Solve 3left(x-2right)^2=4-x^2

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

Solve for x

$$x=1$$

$$x=2$$

Steps Using Factoring By Grouping

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

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

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

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

Subtract 4 from both sides.

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

Subtract 4 from 12 to get 8.

$$3x^{2}-12x+8=-x^{2}$$

Add x^{2} to both sides.

$$3x^{2}-12x+8+x^{2}=0$$

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

$$4x^{2}-12x+8=0$$

Divide both sides by 4.

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

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

$$a+b=-3 ab=1\times 2=2$$

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.

$$a=-2 b=-1$$

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

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

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

$$x\left(x-2\right)-\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x-1\right)$$

To find equation solutions, solve x-2=0 and x-1=0.

$$x=2 x=1$$

Steps Using the Quadratic Formula

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

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

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

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

Subtract 4 from both sides.

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

Subtract 4 from 12 to get 8.

$$3x^{2}-12x+8=-x^{2}$$

Add x^{2} to both sides.

$$3x^{2}-12x+8+x^{2}=0$$

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

$$4x^{2}-12x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -12 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\times 8}}{2\times 4}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 4\times 8}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-12\right)±\sqrt{144-16\times 8}}{2\times 4}$$

Multiply -16 times 8.

$$x=\frac{-\left(-12\right)±\sqrt{144-128}}{2\times 4}$$

Add 144 to -128.

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

Take the square root of 16.

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

The opposite of -12 is 12.

$$x=\frac{12±4}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{12±4}{8}$$

Now solve the equation x=\frac{12±4}{8} when ± is plus. Add 12 to 4.

$$x=\frac{16}{8}$$

Divide 16 by 8.

$$x=2$$

Now solve the equation x=\frac{12±4}{8} when ± is minus. Subtract 4 from 12.

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

Divide 8 by 8.

$$x=1$$

The equation is now solved.

$$x=2 x=1$$

Steps for Completing the Square

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

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

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

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

Add x^{2} to both sides.

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

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

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

Subtract 12 from both sides.

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

Subtract 12 from 4 to get -8.

$$4x^{2}-12x=-8$$

Divide both sides by 4.

$$\frac{4x^{2}-12x}{4}=-\frac{8}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{12}{4}\right)x=-\frac{8}{4}$$

Divide -12 by 4.

$$x^{2}-3x=-\frac{8}{4}$$

Divide -8 by 4.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

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

Add -2 to \frac{9}{4}.

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

Factor x^{2}-3x+\frac{9}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=2 x=1$$