Solve x+1=sqrt{2x+5}

$$x + 1 = \sqrt { 2 x + 5 }$$

Solve for x

$$x=2$$

Solution Steps

Square both sides of the equation.

$$\left(x+1\right)^{2}=\left(\sqrt{2x+5}\right)^{2}$$

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

$$x^{2}+2x+1=\left(\sqrt{2x+5}\right)^{2}$$

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

$$x^{2}+2x+1=2x+5$$

Subtract 2x from both sides.

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

Combine 2x and -2x to get 0.

$$x^{2}+1=5$$

Subtract 5 from both sides.

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

Subtract 5 from 1 to get -4.

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

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

$$\left(x-2\right)\left(x+2\right)=0$$

To find equation solutions, solve x-2=0 and x+2=0.

$$x=2 x=-2$$

Substitute 2 for x in the equation x+1=\sqrt{2x+5}.

$$2+1=\sqrt{2\times 2+5}$$

Simplify. The value x=2 satisfies the equation.

$$3=3$$

Substitute -2 for x in the equation x+1=\sqrt{2x+5}.

$$-2+1=\sqrt{2\left(-2\right)+5}$$

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

$$-1=1$$

Equation x+1=\sqrt{2x+5} has a unique solution.

$$x=2$$