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

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

Solve for x

$$x=1$$

$$x=-\frac{1}{2}=-0.5$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

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

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

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

$$-2x+3=9-6\sqrt{2x+2}+2x+2$$

Add 9 and 2 to get 11.

$$-2x+3=11-6\sqrt{2x+2}+2x$$

Subtract 11+2x from both sides of the equation.

$$-2x+3-\left(11+2x\right)=-6\sqrt{2x+2}$$

To find the opposite of 11+2x, find the opposite of each term.

$$-2x+3-11-2x=-6\sqrt{2x+2}$$

Subtract 11 from 3 to get -8.

$$-2x-8-2x=-6\sqrt{2x+2}$$

Combine -2x and -2x to get -4x.

$$-4x-8=-6\sqrt{2x+2}$$

Square both sides of the equation.

$$\left(-4x-8\right)^{2}=\left(-6\sqrt{2x+2}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4x-8\right)^{2}.

$$16x^{2}+64x+64=\left(-6\sqrt{2x+2}\right)^{2}$$

Expand \left(-6\sqrt{2x+2}\right)^{2}.

$$16x^{2}+64x+64=\left(-6\right)^{2}\left(\sqrt{2x+2}\right)^{2}$$

Calculate -6 to the power of 2 and get 36.

$$16x^{2}+64x+64=36\left(\sqrt{2x+2}\right)^{2}$$

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

$$16x^{2}+64x+64=36\left(2x+2\right)$$

Use the distributive property to multiply 36 by 2x+2.

$$16x^{2}+64x+64=72x+72$$

Subtract 72x from both sides.

$$16x^{2}+64x+64-72x=72$$

Combine 64x and -72x to get -8x.

$$16x^{2}-8x+64=72$$

Subtract 72 from both sides.

$$16x^{2}-8x+64-72=0$$

Subtract 72 from 64 to get -8.

$$16x^{2}-8x-8=0$$

Divide both sides by 8.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=-1 ab=2\left(-1\right)=-2$$

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.

$$a=-2 b=1$$

Rewrite 2x^{2}-x-1 as \left(2x^{2}-2x\right)+\left(x-1\right).

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

Factor out 2x in 2x^{2}-2x.

$$2x\left(x-1\right)+x-1$$

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

$$\left(x-1\right)\left(2x+1\right)$$

To find equation solutions, solve x-1=0 and 2x+1=0.

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

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

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

Simplify. The value x=1 satisfies the equation.

$$3=3$$

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

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

Simplify. The value x=-\frac{1}{2} satisfies the equation.

$$3=3$$

List all solutions of \sqrt{3-2x}=-\sqrt{2x+2}+3.

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