Solve 6x=sqrt{24+12x}

$$6x= \sqrt{ 24+12x }$$

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

$$\left(6x\right)^{2}=\left(\sqrt{24+12x}\right)^{2}$$

Expand \left(6x\right)^{2}.

$$6^{2}x^{2}=\left(\sqrt{24+12x}\right)^{2}$$

Calculate 6 to the power of 2 and get 36.

$$36x^{2}=\left(\sqrt{24+12x}\right)^{2}$$

Calculate \sqrt{24+12x} to the power of 2 and get 24+12x.

$$36x^{2}=24+12x$$

Subtract 24 from both sides.

$$36x^{2}-24=12x$$

Subtract 12x from both sides.

$$36x^{2}-24-12x=0$$

Divide both sides by 12.

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

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

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

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

$$a+b=-1 ab=3\left(-2\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

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

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

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

$$3x\left(x-1\right)+2\left(x-1\right)$$

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

$$\left(x-1\right)\left(3x+2\right)$$

To find equation solutions, solve x-1=0 and 3x+2=0.

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

Substitute 1 for x in the equation 6x=\sqrt{24+12x}.

$$6\times 1=\sqrt{24+12\times 1}$$

Simplify. The value x=1 satisfies the equation.

$$6=6$$

Substitute -\frac{2}{3} for x in the equation 6x=\sqrt{24+12x}.

$$6\left(-\frac{2}{3}\right)=\sqrt{24+12\left(-\frac{2}{3}\right)}$$

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

$$-4=4$$

Equation 6x=\sqrt{12x+24} has a unique solution.

$$x=1$$