Solve sqrt{x}=-2x+1

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

Solve for x

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

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract 4x^{2} from both sides.

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

Add 4x to both sides.

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

Combine x and 4x to get 5x.

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

Subtract 1 from both sides.

$$5x-4x^{2}-1=0$$

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

$$-4x^{2}+5x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=5 ab=-4\left(-1\right)=4$$

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 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 5.

$$a=4 b=1$$

Rewrite -4x^{2}+5x-1 as \left(-4x^{2}+4x\right)+\left(x-1\right).

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

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

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

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

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

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

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

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

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

Simplify. The value x=1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$1=-1$$

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

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

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

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

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

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