Solve sqrt{x+1}=1+sqrt{4-x}

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

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

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

Add 1 and 4 to get 5.

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

Subtract 5-x from both sides of the equation.

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

To find the opposite of 5-x, find the opposite of each term.

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

Subtract 5 from 1 to get -4.

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

Combine x and x to get 2x.

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

Square both sides of the equation.

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

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

$$4x^{2}-16x+16=\left(2\sqrt{4-x}\right)^{2}$$

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

$$4x^{2}-16x+16=2^{2}\left(\sqrt{4-x}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-16x+16=4\left(\sqrt{4-x}\right)^{2}$$

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

$$4x^{2}-16x+16=4\left(4-x\right)$$

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

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

Subtract 16 from both sides.

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

Subtract 16 from 16 to get 0.

$$4x^{2}-16x=-4x$$

Add 4x to both sides.

$$4x^{2}-16x+4x=0$$

Combine -16x and 4x to get -12x.

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

Factor out x.

$$x\left(4x-12\right)=0$$

To find equation solutions, solve x=0 and 4x-12=0.

$$x=0 x=3$$

Substitute 0 for x in the equation \sqrt{x+1}=1+\sqrt{4-x}.

$$\sqrt{0+1}=1+\sqrt{4-0}$$

Simplify. The value x=0 does not satisfy the equation.

$$1=3$$

Substitute 3 for x in the equation \sqrt{x+1}=1+\sqrt{4-x}.

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

Simplify. The value x=3 satisfies the equation.

$$2=2$$

Equation \sqrt{x+1}=\sqrt{4-x}+1 has a unique solution.

$$x=3$$