Solve frac{sqrt{6x+4}-1}{4}=x

$$\frac { \sqrt { 6 x + 4 } - 1 } { 4 } = x$$

Solve for x

$$x=\frac{3}{8}=0.375$$

Solution Steps

Subtract x from both sides.

$$\frac{\sqrt{6x+4}-1}{4}-x=0$$

Multiply both sides of the equation by 4.

$$\sqrt{6x+4}-1-4x=0$$

Reorder the terms.

$$-4x+\sqrt{6x+4}-1=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$-4x+\sqrt{6x+4}=1$$

Subtract -4x from both sides of the equation.

$$\sqrt{6x+4}=1+4x$$

Square both sides of the equation.

$$\left(\sqrt{6x+4}\right)^{2}=\left(1+4x\right)^{2}$$

Calculate \sqrt{6x+4} to the power of 2 and get 6x+4.

$$6x+4=\left(1+4x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+4x\right)^{2}.

$$6x+4=1+8x+16x^{2}$$

Subtract 8x from both sides.

$$6x+4-8x=1+16x^{2}$$

Combine 6x and -8x to get -2x.

$$-2x+4=1+16x^{2}$$

Subtract 16x^{2} from both sides.

$$-2x+4-16x^{2}=1$$

Subtract 1 from both sides.

$$-2x+4-16x^{2}-1=0$$

Subtract 1 from 4 to get 3.

$$-2x+3-16x^{2}=0$$

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

$$-16x^{2}-2x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -16x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-16\times 3=-48$$

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 -48.

$$1,-48 2,-24 3,-16 4,-12 6,-8$$

Calculate the sum for each pair.

$$1-48=-47 2-24=-22 3-16=-13 4-12=-8 6-8=-2$$

The solution is the pair that gives sum -2.

$$a=6 b=-8$$

Rewrite -16x^{2}-2x+3 as \left(-16x^{2}+6x\right)+\left(-8x+3\right).

$$\left(-16x^{2}+6x\right)+\left(-8x+3\right)$$

Factor out 2x in -16x^{2}+6x.

$$2x\left(-8x+3\right)-8x+3$$

Factor out common term -8x+3 by using distributive property.

$$\left(-8x+3\right)\left(2x+1\right)$$

To find equation solutions, solve -8x+3=0 and 2x+1=0.

$$x=\frac{3}{8} x=-\frac{1}{2}$$

Substitute \frac{3}{8} for x in the equation \frac{\sqrt{6x+4}-1}{4}=x.

$$\frac{\sqrt{6\times \frac{3}{8}+4}-1}{4}=\frac{3}{8}$$

Simplify. The value x=\frac{3}{8} satisfies the equation.

$$\frac{3}{8}=\frac{3}{8}$$

Substitute -\frac{1}{2} for x in the equation \frac{\sqrt{6x+4}-1}{4}=x.

$$\frac{\sqrt{6\left(-\frac{1}{2}\right)+4}-1}{4}=-\frac{1}{2}$$

Simplify. The value x=-\frac{1}{2} does not satisfy the equation.

$$0=-\frac{1}{2}$$

Equation \sqrt{6x+4}=4x+1 has a unique solution.

$$x=\frac{3}{8}$$