Solve 2x=sqrt{1-3x}

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

Solve for x

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

Solution Steps

Square both sides of the equation.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

Calculate \sqrt{1-3x} to the power of 2 and get 1-3x.

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

Subtract 1 from both sides.

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

Add 3x to both sides.

$$4x^{2}-1+3x=0$$

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

$$4x^{2}+3x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=3 ab=4\left(-1\right)=-4$$

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 -4.

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

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=-1 b=4$$

Rewrite 4x^{2}+3x-1 as \left(4x^{2}-x\right)+\left(4x-1\right).

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

Factor out x in 4x^{2}-x.

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

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

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

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

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

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

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

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

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

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

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

Simplify. The value x=-1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-2=2$$

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

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