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

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

Solve for x

$$x=2$$

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

Solution Steps

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3x.

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

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

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

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

$$3x^{3}-12x^{2}+12x=\left(x-2\right)^{2}$$

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

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

Subtract x^{2} from both sides.

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

Combine -12x^{2} and -x^{2} to get -13x^{2}.

$$3x^{3}-13x^{2}+12x=-4x+4$$

Add 4x to both sides.

$$3x^{3}-13x^{2}+12x+4x=4$$

Combine 12x and 4x to get 16x.

$$3x^{3}-13x^{2}+16x=4$$

Subtract 4 from both sides.

$$3x^{3}-13x^{2}+16x-4=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -4 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{3}-13x^{2}+16x-4 by x-2 to get 3x^{2}-7x+2. Solve the equation where the result equals to 0.

$$3x^{2}-7x+2=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 3 for a, -7 for b, and 2 for c in the quadratic formula.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 3\times 2}}{2\times 3}$$

Do the calculations.

$$x=\frac{7±5}{6}$$

Solve the equation 3x^{2}-7x+2=0 when ± is plus and when ± is minus.

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

List all found solutions.

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