Solve 1+2sqrt{5x^2-5x+1}=6x

$$1+2 \sqrt{ 5 { x }^{ 2 } -5x+1 } =6x$$

Solve for x

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

Solution Steps

Subtract 1 from both sides of the equation.

$$2\sqrt{5x^{2}-5x+1}=6x-1$$

Square both sides of the equation.

$$\left(2\sqrt{5x^{2}-5x+1}\right)^{2}=\left(6x-1\right)^{2}$$

Expand \left(2\sqrt{5x^{2}-5x+1}\right)^{2}.

$$2^{2}\left(\sqrt{5x^{2}-5x+1}\right)^{2}=\left(6x-1\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{5x^{2}-5x+1}\right)^{2}=\left(6x-1\right)^{2}$$

Calculate \sqrt{5x^{2}-5x+1} to the power of 2 and get 5x^{2}-5x+1.

$$4\left(5x^{2}-5x+1\right)=\left(6x-1\right)^{2}$$

Use the distributive property to multiply 4 by 5x^{2}-5x+1.

$$20x^{2}-20x+4=\left(6x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6x-1\right)^{2}.

$$20x^{2}-20x+4=36x^{2}-12x+1$$

Subtract 36x^{2} from both sides.

$$20x^{2}-20x+4-36x^{2}=-12x+1$$

Combine 20x^{2} and -36x^{2} to get -16x^{2}.

$$-16x^{2}-20x+4=-12x+1$$

Add 12x to both sides.

$$-16x^{2}-20x+4+12x=1$$

Combine -20x and 12x to get -8x.

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

Subtract 1 from both sides.

$$-16x^{2}-8x+4-1=0$$

Subtract 1 from 4 to get 3.

$$-16x^{2}-8x+3=0$$

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.

$$a+b=-8 ab=-16\times 3=-48$$

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 -8.

$$a=4 b=-12$$

Rewrite -16x^{2}-8x+3 as \left(-16x^{2}+4x\right)+\left(-12x+3\right).

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

Factor out 4x in the first and 3 in the second group.

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

Factor out common term -4x+1 by using distributive property.

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

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

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

Substitute \frac{1}{4} for x in the equation 1+2\sqrt{5x^{2}-5x+1}=6x.

$$1+2\sqrt{5\times \left(\frac{1}{4}\right)^{2}-5\times \frac{1}{4}+1}=6\times \frac{1}{4}$$

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

$$\frac{3}{2}=\frac{3}{2}$$

Substitute -\frac{3}{4} for x in the equation 1+2\sqrt{5x^{2}-5x+1}=6x.

$$1+2\sqrt{5\left(-\frac{3}{4}\right)^{2}-5\left(-\frac{3}{4}\right)+1}=6\left(-\frac{3}{4}\right)$$

Simplify. The value x=-\frac{3}{4} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{13}{2}=-\frac{9}{2}$$

Equation 2\sqrt{5x^{2}-5x+1}=6x-1 has a unique solution.

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