Solve sqrt{16x-3}-4x=0

$$\sqrt { 16 x - 3 } - 4 x = 0$$

Solve for x

$$x=\frac{3}{4}=0.75$$

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

Solution Steps

Subtract -4x from both sides of the equation.

$$\sqrt{16x-3}=4x$$

Square both sides of the equation.

$$\left(\sqrt{16x-3}\right)^{2}=\left(4x\right)^{2}$$

Calculate \sqrt{16x-3} to the power of 2 and get 16x-3.

$$16x-3=\left(4x\right)^{2}$$

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

$$16x-3=4^{2}x^{2}$$

Calculate 4 to the power of 2 and get 16.

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

Subtract 16x^{2} from both sides.

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

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

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

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.

$$a+b=16 ab=-16\left(-3\right)=48$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both 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.

$$1+48=49 2+24=26 3+16=19 4+12=16 6+8=14$$

The solution is the pair that gives sum 16.

$$a=12 b=4$$

Rewrite -16x^{2}+16x-3 as \left(-16x^{2}+12x\right)+\left(4x-3\right).

$$\left(-16x^{2}+12x\right)+\left(4x-3\right)$$

Factor out -4x in -16x^{2}+12x.

$$-4x\left(4x-3\right)+4x-3$$

Factor out common term 4x-3 by using distributive property.

$$\left(4x-3\right)\left(-4x+1\right)$$

To find equation solutions, solve 4x-3=0 and -4x+1=0.

$$x=\frac{3}{4} x=\frac{1}{4}$$

Substitute \frac{3}{4} for x in the equation \sqrt{16x-3}-4x=0.

$$\sqrt{16\times \frac{3}{4}-3}-4\times \frac{3}{4}=0$$

Simplify. The value x=\frac{3}{4} satisfies the equation.

$$0=0$$

Substitute \frac{1}{4} for x in the equation \sqrt{16x-3}-4x=0.

$$\sqrt{16\times \frac{1}{4}-3}-4\times \frac{1}{4}=0$$

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

$$0=0$$

List all solutions of \sqrt{16x-3}=4x.

$$x=\frac{3}{4} x=\frac{1}{4}$$