Solve 4x=sqrt{12x-2}

$$4 x = \sqrt { 12 x - 2 }$$

Solve for x

$$x=\frac{1}{2}=0.5$$

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

Solution Steps

Square both sides of the equation.

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

Expand \left(4x\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$16x^{2}=\left(\sqrt{12x-2}\right)^{2}$$

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

$$16x^{2}=12x-2$$

Subtract 12x from both sides.

$$16x^{2}-12x=-2$$

Add 2 to both sides.

$$16x^{2}-12x+2=0$$

Divide both sides by 2.

$$8x^{2}-6x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-6 ab=8\times 1=8$$

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 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -6.

$$a=-4 b=-2$$

Rewrite 8x^{2}-6x+1 as \left(8x^{2}-4x\right)+\left(-2x+1\right).

$$\left(8x^{2}-4x\right)+\left(-2x+1\right)$$

Factor out 4x in the first and -1 in the second group.

$$4x\left(2x-1\right)-\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(4x-1\right)$$

To find equation solutions, solve 2x-1=0 and 4x-1=0.

$$x=\frac{1}{2} x=\frac{1}{4}$$

Substitute \frac{1}{2} for x in the equation 4x=\sqrt{12x-2}.

$$4\times \frac{1}{2}=\sqrt{12\times \frac{1}{2}-2}$$

Simplify. The value x=\frac{1}{2} satisfies the equation.

$$2=2$$

Substitute \frac{1}{4} for x in the equation 4x=\sqrt{12x-2}.

$$4\times \frac{1}{4}=\sqrt{12\times \frac{1}{4}-2}$$

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

$$1=1$$

List all solutions of 4x=\sqrt{12x-2}.

$$x=\frac{1}{2} x=\frac{1}{4}$$