Solve sqrt{3+x}=1/2x+2/2

$$\sqrt{ 3+x } = \frac{ 1 }{ 2 } x+ \frac{ 2 }{ 2 }$$

Solve for x

$$x=2\sqrt{2}\approx 2.828427125$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3+x}\right)^{2}=\left(\frac{1}{2}x+\frac{2}{2}\right)^{2}$$

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

$$3+x=\left(\frac{1}{2}x+\frac{2}{2}\right)^{2}$$

Divide 2 by 2 to get 1.

$$3+x=\left(\frac{1}{2}x+1\right)^{2}$$

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

$$3+x=\frac{1}{4}x^{2}+x+1$$

Subtract \frac{1}{4}x^{2} from both sides.

$$3+x-\frac{1}{4}x^{2}=x+1$$

Subtract x from both sides.

$$3+x-\frac{1}{4}x^{2}-x=1$$

Combine x and -x to get 0.

$$3-\frac{1}{4}x^{2}=1$$

Subtract 3 from both sides.

$$-\frac{1}{4}x^{2}=1-3$$

Subtract 3 from 1 to get -2.

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

Multiply both sides by -4, the reciprocal of -\frac{1}{4}.

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

Multiply -2 and -4 to get 8.

$$x^{2}=8$$

Take the square root of both sides of the equation.

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

Substitute 2\sqrt{2} for x in the equation \sqrt{3+x}=\frac{1}{2}x+\frac{2}{2}.

$$\sqrt{3+2\sqrt{2}}=\frac{1}{2}\times 2\sqrt{2}+\frac{2}{2}$$

Simplify. The value x=2\sqrt{2} satisfies the equation.

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

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

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

Simplify. The value x=-2\sqrt{2} does not satisfy the equation because the left and the right hand side have opposite signs.

$$2^{\frac{1}{2}}-1=-2^{\frac{1}{2}}+1$$

Equation \sqrt{x+3}=\frac{x}{2}+1 has a unique solution.

$$x=2\sqrt{2}$$