Solve (3x-1)^2=4(1-x)^2

$$( 3 x - 1 ) ^ { 2 } = 4 ( 1 - x ) ^ { 2 }$$

Solve for x

$$x=-1$$

$$x=\frac{3}{5}=0.6$$

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}.

$$9x^{2}-6x+1=4\left(1-x\right)^{2}$$

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

$$9x^{2}-6x+1=4\left(1-2x+x^{2}\right)$$

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

$$9x^{2}-6x+1=4-8x+4x^{2}$$

Subtract 4 from both sides.

$$9x^{2}-6x+1-4=-8x+4x^{2}$$

Subtract 4 from 1 to get -3.

$$9x^{2}-6x-3=-8x+4x^{2}$$

Add 8x to both sides.

$$9x^{2}-6x-3+8x=4x^{2}$$

Combine -6x and 8x to get 2x.

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

Subtract 4x^{2} from both sides.

$$9x^{2}+2x-3-4x^{2}=0$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}+2x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=2 ab=5\left(-3\right)=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -15.

$$-1,15 -3,5$$

Calculate the sum for each pair.

$$-1+15=14 -3+5=2$$

The solution is the pair that gives sum 2.

$$a=-3 b=5$$

Rewrite 5x^{2}+2x-3 as \left(5x^{2}-3x\right)+\left(5x-3\right).

$$\left(5x^{2}-3x\right)+\left(5x-3\right)$$

Factor out x in 5x^{2}-3x.

$$x\left(5x-3\right)+5x-3$$

Factor out common term 5x-3 by using distributive property.

$$\left(5x-3\right)\left(x+1\right)$$

To find equation solutions, solve 5x-3=0 and x+1=0.

$$x=\frac{3}{5} x=-1$$

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}.

$$9x^{2}-6x+1=4\left(1-x\right)^{2}$$

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

$$9x^{2}-6x+1=4\left(1-2x+x^{2}\right)$$

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

$$9x^{2}-6x+1=4-8x+4x^{2}$$

Subtract 4 from both sides.

$$9x^{2}-6x+1-4=-8x+4x^{2}$$

Subtract 4 from 1 to get -3.

$$9x^{2}-6x-3=-8x+4x^{2}$$

Add 8x to both sides.

$$9x^{2}-6x-3+8x=4x^{2}$$

Combine -6x and 8x to get 2x.

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

Subtract 4x^{2} from both sides.

$$9x^{2}+2x-3-4x^{2}=0$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}+2x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 5\left(-3\right)}}{2\times 5}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 5\left(-3\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-2±\sqrt{4-20\left(-3\right)}}{2\times 5}$$

Multiply -20 times -3.

$$x=\frac{-2±\sqrt{4+60}}{2\times 5}$$

Add 4 to 60.

$$x=\frac{-2±\sqrt{64}}{2\times 5}$$

Take the square root of 64.

$$x=\frac{-2±8}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-2±8}{10}$$

Now solve the equation x=\frac{-2±8}{10} when ± is plus. Add -2 to 8.

$$x=\frac{6}{10}$$

Reduce the fraction \frac{6}{10} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-2±8}{10} when ± is minus. Subtract 8 from -2.

$$x=-\frac{10}{10}$$

Divide -10 by 10.

$$x=-1$$

The equation is now solved.

$$x=\frac{3}{5} x=-1$$

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}.

$$9x^{2}-6x+1=4\left(1-x\right)^{2}$$

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

$$9x^{2}-6x+1=4\left(1-2x+x^{2}\right)$$

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

$$9x^{2}-6x+1=4-8x+4x^{2}$$

Add 8x to both sides.

$$9x^{2}-6x+1+8x=4+4x^{2}$$

Combine -6x and 8x to get 2x.

$$9x^{2}+2x+1=4+4x^{2}$$

Subtract 4x^{2} from both sides.

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

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}+2x+1=4$$

Subtract 1 from both sides.

$$5x^{2}+2x=4-1$$

Subtract 1 from 4 to get 3.

$$5x^{2}+2x=3$$

Divide both sides by 5.

$$\frac{5x^{2}+2x}{5}=\frac{3}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{2}{5}x=\frac{3}{5}$$

Divide \frac{2}{5}, the coefficient of the x term, by 2 to get \frac{1}{5}. Then add the square of \frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=\frac{3}{5}+\frac{1}{25}$$

Add \frac{3}{5} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=\frac{16}{25}$$

Factor x^{2}+\frac{2}{5}x+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{5}\right)^{2}=\frac{16}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{5}\right)^{2}}=\sqrt{\frac{16}{25}}$$

Simplify.

$$x+\frac{1}{5}=\frac{4}{5} x+\frac{1}{5}=-\frac{4}{5}$$

Subtract \frac{1}{5} from both sides of the equation.

$$x=\frac{3}{5} x=-1$$