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

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

Solve for x

$$x=-\frac{1}{5}=-0.2$$

$$x=\frac{5}{7}\approx 0.714285714$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9=4\left(3x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

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

Use the distributive property to multiply 4 by 9x^{2}-6x+1.

$$x^{2}-6x+9=36x^{2}-24x+4$$

Subtract 36x^{2} from both sides.

$$x^{2}-6x+9-36x^{2}=-24x+4$$

Combine x^{2} and -36x^{2} to get -35x^{2}.

$$-35x^{2}-6x+9=-24x+4$$

Add 24x to both sides.

$$-35x^{2}-6x+9+24x=4$$

Combine -6x and 24x to get 18x.

$$-35x^{2}+18x+9=4$$

Subtract 4 from both sides.

$$-35x^{2}+18x+9-4=0$$

Subtract 4 from 9 to get 5.

$$-35x^{2}+18x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -35x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=18 ab=-35\times 5=-175$$

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 -175.

$$-1,175 -5,35 -7,25$$

Calculate the sum for each pair.

$$-1+175=174 -5+35=30 -7+25=18$$

The solution is the pair that gives sum 18.

$$a=25 b=-7$$

Rewrite -35x^{2}+18x+5 as \left(-35x^{2}+25x\right)+\left(-7x+5\right).

$$\left(-35x^{2}+25x\right)+\left(-7x+5\right)$$

Factor out 5x in -35x^{2}+25x.

$$5x\left(-7x+5\right)-7x+5$$

Factor out common term -7x+5 by using distributive property.

$$\left(-7x+5\right)\left(5x+1\right)$$

To find equation solutions, solve -7x+5=0 and 5x+1=0.

$$x=\frac{5}{7} x=-\frac{1}{5}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9=4\left(3x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

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

Use the distributive property to multiply 4 by 9x^{2}-6x+1.

$$x^{2}-6x+9=36x^{2}-24x+4$$

Subtract 36x^{2} from both sides.

$$x^{2}-6x+9-36x^{2}=-24x+4$$

Combine x^{2} and -36x^{2} to get -35x^{2}.

$$-35x^{2}-6x+9=-24x+4$$

Add 24x to both sides.

$$-35x^{2}-6x+9+24x=4$$

Combine -6x and 24x to get 18x.

$$-35x^{2}+18x+9=4$$

Subtract 4 from both sides.

$$-35x^{2}+18x+9-4=0$$

Subtract 4 from 9 to get 5.

$$-35x^{2}+18x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -35 for a, 18 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\left(-35\right)\times 5}}{2\left(-35\right)}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\left(-35\right)\times 5}}{2\left(-35\right)}$$

Multiply -4 times -35.

$$x=\frac{-18±\sqrt{324+140\times 5}}{2\left(-35\right)}$$

Multiply 140 times 5.

$$x=\frac{-18±\sqrt{324+700}}{2\left(-35\right)}$$

Add 324 to 700.

$$x=\frac{-18±\sqrt{1024}}{2\left(-35\right)}$$

Take the square root of 1024.

$$x=\frac{-18±32}{2\left(-35\right)}$$

Multiply 2 times -35.

$$x=\frac{-18±32}{-70}$$

Now solve the equation x=\frac{-18±32}{-70} when ± is plus. Add -18 to 32.

$$x=\frac{14}{-70}$$

Reduce the fraction \frac{14}{-70} to lowest terms by extracting and canceling out 14.

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

Now solve the equation x=\frac{-18±32}{-70} when ± is minus. Subtract 32 from -18.

$$x=-\frac{50}{-70}$$

Reduce the fraction \frac{-50}{-70} to lowest terms by extracting and canceling out 10.

$$x=\frac{5}{7}$$

The equation is now solved.

$$x=-\frac{1}{5} x=\frac{5}{7}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9=4\left(3x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

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

Use the distributive property to multiply 4 by 9x^{2}-6x+1.

$$x^{2}-6x+9=36x^{2}-24x+4$$

Subtract 36x^{2} from both sides.

$$x^{2}-6x+9-36x^{2}=-24x+4$$

Combine x^{2} and -36x^{2} to get -35x^{2}.

$$-35x^{2}-6x+9=-24x+4$$

Add 24x to both sides.

$$-35x^{2}-6x+9+24x=4$$

Combine -6x and 24x to get 18x.

$$-35x^{2}+18x+9=4$$

Subtract 9 from both sides.

$$-35x^{2}+18x=4-9$$

Subtract 9 from 4 to get -5.

$$-35x^{2}+18x=-5$$

Divide both sides by -35.

$$\frac{-35x^{2}+18x}{-35}=-\frac{5}{-35}$$

Dividing by -35 undoes the multiplication by -35.

$$x^{2}+\frac{18}{-35}x=-\frac{5}{-35}$$

Divide 18 by -35.

$$x^{2}-\frac{18}{35}x=-\frac{5}{-35}$$

Reduce the fraction \frac{-5}{-35} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{18}{35}x=\frac{1}{7}$$

Divide -\frac{18}{35}, the coefficient of the x term, by 2 to get -\frac{9}{35}. Then add the square of -\frac{9}{35} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{18}{35}x+\left(-\frac{9}{35}\right)^{2}=\frac{1}{7}+\left(-\frac{9}{35}\right)^{2}$$

Square -\frac{9}{35} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{18}{35}x+\frac{81}{1225}=\frac{1}{7}+\frac{81}{1225}$$

Add \frac{1}{7} to \frac{81}{1225} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{18}{35}x+\frac{81}{1225}=\frac{256}{1225}$$

Factor x^{2}-\frac{18}{35}x+\frac{81}{1225}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{35}\right)^{2}=\frac{256}{1225}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{35}\right)^{2}}=\sqrt{\frac{256}{1225}}$$

Simplify.

$$x-\frac{9}{35}=\frac{16}{35} x-\frac{9}{35}=-\frac{16}{35}$$

Add \frac{9}{35} to both sides of the equation.

$$x=\frac{5}{7} x=-\frac{1}{5}$$