Solve 2x+1=sqrt{3-3x}

$$2x+1= \sqrt{ 3-3x }$$

Solve for x

$$x=\frac{1}{4}=0.25$$

Solution Steps

Square both sides of the equation.

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

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

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

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

$$4x^{2}+4x+1=3-3x$$

Subtract 3 from both sides.

$$4x^{2}+4x+1-3=-3x$$

Subtract 3 from 1 to get -2.

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

Add 3x to both sides.

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

Combine 4x and 3x to get 7x.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=7 ab=4\left(-2\right)=-8$$

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 -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 7.

$$a=-1 b=8$$

Rewrite 4x^{2}+7x-2 as \left(4x^{2}-x\right)+\left(8x-2\right).

$$\left(4x^{2}-x\right)+\left(8x-2\right)$$

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

$$x\left(4x-1\right)+2\left(4x-1\right)$$

Factor out common term 4x-1 by using distributive property.

$$\left(4x-1\right)\left(x+2\right)$$

To find equation solutions, solve 4x-1=0 and x+2=0.

$$x=\frac{1}{4} x=-2$$

Substitute \frac{1}{4} for x in the equation 2x+1=\sqrt{3-3x}.

$$2\times \frac{1}{4}+1=\sqrt{3-3\times \frac{1}{4}}$$

Simplify. The value x=\frac{1}{4} satisfies the equation.

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

Substitute -2 for x in the equation 2x+1=\sqrt{3-3x}.

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

Simplify. The value x=-2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-3=3$$

Equation 2x+1=\sqrt{3-3x} has a unique solution.

$$x=\frac{1}{4}$$