{2}-4x+4.

$$x^{2}-4x=-3x^{2}+12x-12$$

Add 3x^{2} to both sides.

$$x^{2}-4x+3x^{2}=12x-12$$

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

$$4x^{2}-4x=12x-12$$

Subtract 12x from both sides.

$$4x^{2}-4x-12x=-12$$

Combine -4x and -12x to get -16x.

$$4x^{2}-16x=-12$$

Divide both sides by 4.

$$\frac{4x^{2}-16x}{4}=-\frac{12}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{16}{4}\right)x=-\frac{12}{4}$$

Divide -16 by 4.

$$x^{2}-4x=-\frac{12}{4}$$

Divide -12 by 4.

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

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

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

Add -3 to 4.

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

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=1 x-2=-1$$

Add 2 to both sides of the equation.

$$x=3 x=1$$