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

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

Solve for x

$$x=-1$$

Solution Steps

Subtract \sqrt{2x+3} from both sides of the equation.

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

Square both sides of the equation.

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

Calculate \sqrt{x+1} to the power of 2 and get x+1.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-\sqrt{2x+3}\right)^{2}.

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

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

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

Add 1 and 3 to get 4.

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

Subtract 4+2x from both sides of the equation.

$$x+1-\left(4+2x\right)=-2\sqrt{2x+3}$$

To find the opposite of 4+2x, find the opposite of each term.

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

Subtract 4 from 1 to get -3.

$$x-3-2x=-2\sqrt{2x+3}$$

Combine x and -2x to get -x.

$$-x-3=-2\sqrt{2x+3}$$

Square both sides of the equation.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-x-3\right)^{2}.

$$x^{2}+6x+9=\left(-2\sqrt{2x+3}\right)^{2}$$

Expand \left(-2\sqrt{2x+3}\right)^{2}.

$$x^{2}+6x+9=\left(-2\right)^{2}\left(\sqrt{2x+3}\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$x^{2}+6x+9=4\left(\sqrt{2x+3}\right)^{2}$$

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

$$x^{2}+6x+9=4\left(2x+3\right)$$

Use the distributive property to multiply 4 by 2x+3.

$$x^{2}+6x+9=8x+12$$

Subtract 8x from both sides.

$$x^{2}+6x+9-8x=12$$

Combine 6x and -8x to get -2x.

$$x^{2}-2x+9=12$$

Subtract 12 from both sides.

$$x^{2}-2x+9-12=0$$

Subtract 12 from 9 to get -3.

$$x^{2}-2x-3=0$$

To solve the equation, factor x^{2}-2x-3 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-2 ab=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-3 b=1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-3\right)\left(x+1\right)$$

To find equation solutions, solve x-3=0 and x+1=0.

$$x=3 x=-1$$

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

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

Simplify. The value x=3 does not satisfy the equation.

$$5=1$$

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

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

Simplify. The value x=-1 satisfies the equation.

$$1=1$$

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

$$x=-1$$