Solve The Equation By Completing The Square

Solve The Equation By Completing The Square1 5n210n15solve By

Solve the equation by completing the square: 5n² = -10n + 15. Then, solve the equation by using the quadratic formula: 3x² = -10x - 8.

Paper For Above instruction

In this paper, we explore the methods of solving quadratic equations, specifically focusing on completing the square and using the quadratic formula. Additionally, we examine the application of algebra skills in business calculus, analyze graph representations involving sales data, and compare exponential and logarithmic functions for investment purposes. The integration section reviews similarities in the methods of differentiation and integration, emphasizing the interconnectedness of calculus concepts.

Solving Quadratic Equations

Quadratic equations are fundamental in algebra, and solving them can be approached through various methods including completing the square and the quadratic formula. The ability to choose the appropriate technique depends on the specific form of the equation and the context of the problem.

Completing the Square: Solving 5n² = -10n + 15

First, rewrite the equation to set it equal to zero:

5n² + 10n - 15 = 0

Divide through by 5 to simplify:

n² + 2n - 3 = 0

Next, complete the square for the quadratic expression. Take half of the coefficient of n (which is 2), square it, and add it to both sides:

(n + 1)² - 1 - 3 = 0

Rewrite:

(n + 1)² = 4

Take the square root of both sides:

n + 1 = ±2

Solution for n:

  • n + 1 = 2 → n = 1
  • n + 1 = -2 → n = -3

Hence, the solutions are n = 1 and n = -3.

Using the Quadratic Formula: Solving 3x² = -10x - 8

Rewrite the equation in standard form:

3x² + 10x + 8 = 0

Identify coefficients: a = 3, b = 10, c = 8

Apply the quadratic formula:

x = [-b ± √(b² - 4ac)] / 2a

Compute the discriminant:

Δ = 10² - 4(3)(8) = 100 - 96 = 4

Calculate the roots:

  • x = [-10 + √4] / (2*3) = [-10 + 2] / 6 = -8 / 6 = -4/3
  • x = [-10 - √4] / 6 = [-10 - 2] / 6 = -12 / 6 = -2

Solutions are x = -4/3 and x = -2.

Algebra in Business Calculus

One key algebra topic used in solving the above equations is factoring and completing the square. Factoring transforms quadratic equations into simpler linear factors, enabling straightforward solutions. Completing the square rewrites quadratics into perfect square binomials, facilitating the extraction of solutions or graphing.

For example, in solving the first equation, completing the square restructured the quadratic, making it easier to find roots without the quadratic formula. In business calculus, such skills are essential in modeling demand functions, cost functions, and revenue models where quadratic forms frequently appear.

Application of Algebra in Business Contexts

A typical business calculus problem involving algebra is optimizing profit by modeling cost and revenue functions. Suppose a company’s cost function is quadratic due to increasing marginal costs, and the revenue function depends on the quantity sold. To find the production level that maximizes profit, algebraic manipulation—including factoring, completing the square, or using the quadratic formula—is necessary. Such techniques are crucial in determining optimal output levels, pricing strategies, and break-even points.

For instance, if the profit function P(x) = -2x² + 40x - 50, finding the vertex (maximum point) involves completing the square or applying the quadratic formula to find critical points, illustrating how algebra influences business decision-making.

Graphical Analysis of Sales Data and Decision-Making

The graphs depict different sales trends over weeks. If the first graph shows a steady increase in sales and the second depicts a sharper rise after an initial lag, a store manager can interpret these patterns to adjust inventory orders. During week 9, based on the trend, the manager might order more of the item represented by the steeper graph to meet anticipated higher demand, optimizing inventory without overstocking.

Logarithmic and Exponential Functions in Investment

Comparison for Short-Term Investment (5 years)

When evaluating which investment is better over five years, understanding compound interest formulas is vital. Option 1 involves compound interest compounded quarterly:

A = P(1 + r/n)^{nt}

where P=2000, r=0.06, n=4 (quarterly), t=5

Calculating the amount:

A = 2000(1 + 0.06/4)^{4*5} ≈ 2000(1 + 0.015)^{20} ≈ 2000(1.015)^{20} ≈ 2000(1.3469) ≈ $2693.80

Option 2 is continuous compounding:

A = Pe^{rt} = 2000 e^{0.065} ≈ 2000 e^{0.3} ≈ 2000 1.3499 ≈ $2699.80

Thus, continuous compounding yields slightly more, making it the better option for a 5-year investment.

Long-Term Investment for College Savings

Suppose two accounts: Account 1 (monthly compounding at 2%) and Account 2 (annual at 4%), with investments of $2500 and $2000 respectively, over 17 years.

Account 1:

A = 2500(1 + 0.02/12)^{1217} ≈ 2500(1 + 0.0016667)^{204} ≈ 2500(1.0016667)^{204} ≈ 2500 1.3923 ≈ $3480.75

Account 2:

A = 2000(1 + 0.04)^{17} ≈ 2000 (1.04)^{17} ≈ 2000 2.0314 ≈ $4062.80

This indicates that the account with 4% compounded annually produces more interest, making it more advantageous for long-term savings.

Similarities Between Methods of Integration and Differentiation

Both integration and differentiation are inverse processes, reflecting fundamental concepts in calculus. A significant similarity is the application of the power rule: in differentiation, the power rule states that d/dx[x^n] = nx^{n-1}, while the basic rule for integration is ∫x^n dx = x^{n+1}/(n+1) + C, emphasizing the inverse relationship. Additionally, both methods rely heavily on the ability to manipulate algebraic expressions, utilize substitution techniques, and understand the behavior of functions. For example, u-substitution in integration mirrors the chain rule in differentiation, allowing for the transformation of complex functions into simpler forms.

Understanding these similarities enhances comprehension of calculus as a unified mathematical framework, supporting advanced applications across science, engineering, and economics.

Conclusion

Solving quadratic equations through different methods highlights the versatility of algebra in problem-solving. Applying these techniques within business calculus enables informed decision-making and optimization. Graphical analysis of sales trends informs inventory management, while understanding exponential and logarithmic functions aids in investment planning. Recognizing the parallels between integration and differentiation underscores the interconnectedness of calculus concepts, fostering deeper mathematical understanding.

References

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