Solve 1-2x+sqrt{5x-1}=0

$$1 - 2 x + \sqrt { 5 x - 1 } = 0$$

Solve for x

$$x=2$$

Solution Steps

Subtract 1-2x from both sides of the equation.

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

To find the opposite of 1-2x, find the opposite of each term.

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

The opposite of -2x is 2x.

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

Square both sides of the equation.

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

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

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

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

$$5x-1=1-4x+4x^{2}$$

Subtract 1 from both sides.

$$5x-1-1=-4x+4x^{2}$$

Subtract 1 from -1 to get -2.

$$5x-2=-4x+4x^{2}$$

Add 4x to both sides.

$$5x-2+4x=4x^{2}$$

Combine 5x and 4x to get 9x.

$$9x-2=4x^{2}$$

Subtract 4x^{2} from both sides.

$$9x-2-4x^{2}=0$$

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

$$-4x^{2}+9x-2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=9 ab=-4\left(-2\right)=8$$

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 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 9.

$$a=8 b=1$$

Rewrite -4x^{2}+9x-2 as \left(-4x^{2}+8x\right)+\left(x-2\right).

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

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

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

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

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

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

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

Substitute 2 for x in the equation 1-2x+\sqrt{5x-1}=0.

$$1-2\times 2+\sqrt{5\times 2-1}=0$$

Simplify. The value x=2 satisfies the equation.

$$0=0$$

Substitute \frac{1}{4} for x in the equation 1-2x+\sqrt{5x-1}=0.

$$1-2\times \frac{1}{4}+\sqrt{5\times \frac{1}{4}-1}=0$$

Simplify. The value x=\frac{1}{4} does not satisfy the equation.

$$1=0$$

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

$$x=2$$