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

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

Solve for x

$$x=4$$

Solution Steps

Square both sides of the equation.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

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

$$4x+20=\left(x+2\right)^{2}$$

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

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

Subtract x^{2} from both sides.

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

Subtract 4x from both sides.

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

Combine 4x and -4x to get 0.

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

Subtract 20 from both sides.

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

Subtract 20 from 4 to get -16.

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

Divide both sides by -1.

$$x^{2}=\frac{-16}{-1}$$

Fraction \frac{-16}{-1} can be simplified to 16 by removing the negative sign from both the numerator and the denominator.

$$x^{2}=16$$

Take the square root of both sides of the equation.

$$x=4 x=-4$$

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

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

Simplify. The value x=4 satisfies the equation.

$$6=6$$

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

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

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

$$2=-2$$

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

$$x=4$$