Solve sqrt{4-x}+sqrt{1+x}=3

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

Solve for x

$$x=3$$

$$x=0$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

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

$$4-x=9-6\sqrt{1+x}+\left(\sqrt{1+x}\right)^{2}$$

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

$$4-x=9-6\sqrt{1+x}+1+x$$

Add 9 and 1 to get 10.

$$4-x=10-6\sqrt{1+x}+x$$

Subtract 10+x from both sides of the equation.

$$4-x-\left(10+x\right)=-6\sqrt{1+x}$$

To find the opposite of 10+x, find the opposite of each term.

$$4-x-10-x=-6\sqrt{1+x}$$

Subtract 10 from 4 to get -6.

$$-6-x-x=-6\sqrt{1+x}$$

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

$$-6-2x=-6\sqrt{1+x}$$

Square both sides of the equation.

$$\left(-6-2x\right)^{2}=\left(-6\sqrt{1+x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-6-2x\right)^{2}.

$$36+24x+4x^{2}=\left(-6\sqrt{1+x}\right)^{2}$$

Expand \left(-6\sqrt{1+x}\right)^{2}.

$$36+24x+4x^{2}=\left(-6\right)^{2}\left(\sqrt{1+x}\right)^{2}$$

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

$$36+24x+4x^{2}=36\left(\sqrt{1+x}\right)^{2}$$

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

$$36+24x+4x^{2}=36\left(1+x\right)$$

Use the distributive property to multiply 36 by 1+x.

$$36+24x+4x^{2}=36+36x$$

Subtract 36 from both sides.

$$36+24x+4x^{2}-36=36x$$

Subtract 36 from 36 to get 0.

$$24x+4x^{2}=36x$$

Subtract 36x from both sides.

$$24x+4x^{2}-36x=0$$

Combine 24x and -36x to get -12x.

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

Factor out x.

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

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

$$x=0 x=3$$

Substitute 0 for x in the equation \sqrt{4-x}+\sqrt{1+x}=3.

$$\sqrt{4-0}+\sqrt{1+0}=3$$

Simplify. The value x=0 satisfies the equation.

$$3=3$$

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

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

Simplify. The value x=3 satisfies the equation.

$$3=3$$

List all solutions of \sqrt{4-x}=-\sqrt{x+1}+3.

$$x=0 x=3$$