Solve 2x=sqrt{12-8x}

$$2 x = \sqrt { 12 - 8 x }$$

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

$$\left(2x\right)^{2}=\left(\sqrt{12-8x}\right)^{2}$$

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

$$2^{2}x^{2}=\left(\sqrt{12-8x}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}=\left(\sqrt{12-8x}\right)^{2}$$

Calculate \sqrt{12-8x} to the power of 2 and get 12-8x.

$$4x^{2}=12-8x$$

Subtract 12 from both sides.

$$4x^{2}-12=-8x$$

Add 8x to both sides.

$$4x^{2}-12+8x=0$$

Divide both sides by 4.

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

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

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

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

$$a+b=2 ab=1\left(-3\right)=-3$$

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. The only such pair is the system solution.

$$a=-1 b=3$$

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

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

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

$$x\left(x-1\right)+3\left(x-1\right)$$

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

$$\left(x-1\right)\left(x+3\right)$$

To find equation solutions, solve x-1=0 and x+3=0.

$$x=1 x=-3$$

Substitute 1 for x in the equation 2x=\sqrt{12-8x}.

$$2\times 1=\sqrt{12-8}$$

Simplify. The value x=1 satisfies the equation.

$$2=2$$

Substitute -3 for x in the equation 2x=\sqrt{12-8x}.

$$2\left(-3\right)=\sqrt{12-8\left(-3\right)}$$

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{12-8x} has a unique solution.

$$x=1$$