Solve sqrt{3(2-x)}=2x-(4-x)

$$\sqrt { 3 ( 2 - x ) } = 2 x - ( 4 - x )$$

Solve for x

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Solution Steps

Square both sides of the equation.

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

Use the distributive property to multiply 3 by 2-x.

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

Calculate \sqrt{6-3x} to the power of 2 and get 6-3x.

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

To find the opposite of 4-x, find the opposite of each term.

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

The opposite of -x is x.

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

Combine 2x and x to get 3x.

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

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

$$6-3x=9x^{2}-24x+16$$

Subtract 9x^{2} from both sides.

$$6-3x-9x^{2}=-24x+16$$

Add 24x to both sides.

$$6-3x-9x^{2}+24x=16$$

Combine -3x and 24x to get 21x.

$$6+21x-9x^{2}=16$$

Subtract 16 from both sides.

$$6+21x-9x^{2}-16=0$$

Subtract 16 from 6 to get -10.

$$-10+21x-9x^{2}=0$$

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

$$-9x^{2}+21x-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -9x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=21 ab=-9\left(-10\right)=90$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 90.

$$1,90 2,45 3,30 5,18 6,15 9,10$$

Calculate the sum for each pair.

$$1+90=91 2+45=47 3+30=33 5+18=23 6+15=21 9+10=19$$

The solution is the pair that gives sum 21.

$$a=15 b=6$$

Rewrite -9x^{2}+21x-10 as \left(-9x^{2}+15x\right)+\left(6x-10\right).

$$\left(-9x^{2}+15x\right)+\left(6x-10\right)$$

Factor out -3x in the first and 2 in the second group.

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

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

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

To find equation solutions, solve 3x-5=0 and -3x+2=0.

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

Substitute \frac{5}{3} for x in the equation \sqrt{3\left(2-x\right)}=2x-\left(4-x\right).

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

Simplify. The value x=\frac{5}{3} satisfies the equation.

$$1=1$$

Substitute \frac{2}{3} for x in the equation \sqrt{3\left(2-x\right)}=2x-\left(4-x\right).

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

Simplify. The value x=\frac{2}{3} does not satisfy the equation because the left and the right hand side have opposite signs.

$$2=-2$$

Equation \sqrt{3\left(2-x\right)}=-\left(4-x\right)+2x has a unique solution.

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