Solve x+2=2sqrt{x+5}

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

Solve for x

$$x=4$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply 4 by x+5.

$$x^{2}+4x+4=4x+20$$

Subtract 4x from both sides.

$$x^{2}+4x+4-4x=20$$

Combine 4x and -4x to get 0.

$$x^{2}+4=20$$

Subtract 20 from both sides.

$$x^{2}+4-20=0$$

Subtract 20 from 4 to get -16.

$$x^{2}-16=0$$

Consider x^{2}-16. Rewrite x^{2}-16 as x^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-4\right)\left(x+4\right)=0$$

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

$$x=4 x=-4$$

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

$$4+2=2\sqrt{4+5}$$

Simplify. The value x=4 satisfies the equation.

$$6=6$$

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

$$-4+2=2\sqrt{-4+5}$$

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

$$-2=2$$

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

$$x=4$$