Solve x-6=sqrt{3x}

$$x - 6 = \sqrt { 3 x }$$

Solve for x

$$x=12$$

Solution Steps

Square both sides of the equation.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$x^{2}-12x+36=\left(\sqrt{3x}\right)^{2}$$

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

$$x^{2}-12x+36=3x$$

Subtract 3x from both sides.

$$x^{2}-12x+36-3x=0$$

Combine -12x and -3x to get -15x.

$$x^{2}-15x+36=0$$

To solve the equation, factor x^{2}-15x+36 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-15 ab=36$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -15.

$$a=-12 b=-3$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-12\right)\left(x-3\right)$$

To find equation solutions, solve x-12=0 and x-3=0.

$$x=12 x=3$$

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

$$12-6=\sqrt{3\times 12}$$

Simplify. The value x=12 satisfies the equation.

$$6=6$$

Substitute 3 for x in the equation x-6=\sqrt{3x}.

$$3-6=\sqrt{3\times 3}$$

Simplify. The value x=3 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-3=3$$

Equation x-6=\sqrt{3x} has a unique solution.

$$x=12$$