Solve sqrt{4x}=x+1

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

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract x^{2} from both sides.

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

Subtract 2x from both sides.

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

Combine 4x and -2x to get 2x.

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

Subtract 1 from both sides.

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

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

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

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

$$a+b=2 ab=-\left(-1\right)=1$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=1$$

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

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

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

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

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

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

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

$$x=1 x=1$$

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

$$\sqrt{4\times 1}=1+1$$

Simplify. The value x=1 satisfies the equation.

$$2=2$$

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

$$\sqrt{4\times 1}=1+1$$

Simplify. The value x=1 satisfies the equation.

$$2=2$$

List all solutions of \sqrt{4x}=x+1.

$$x=1 x=1$$