Solve sqrt{5-x}+sqrt{x-3}=2

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

Solve for x

$$x=4$$

Solution Steps

Subtract \sqrt{x-3} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

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

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

Subtract 3 from 4 to get 1.

$$5-x=1-4\sqrt{x-3}+x$$

Subtract 1+x from both sides of the equation.

$$5-x-\left(1+x\right)=-4\sqrt{x-3}$$

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

$$5-x-1-x=-4\sqrt{x-3}$$

Subtract 1 from 5 to get 4.

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

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

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

Square both sides of the equation.

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

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

$$16-16x+4x^{2}=\left(-4\sqrt{x-3}\right)^{2}$$

Expand \left(-4\sqrt{x-3}\right)^{2}.

$$16-16x+4x^{2}=\left(-4\right)^{2}\left(\sqrt{x-3}\right)^{2}$$

Calculate -4 to the power of 2 and get 16.

$$16-16x+4x^{2}=16\left(\sqrt{x-3}\right)^{2}$$

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

$$16-16x+4x^{2}=16\left(x-3\right)$$

Use the distributive property to multiply 16 by x-3.

$$16-16x+4x^{2}=16x-48$$

Subtract 16x from both sides.

$$16-16x+4x^{2}-16x=-48$$

Combine -16x and -16x to get -32x.

$$16-32x+4x^{2}=-48$$

Add 48 to both sides.

$$16-32x+4x^{2}+48=0$$

Add 16 and 48 to get 64.

$$64-32x+4x^{2}=0$$

Divide both sides by 4.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 16=16$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -8.

$$a=-4 b=-4$$

Rewrite x^{2}-8x+16 as \left(x^{2}-4x\right)+\left(-4x+16\right).

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

Factor out x in the first and -4 in the second group.

$$x\left(x-4\right)-4\left(x-4\right)$$

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

$$\left(x-4\right)\left(x-4\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve x-4=0.

$$x=4$$

Substitute 4 for x in the equation \sqrt{5-x}+\sqrt{x-3}=2.

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

Simplify. The value x=4 satisfies the equation.

$$2=2$$

Equation \sqrt{5-x}=-\sqrt{x-3}+2 has a unique solution.

$$x=4$$