Math Problem Statement

Solve the equation: sqrt(4 - x) + sqrt(x - 2) = 2

Solution

The equation given is:

4x+x2=2\sqrt{4 - x} + \sqrt{x - 2} = 2

To solve this, let's proceed step by step:

Step 1: Isolate one square root term.

Move one square root to the other side:

4x=2x2\sqrt{4 - x} = 2 - \sqrt{x - 2}

Step 2: Square both sides.

Square both sides to eliminate the square root on the left-hand side:

(4x)2=(2x2)2( \sqrt{4 - x} )^2 = (2 - \sqrt{x - 2})^2

Simplifying:

4x=(2x2)24 - x = (2 - \sqrt{x - 2})^2

Step 3: Expand the right-hand side.

Now expand the square on the right-hand side:

4x=44x2+(x2)4 - x = 4 - 4\sqrt{x - 2} + (x - 2)

Simplifying further:

4x=2+x4x24 - x = 2 + x - 4\sqrt{x - 2}

Step 4: Isolate the square root term again.

Move all terms except the square root to one side:

4x(2+x)=4x24 - x - (2 + x) = -4\sqrt{x - 2}

Simplify:

22x=4x22 - 2x = -4\sqrt{x - 2}

Step 5: Divide both sides by -4.

To simplify:

22x4=x2\frac{2 - 2x}{-4} = \sqrt{x - 2}

This simplifies to:

x12=x2\frac{x - 1}{2} = \sqrt{x - 2}

Step 6: Square both sides again.

Square both sides to eliminate the square root:

(x12)2=(x2)\left( \frac{x - 1}{2} \right)^2 = (x - 2)

Simplifying:

(x1)24=x2\frac{(x - 1)^2}{4} = x - 2

Step 7: Clear the denominator.

Multiply both sides by 4:

(x1)2=4(x2)(x - 1)^2 = 4(x - 2)

Step 8: Expand and solve.

Expand both sides:

x22x+1=4x8x^2 - 2x + 1 = 4x - 8

Now move all terms to one side:

x26x+9=0x^2 - 6x + 9 = 0

Step 9: Factor the quadratic.

The quadratic factors as:

(x3)2=0(x - 3)^2 = 0

Thus, x=3x = 3.

Step 10: Check the solution.

Substitute x=3x = 3 into the original equation to check:

43+32=1+1=1+1=2\sqrt{4 - 3} + \sqrt{3 - 2} = \sqrt{1} + \sqrt{1} = 1 + 1 = 2

The solution is correct.

Final Answer:

x=3x = 3

Would you like more details on any of these steps? Here are five related questions you might find useful:

  1. How do you handle solving equations with multiple square roots?
  2. Can you explain the reasoning behind squaring both sides in more detail?
  3. Are there other methods to solve equations like this one?
  4. What are common pitfalls when squaring both sides of an equation?
  5. How can you check if a solution is extraneous?

Tip: Always check for extraneous solutions when squaring both sides of an equation, as squaring can sometimes introduce invalid results.

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Math Problem Analysis

Mathematical Concepts

Algebra
Square roots
Quadratic equations

Formulas

Square root properties
(a + b)^2 = a^2 + 2ab + b^2

Theorems

Factoring quadratic equations

Suitable Grade Level

Grades 9-12