Solve sqrt{x+5}+sqrt{-4-x}=sqrt{x+5-4-x}

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

Solve for x

$$x=-4$$

$$x=-5$$

Solution Steps

Subtract \sqrt{-4-x} from both sides of the equation.

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

Subtract 4 from 5 to get 1.

$$\sqrt{x+5}=\sqrt{x+1-x}-\sqrt{-4-x}$$

Combine x and -x to get 0.

$$\sqrt{x+5}=\sqrt{1}-\sqrt{-4-x}$$

Calculate the square root of 1 and get 1.

$$\sqrt{x+5}=1-\sqrt{-4-x}$$

Square both sides of the equation.

$$\left(\sqrt{x+5}\right)^{2}=\left(1-\sqrt{-4-x}\right)^{2}$$

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

$$x+5=\left(1-\sqrt{-4-x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-\sqrt{-4-x}\right)^{2}.

$$x+5=1-2\sqrt{-4-x}+\left(\sqrt{-4-x}\right)^{2}$$

Calculate \sqrt{-4-x} to the power of 2 and get -4-x.

$$x+5=1-2\sqrt{-4-x}-4-x$$

Subtract 4 from 1 to get -3.

$$x+5=-3-2\sqrt{-4-x}-x$$

Subtract -3-x from both sides of the equation.

$$x+5-\left(-3-x\right)=-2\sqrt{-4-x}$$

To find the opposite of -3-x, find the opposite of each term.

$$x+5+3+x=-2\sqrt{-4-x}$$

Add 5 and 3 to get 8.

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

Combine x and x to get 2x.

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

Square both sides of the equation.

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

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

$$4x^{2}+32x+64=\left(-2\sqrt{-4-x}\right)^{2}$$

Expand \left(-2\sqrt{-4-x}\right)^{2}.

$$4x^{2}+32x+64=\left(-2\right)^{2}\left(\sqrt{-4-x}\right)^{2}$$

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

$$4x^{2}+32x+64=4\left(\sqrt{-4-x}\right)^{2}$$

Calculate \sqrt{-4-x} to the power of 2 and get -4-x.

$$4x^{2}+32x+64=4\left(-4-x\right)$$

Use the distributive property to multiply 4 by -4-x.

$$4x^{2}+32x+64=-16-4x$$

Add 4x to both sides.

$$4x^{2}+32x+64+4x=-16$$

Combine 32x and 4x to get 36x.

$$4x^{2}+36x+64=-16$$

Add 16 to both sides.

$$4x^{2}+36x+64+16=0$$

Add 64 and 16 to get 80.

$$4x^{2}+36x+80=0$$

Divide both sides by 4.

$$x^{2}+9x+20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=9 ab=1\times 20=20$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 20.

$$1,20 2,10 4,5$$

Calculate the sum for each pair.

$$1+20=21 2+10=12 4+5=9$$

The solution is the pair that gives sum 9.

$$a=4 b=5$$

Rewrite x^{2}+9x+20 as \left(x^{2}+4x\right)+\left(5x+20\right).

$$\left(x^{2}+4x\right)+\left(5x+20\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(x+4\right)+5\left(x+4\right)$$

Factor out common term x+4 by using distributive property.

$$\left(x+4\right)\left(x+5\right)$$

To find equation solutions, solve x+4=0 and x+5=0.

$$x=-4 x=-5$$

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

$$\sqrt{-4+5}+\sqrt{-4-\left(-4\right)}=\sqrt{-4+5-4-\left(-4\right)}$$

Simplify. The value x=-4 satisfies the equation.

$$1=1$$

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

$$\sqrt{-5+5}+\sqrt{-4-\left(-5\right)}=\sqrt{-5+5-4-\left(-5\right)}$$

Simplify. The value x=-5 satisfies the equation.

$$1=1$$

List all solutions of \sqrt{x+5}=-\sqrt{-x-4}+1.

$$x=-4 x=-5$$