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

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

Solve for x

$$x=1$$

Solution Steps

Subtract -\sqrt{x-1} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

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

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

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

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

Combine 2x and x to get 3x.

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

Subtract 1 from 2 to get 1.

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

Subtract 3x+1 from both sides of the equation.

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

To find the opposite of 3x+1, find the opposite of each term.

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

Combine x and -3x to get -2x.

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

Subtract 1 from 3 to get 2.

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

Square both sides of the equation.

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

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

$$4x^{2}-8x+4=\left(2\sqrt{2x+2}\sqrt{x-1}\right)^{2}$$

Expand \left(2\sqrt{2x+2}\sqrt{x-1}\right)^{2}.

$$4x^{2}-8x+4=2^{2}\left(\sqrt{2x+2}\right)^{2}\left(\sqrt{x-1}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-8x+4=4\left(\sqrt{2x+2}\right)^{2}\left(\sqrt{x-1}\right)^{2}$$

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

$$4x^{2}-8x+4=4\left(2x+2\right)\left(\sqrt{x-1}\right)^{2}$$

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

$$4x^{2}-8x+4=4\left(2x+2\right)\left(x-1\right)$$

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

$$4x^{2}-8x+4=\left(8x+8\right)\left(x-1\right)$$

Apply the distributive property by multiplying each term of 8x+8 by each term of x-1.

$$4x^{2}-8x+4=8x^{2}-8x+8x-8$$

Combine -8x and 8x to get 0.

$$4x^{2}-8x+4=8x^{2}-8$$

Subtract 8x^{2} from both sides.

$$4x^{2}-8x+4-8x^{2}=-8$$

Combine 4x^{2} and -8x^{2} to get -4x^{2}.

$$-4x^{2}-8x+4=-8$$

Add 8 to both sides.

$$-4x^{2}-8x+4+8=0$$

Add 4 and 8 to get 12.

$$-4x^{2}-8x+12=0$$

Divide both sides by 4.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

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

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.

$$a=1 b=-3$$

Rewrite -x^{2}-2x+3 as \left(-x^{2}+x\right)+\left(-3x+3\right).

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

Factor out x in the first and 3 in the second group.

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

Factor out common term -x+1 by using distributive property.

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

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

$$x=1 x=-3$$

Substitute -3 for x in the equation \sqrt{x+3}-\sqrt{x-1}=\sqrt{2x+2}. The expression \sqrt{-3-1} is undefined because the radicand cannot be negative.

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

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

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

Simplify. The value x=1 satisfies the equation.

$$2=2$$

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

$$x=1$$