Solving Quadratic Equations - A Step-by-Step Guide

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In the realm of mathematics, solving quadratic equations is a fundamental skill with widespread applications in various fields, from physics and engineering to economics and computer science. A quadratic equation is a polynomial equation of the second degree, generally expressed in the standard form:

ax2+bx+c=0ax^2 + bx + c = 0

where a, b, and c are constants, and a is not equal to zero. The solutions to a quadratic equation, also known as roots or zeros, are the values of x that satisfy the equation. These solutions represent the points where the parabola defined by the quadratic equation intersects the x-axis.

This comprehensive guide delves into the intricacies of solving the specific quadratic equation:

−7f2+11f+9=4−3f-7f^2 + 11f + 9 = 4 - 3f

We will explore the step-by-step process of transforming this equation into its standard form, identifying the coefficients, and employing the quadratic formula to determine the solutions, rounded to two decimal places. Understanding these techniques will empower you to tackle a wide array of quadratic equations with confidence.

Transforming to Standard Form: Setting the Stage for Solution

The first crucial step in solving quadratic equations is to rearrange the equation into its standard form. This involves manipulating the equation algebraically to have all terms on one side, leaving zero on the other side. By doing so, we can readily identify the coefficients a, b, and c, which are essential for applying the quadratic formula. Let's apply this principle to our equation:

−7f2+11f+9=4−3f-7f^2 + 11f + 9 = 4 - 3f

To begin, we aim to eliminate the terms on the right-hand side of the equation. We can achieve this by adding 3f to both sides and subtracting 4 from both sides. This process maintains the equality of the equation while strategically moving terms:

−7f2+11f+9+3f−4=4−3f+3f−4-7f^2 + 11f + 9 + 3f - 4 = 4 - 3f + 3f - 4

Now, we simplify both sides of the equation by combining like terms. On the left-hand side, we combine the f terms (11f and 3f) and the constant terms (9 and -4). On the right-hand side, the 4 and -4 cancel each other out, as do the -3f and 3f terms:

−7f2+(11f+3f)+(9−4)=0-7f^2 + (11f + 3f) + (9 - 4) = 0

This simplification yields:

−7f2+14f+5=0-7f^2 + 14f + 5 = 0

We have successfully transformed the original equation into its standard form, where the terms are arranged in descending order of their exponents, and the equation is set equal to zero. This standard form is the foundation for applying the quadratic formula.

Identifying Coefficients: Unveiling the Numerical Values

Now that the quadratic equation is in standard form, the next step is to identify the coefficients a, b, and c. These coefficients are the numerical values that multiply the corresponding terms in the quadratic equation. In the standard form equation:

ax2+bx+c=0ax^2 + bx + c = 0

  • a represents the coefficient of the quadratic term (x²).
  • b represents the coefficient of the linear term (x).
  • c represents the constant term.

Careful identification of these coefficients is crucial for the accurate application of the quadratic formula. Let's revisit our transformed equation:

−7f2+14f+5=0-7f^2 + 14f + 5 = 0

Comparing this equation to the standard form, we can readily identify the coefficients:

  • a = -7 (the coefficient of f²)
  • b = 14 (the coefficient of f)
  • c = 5 (the constant term)

These values will be directly substituted into the quadratic formula in the subsequent step.

Applying the Quadratic Formula: The Key to Unlocking Solutions

The quadratic formula is a powerful tool for solving quadratic equations. It provides a direct method for finding the solutions (roots) of any quadratic equation expressed in standard form. The formula is given by:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

where a, b, and c are the coefficients of the quadratic equation in standard form. The "±" symbol indicates that there are two possible solutions, one obtained by adding the square root term and the other by subtracting it.

To apply the quadratic formula to our equation, we substitute the previously identified coefficient values (a = -7, b = 14, and c = 5) into the formula:

f=−14±142−4(−7)(5)2(−7)f = \frac{-14 \pm \sqrt{14^2 - 4(-7)(5)}}{2(-7)}

This substitution sets the stage for simplifying the expression and finding the solutions for f.

Simplifying the Expression: Step-by-Step Calculation

After substituting the coefficients into the quadratic formula, the next step is to simplify the expression. This involves performing the arithmetic operations within the formula, following the order of operations (PEMDAS/BODMAS). Let's continue with our calculation:

f=−14±142−4(−7)(5)2(−7)f = \frac{-14 \pm \sqrt{14^2 - 4(-7)(5)}}{2(-7)}

First, we calculate the square of 14: 14² = 196. Then, we multiply the terms under the square root: 4 * (-7) * 5 = -140. Note that the negative signs must be carefully accounted for:

f=−14±196−(−140)−14f = \frac{-14 \pm \sqrt{196 - (-140)}}{-14}

Next, we simplify the expression under the square root by subtracting the negative value: 196 - (-140) = 196 + 140 = 336:

f=−14±336−14f = \frac{-14 \pm \sqrt{336}}{-14}

Now, we calculate the square root of 336. The square root of 336 is approximately 18.33. We will round this to two decimal places later, but for now, we keep the full value to maintain accuracy:

f=−14±18.33−14f = \frac{-14 \pm 18.33}{-14}

Finally, we multiply 2 by -7 in the denominator: 2 * (-7) = -14:

f=−14±18.33−14f = \frac{-14 \pm 18.33}{-14}

This simplified expression sets the stage for calculating the two possible solutions for f.

Calculating the Solutions: Finding the Roots

With the simplified expression from the quadratic formula, we can now calculate the two solutions for f. We have two cases to consider, one with the addition of the square root term and one with the subtraction:

Case 1: Addition

f1=−14+18.33−14f_1 = \frac{-14 + 18.33}{-14}

First, we add the terms in the numerator: -14 + 18.33 = 4.33.

f1=4.33−14f_1 = \frac{4.33}{-14}

Then, we divide 4.33 by -14:

f1≈−0.31f_1 \approx -0.31

Case 2: Subtraction

f2=−14−18.33−14f_2 = \frac{-14 - 18.33}{-14}

First, we subtract the terms in the numerator: -14 - 18.33 = -32.33.

f2=−32.33−14f_2 = \frac{-32.33}{-14}

Then, we divide -32.33 by -14:

f2≈2.31f_2 \approx 2.31

Rounding to Two Decimal Places: Presenting the Final Answer

The problem statement requires us to provide the solutions rounded to two decimal places. We have already calculated the approximate values of the solutions:

  • f₁ ≈ -0.31
  • f₂ ≈ 2.31

Both solutions are already expressed to two decimal places, so no further rounding is necessary. Therefore, the solutions to the quadratic equation -7f² + 11f + 9 = 4 - 3f, rounded to two decimal places, are approximately -0.31 and 2.31.

In conclusion, by transforming the equation into standard form, identifying the coefficients, applying the quadratic formula, simplifying the expression, and calculating the solutions, we have successfully solved the given quadratic equation. This step-by-step approach provides a clear and systematic method for tackling quadratic equations, empowering you to find their solutions with confidence and accuracy.

Solving quadratic equations requires a systematic approach. We've successfully navigated through the process for the equation -7f² + 11f + 9 = 4 - 3f. Let's recap the key steps:

  1. Transform to Standard Form: Rearrange the equation into the form ax² + bx + c = 0. This involves moving all terms to one side, simplifying, and arranging in descending order of exponents.

  2. Identify Coefficients: Determine the values of a, b, and c from the standard form equation. These values are crucial for applying the quadratic formula.

  3. Apply the Quadratic Formula: Substitute the coefficients a, b, and c into the quadratic formula:

    x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

  4. Simplify the Expression: Carefully perform the arithmetic operations within the formula, following the order of operations (PEMDAS/BODMAS). This includes calculating the square, multiplication, subtraction, and square root.

  5. Calculate the Solutions: Separate the "±" into two cases, one with addition and one with subtraction. Calculate the two possible values of x (the solutions or roots).

  6. Round to the Specified Decimal Places: If required, round the solutions to the specified number of decimal places.

By following these steps diligently, you can confidently solve a wide range of quadratic equations.

Solving quadratic equations isn't just an abstract mathematical exercise; it's a fundamental skill with numerous practical applications in various fields. Understanding how to solve these equations allows us to model and solve real-world problems across different disciplines.

Physics

In physics, quadratic equations are frequently used to describe projectile motion. For example, the height of a projectile launched into the air can be modeled by a quadratic equation, where the variable represents time. By solving quadratic equations, physicists can determine the time it takes for the projectile to reach a certain height, the maximum height it reaches, and the total time it spends in the air. These calculations are essential in fields like ballistics, sports science, and aerospace engineering.

Engineering

Engineers use quadratic equations in various applications, such as designing structures, analyzing circuits, and optimizing processes. For example, the load-bearing capacity of a beam can be related to its dimensions through a quadratic equation. Solving quadratic equations allows engineers to determine the dimensions required for a beam to support a specific load safely. Similarly, in electrical engineering, quadratic equations can model the behavior of circuits, helping engineers design efficient and reliable systems.

Economics

In economics, quadratic equations can be used to model supply and demand curves. The equilibrium price and quantity in a market occur where the supply and demand curves intersect, which often involves solving quadratic equations. Economists can use these solutions to analyze market trends, predict price fluctuations, and make informed decisions about resource allocation.

Computer Science

Quadratic equations also find applications in computer science, particularly in areas like computer graphics and optimization algorithms. For example, the intersection of lines and curves, a fundamental operation in computer graphics, can often be determined by solving quadratic equations. Additionally, some optimization algorithms use quadratic functions to model the objective function, and finding the minimum or maximum of these functions involves solving quadratic equations.

Everyday Life

Beyond these technical fields, quadratic equations even appear in everyday life situations. For example, if you're planning a garden and want to enclose a rectangular area with a fixed amount of fencing, you might use a quadratic equation to determine the dimensions that maximize the area. Similarly, understanding quadratic relationships can help in making financial decisions, such as calculating loan payments or investment returns.

The ability to solve quadratic equations provides a powerful tool for understanding and solving problems in a wide range of disciplines. From physics and engineering to economics and computer science, the applications are numerous and impactful. This skill empowers individuals to analyze situations, make predictions, and design solutions in both professional and personal contexts.

In this comprehensive guide, we've explored the process of solving quadratic equations, focusing on the specific example of -7f² + 11f + 9 = 4 - 3f. We've delved into the step-by-step methodology, encompassing transforming the equation to standard form, identifying coefficients, applying the quadratic formula, simplifying the expression, calculating the solutions, and rounding to the required decimal places. The solutions we found, approximately -0.31 and 2.31, represent the values of f that satisfy the given equation.

Furthermore, we've highlighted the broader significance of quadratic equations and their far-reaching applications across various fields, including physics, engineering, economics, computer science, and even everyday life scenarios. Understanding how to solve these equations provides a valuable tool for modeling real-world phenomena, making informed decisions, and solving practical problems.

The ability to solve quadratic equations empowers individuals to tackle complex challenges and gain a deeper understanding of the world around them. Whether you're a student learning the fundamentals of mathematics, a professional applying these concepts in your field, or simply someone seeking to enhance your problem-solving skills, mastering this technique will undoubtedly prove beneficial.

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