Math Problem Statement

2x(4x)123(4x)=0\left(4-x\right)^{-\frac12}-3\left(\sqrt{4-x}\right)=0

Solution

We are given the equation:

2x(4x)123(4x)=02x \left(4 - x\right)^{-\frac{1}{2}} - 3\left(\sqrt{4 - x}\right) = 0

Step 1: Substitute a new variable

Let's simplify this by substituting y=4xy = \sqrt{4 - x}, so y2=4xy^2 = 4 - x and x=4y2x = 4 - y^2.

Substitute these into the equation:

2(4y2)y13y=02(4 - y^2) y^{-1} - 3y = 0

Step 2: Simplify the equation

Now, simplify the terms:

2(4y2)1y3y=02(4 - y^2) \cdot \frac{1}{y} - 3y = 0

2(4y2)y3y=0\frac{2(4 - y^2)}{y} - 3y = 0

Expand the first term:

82y2y3y=0\frac{8 - 2y^2}{y} - 3y = 0

Step 3: Eliminate the fraction

Multiply the entire equation by yy to eliminate the denominator:

82y23y2=08 - 2y^2 - 3y^2 = 0

Step 4: Combine like terms

Simplify the quadratic expression:

85y2=08 - 5y^2 = 0

Step 5: Solve for y2y^2

Rearrange the equation:

5y2=85y^2 = 8

y2=85y^2 = \frac{8}{5}

Step 6: Solve for yy

Take the square root of both sides:

y=±85=±2105y = \pm \sqrt{\frac{8}{5}} = \pm \frac{2\sqrt{10}}{5}

Step 7: Solve for xx

Recall that y=4xy = \sqrt{4 - x}. Substituting y=±2105y = \pm \frac{2\sqrt{10}}{5}:

4x=±2105\sqrt{4 - x} = \pm \frac{2\sqrt{10}}{5}

Square both sides:

4x=854 - x = \frac{8}{5}

Solve for xx:

x=485=20585=125x = 4 - \frac{8}{5} = \frac{20}{5} - \frac{8}{5} = \frac{12}{5}

Thus, the solution is:

x=125x = \frac{12}{5}

Let me know if you want more details or further explanation.

Here are 5 related questions:

  1. How does substitution help simplify complex equations?
  2. What is the importance of eliminating fractions in an equation?
  3. Why do we square both sides when solving equations involving square roots?
  4. What does it mean for a solution to have both positive and negative square roots?
  5. How can we verify the solution by substituting it back into the original equation?

Tip: When solving equations involving square roots, always check if both the positive and negative roots provide valid solutions in the context of the problem.

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Math Problem Analysis

Mathematical Concepts

Substitution
Algebra
Square Roots
Quadratic Equations

Formulas

y = \sqrt{4 - x}, y^2 = 4 - x
Quadratic equation: ax^2 + bx + c = 0
Solving square roots: \(y = \pm \frac{2\sqrt{10}}{5}\)

Theorems

Substitution method for simplifying complex equations
Square root theorem for solving quadratic expressions

Suitable Grade Level

Grade 10-12