Solve 2x=sqrt{((1-x)(6-x))}

$$2x= \sqrt{ ((1-x)(6-x)) }$$

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

Solution Steps

Square both sides of the equation.

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

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Use the distributive property to multiply 1-x by 6-x and combine like terms.

$$4x^{2}=\left(\sqrt{6-7x+x^{2}}\right)^{2}$$

Calculate \sqrt{6-7x+x^{2}} to the power of 2 and get 6-7x+x^{2}.

$$4x^{2}=6-7x+x^{2}$$

Subtract 6 from both sides.

$$4x^{2}-6=-7x+x^{2}$$

Add 7x to both sides.

$$4x^{2}-6+7x=x^{2}$$

Subtract x^{2} from both sides.

$$4x^{2}-6+7x-x^{2}=0$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-6+7x=0$$

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

$$3x^{2}+7x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=7 ab=3\left(-6\right)=-18$$

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 -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 7.

$$a=-2 b=9$$

Rewrite 3x^{2}+7x-6 as \left(3x^{2}-2x\right)+\left(9x-6\right).

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

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

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

Factor out common term 3x-2 by using distributive property.

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

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

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

Substitute \frac{2}{3} for x in the equation 2x=\sqrt{\left(1-x\right)\left(6-x\right)}.

$$2\times \frac{2}{3}=\sqrt{\left(1-\frac{2}{3}\right)\left(6-\frac{2}{3}\right)}$$

Simplify. The value x=\frac{2}{3} satisfies the equation.

$$\frac{4}{3}=\frac{4}{3}$$

Substitute -3 for x in the equation 2x=\sqrt{\left(1-x\right)\left(6-x\right)}.

$$2\left(-3\right)=\sqrt{\left(1-\left(-3\right)\right)\left(6-\left(-3\right)\right)}$$

Simplify. The value x=-3 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-6=6$$

Equation 2x=\sqrt{\left(1-x\right)\left(6-x\right)} has a unique solution.

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