Solve sqrt{left(2x-1right)^2}left(3-xright)=left(1-2xright)sqrt{x-x}

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

Solve for x

$$x=3$$

$$x=\frac{1}{2}=0.5$$

Solution Steps

Square both sides of the equation.

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

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

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

Use the distributive property to multiply \sqrt{4x^{2}-4x+1} by 3-x.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{4x^{2}-4x+1}-\sqrt{4x^{2}-4x+1}x\right)^{2}.

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

Multiply \sqrt{4x^{2}-4x+1} and \sqrt{4x^{2}-4x+1} to get \left(\sqrt{4x^{2}-4x+1}\right)^{2}.

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

Calculate \sqrt{4x^{2}-4x+1} to the power of 2 and get 4x^{2}-4x+1.

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

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

$$36x^{2}-36x+9-6\left(\sqrt{4x^{2}-4x+1}\right)^{2}x+\left(\sqrt{4x^{2}-4x+1}\right)^{2}x^{2}=\left(\left(1-2x\right)\sqrt{x-x}\right)^{2}$$

Calculate \sqrt{4x^{2}-4x+1} to the power of 2 and get 4x^{2}-4x+1.

$$36x^{2}-36x+9-6\left(4x^{2}-4x+1\right)x+\left(\sqrt{4x^{2}-4x+1}\right)^{2}x^{2}=\left(\left(1-2x\right)\sqrt{x-x}\right)^{2}$$

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

$$36x^{2}-36x+9+\left(-24x^{2}+24x-6\right)x+\left(\sqrt{4x^{2}-4x+1}\right)^{2}x^{2}=\left(\left(1-2x\right)\sqrt{x-x}\right)^{2}$$

Use the distributive property to multiply -24x^{2}+24x-6 by x.

$$36x^{2}-36x+9-24x^{3}+24x^{2}-6x+\left(\sqrt{4x^{2}-4x+1}\right)^{2}x^{2}=\left(\left(1-2x\right)\sqrt{x-x}\right)^{2}$$

Combine 36x^{2} and 24x^{2} to get 60x^{2}.

$$60x^{2}-36x+9-24x^{3}-6x+\left(\sqrt{4x^{2}-4x+1}\right)^{2}x^{2}=\left(\left(1-2x\right)\sqrt{x-x}\right)^{2}$$

Combine -36x and -6x to get -42x.

$$60x^{2}-42x+9-24x^{3}+\left(\sqrt{4x^{2}-4x+1}\right)^{2}x^{2}=\left(\left(1-2x\right)\sqrt{x-x}\right)^{2}$$

Calculate \sqrt{4x^{2}-4x+1} to the power of 2 and get 4x^{2}-4x+1.

$$60x^{2}-42x+9-24x^{3}+\left(4x^{2}-4x+1\right)x^{2}=\left(\left(1-2x\right)\sqrt{x-x}\right)^{2}$$

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

$$60x^{2}-42x+9-24x^{3}+4x^{4}-4x^{3}+x^{2}=\left(\left(1-2x\right)\sqrt{x-x}\right)^{2}$$

Combine -24x^{3} and -4x^{3} to get -28x^{3}.

$$60x^{2}-42x+9-28x^{3}+4x^{4}+x^{2}=\left(\left(1-2x\right)\sqrt{x-x}\right)^{2}$$

Combine 60x^{2} and x^{2} to get 61x^{2}.

$$61x^{2}-42x+9-28x^{3}+4x^{4}=\left(\left(1-2x\right)\sqrt{x-x}\right)^{2}$$

Combine x and -x to get 0.

$$61x^{2}-42x+9-28x^{3}+4x^{4}=\left(\left(1-2x\right)\sqrt{0}\right)^{2}$$

Calculate the square root of 0 and get 0.

$$61x^{2}-42x+9-28x^{3}+4x^{4}=\left(\left(1-2x\right)\times 0\right)^{2}$$

Anything times zero gives zero.

$$61x^{2}-42x+9-28x^{3}+4x^{4}=0^{2}$$

Calculate 0 to the power of 2 and get 0.

$$61x^{2}-42x+9-28x^{3}+4x^{4}=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$4x^{4}-28x^{3}+61x^{2}-42x+9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 9 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±\frac{9}{4},±\frac{9}{2},±9,±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{4}-28x^{3}+61x^{2}-42x+9 by x-3 to get 4x^{3}-16x^{2}+13x-3. Solve the equation where the result equals to 0.

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

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

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

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}-16x^{2}+13x-3 by x-3 to get 4x^{2}-4x+1. Solve the equation where the result equals to 0.

$$4x^{2}-4x+1=0$$

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 4 for a, -4 for b, and 1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{4±0}{8}$$

Solutions are the same.

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

List all found solutions.

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

Substitute 3 for x in the equation \sqrt{\left(2x-1\right)^{2}}\left(3-x\right)=\left(1-2x\right)\sqrt{x-x}.

$$\sqrt{\left(2\times 3-1\right)^{2}}\left(3-3\right)=\left(1-2\times 3\right)\sqrt{3-3}$$

Simplify. The value x=3 satisfies the equation.

$$0=0$$

Substitute \frac{1}{2} for x in the equation \sqrt{\left(2x-1\right)^{2}}\left(3-x\right)=\left(1-2x\right)\sqrt{x-x}.

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

Simplify. The value x=\frac{1}{2} satisfies the equation.

$$0=0$$

List all solutions of \left(3-x\right)\sqrt{\left(2x-1\right)^{2}}=\sqrt{x-x}\left(1-2x\right).

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