Solve 2x=sqrt{4x+24}

$$2 x = \sqrt { 4 x + 24 }$$

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

$$\left(2x\right)^{2}=\left(\sqrt{4x+24}\right)^{2}$$

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

$$2^{2}x^{2}=\left(\sqrt{4x+24}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}=\left(\sqrt{4x+24}\right)^{2}$$

Calculate \sqrt{4x+24} to the power of 2 and get 4x+24.

$$4x^{2}=4x+24$$

Subtract 4x from both sides.

$$4x^{2}-4x=24$$

Subtract 24 from both sides.

$$4x^{2}-4x-24=0$$

Divide both sides by 4.

$$x^{2}-x-6=0$$

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

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

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

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

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

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

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

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

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

$$x=3 x=-2$$

Substitute 3 for x in the equation 2x=\sqrt{4x+24}.

$$2\times 3=\sqrt{4\times 3+24}$$

Simplify. The value x=3 satisfies the equation.

$$6=6$$

Substitute -2 for x in the equation 2x=\sqrt{4x+24}.

$$2\left(-2\right)=\sqrt{4\left(-2\right)+24}$$

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

$$-4=4$$

Equation 2x=\sqrt{4x+24} has a unique solution.

$$x=3$$