Math Problem Statement

What is X if ((306×X)× (260×X)+ π× (71× X)× (153× X))× 2+ (70,7530918052× X)²= 400000000

Solution

To solve the equation:

((306×X)×(260×X)+π×(71×X)×(153×X))×2+(70.7530918052×X)2=400000000\left( (306 \times X) \times (260 \times X) + \pi \times (71 \times X) \times (153 \times X) \right) \times 2 + \left( 70.7530918052 \times X \right)^2 = 400000000

Step 1: Simplify the expression

Let's break this down step by step.

  1. First term: (306×X)×(260×X)(306 \times X) \times (260 \times X) (306×X)×(260×X)=306×260×X2=79760×X2(306 \times X) \times (260 \times X) = 306 \times 260 \times X^2 = 79760 \times X^2

  2. Second term: π×(71×X)×(153×X)\pi \times (71 \times X) \times (153 \times X) π×(71×X)×(153×X)=π×71×153×X2=π×10863×X234142.711×X2\pi \times (71 \times X) \times (153 \times X) = \pi \times 71 \times 153 \times X^2 = \pi \times 10863 \times X^2 \approx 34142.711 \times X^2

  3. Third term: (70.7530918052×X)2\left( 70.7530918052 \times X \right)^2 (70.7530918052×X)2=5006.398×X2\left( 70.7530918052 \times X \right)^2 = 5006.398 \times X^2

Step 2: Substitute simplified terms into the equation

Now the equation becomes:

(79760×X2+34142.711×X2)×2+5006.398×X2=400000000\left( 79760 \times X^2 + 34142.711 \times X^2 \right) \times 2 + 5006.398 \times X^2 = 400000000

Simplify the terms inside the parentheses: (79760+34142.711)×X2×2+5006.398×X2=400000000(79760 + 34142.711) \times X^2 \times 2 + 5006.398 \times X^2 = 400000000 113902.711×X2×2+5006.398×X2=400000000113902.711 \times X^2 \times 2 + 5006.398 \times X^2 = 400000000 227805.422×X2+5006.398×X2=400000000227805.422 \times X^2 + 5006.398 \times X^2 = 400000000 232811.82×X2=400000000232811.82 \times X^2 = 400000000

Step 3: Solve for X2X^2

Now, divide both sides by 232811.82:

X2=400000000232811.821717.235X^2 = \frac{400000000}{232811.82} \approx 1717.235

Step 4: Take the square root to find XX

X=1717.23541.45X = \sqrt{1717.235} \approx 41.45

Final Answer:

X41.45X \approx 41.45

Would you like any further explanations or details about the steps?


Related Questions:

  1. How would the equation change if we replaced π\pi with a different constant, such as 3.14?
  2. What happens if the power of XX in the equation changes from X2X^2 to X3X^3?
  3. How do we approach solving similar equations with more terms or powers of XX?
  4. What would happen if the equation were multiplied by a factor of 10 instead of 2?
  5. How can this approach be used to solve equations with more complex algebraic expressions?

Tip: When solving complex algebraic equations, always start by simplifying terms to reduce the equation to a more manageable form.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Pi (π)
Simplification of Expressions

Formulas

(a × X) × (b × X) = ab × X^2
(π × a × X) × (b × X) = πab × X^2
(a × X)^2 = a^2 × X^2

Theorems

Simplification of algebraic expressions
Quadratic formula

Suitable Grade Level

Grades 10-12