Problem 24 - Basic Technical Mathematics with Calculus

Question

$$\text { Solve the given quadratic equations by factoring.}$$ $$4 x^{2}+25=20 x$$

Step-by-Step Solution

Answer

The solution is \( x = \frac{5}{2} \).

Step 1: Rewrite the Equation

Start by writing the equation in standard form, which is: \[ ax^2 + bx + c = 0 \] For the given equation, this involves moving all terms to one side to get: \[ 4x^2 - 20x + 25 = 0 \]

Step 2: Find the Greatest Common Factor

Check each term to see if there is a common factor we can factor out, but in this equation, the coefficients 4, -20, and 25 do not share any common factor other than 1. Thus, we proceed to factoring the quadratic trinomial.

Step 3: Factor the Quadratic Trinomial

Factor the quadratic equation using the standard factoring method, looking for two numbers that multiply to \( ac \) (where \( a = 4 \) and \( c = 25 \)) and add to \( b = -20 \). We need two numbers that multiply to \( 4 \times 25 = 100 \) and add to \( -20 \). These two numbers are \(-10\) and \(-10\). Thus, the equation can be rewritten as: \[ 4x^2 - 10x - 10x + 25 = 0 \]

Step 4: Group Terms and Factor by Grouping

Group the terms to make factoring easier: \[ (4x^2 - 10x) + (-10x + 25) = 0 \] Factor out the common factors in each group: \[ 2x(2x - 5) - 5(2x - 5) = 0 \] Notice that both terms have a common binomial factor \((2x - 5)\). Factor out \((2x - 5)\): \[ (2x - 5)(2x - 5) = 0 \] This can also be written as: \[ (2x - 5)^2 = 0 \]

Step 5: Solve for the Variable

Since \((2x - 5)^2 = 0\), set the inner factor equal to zero:\[ 2x - 5 = 0 \] Solve for \( x \) by adding 5 to both sides and then dividing by 2: \[ 2x = 5 \]\[ x = \frac{5}{2} \]

Step 6: Conclusion

The solution to the quadratic equation is that \( x = \frac{5}{2} \) is a repeated root.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. This formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula works for any quadratic equation in the standard form:\[ ax^2 + bx + c = 0 \]where \( a \), \( b \), and \( c \) are coefficients. It is especially useful when factoring is difficult or impossible. Diving into the formula, the part under the square root symbol, \( b^2 - 4ac \), is known as the discriminant. It tells us about the nature of the roots:

• If the discriminant is positive, we get two distinct real roots.
• If it's zero, we find exactly one real root, a repeated one.
• And if negative, no real roots exist, but instead two complex ones.

Understanding the quadratic formula requires practice, but mastering it opens up a straightforward path to solving quadratics.

Solving Quadratics

Solving quadratics involves finding the values of \( x \) that satisfy the equation. This can be approached in multiple ways, including:

• Factoring: When the quadratic expression can be rewritten as a product of two binomials. This is often the easiest method when applicable.
• Quadratic Formula: As discussed, it applies universally to any quadratic equation.
• Completing the Square: This method involves rearranging the equation to reveal a perfect square trinomial, making it easier to solve.

In our given example, \( 4x^2 - 20x + 25 = 0 \), the equation was solved by factoring. The quadratic was expressed as \((2x - 5)^2 = 0\), indicating a repeated root.Solving quadratics is a crucial skill, as these equations appear in various mathematical and real-world problems.

Algebraic Factoring

Algebraic factoring transforms a quadratic equation into a simpler expression that reveals its roots more easily. When factoring quadratics, we look for two numbers that multiply to the product of \( a \) and \( c \) (from \( ax^2 + bx + c \)) and add up to \( b \).For example, in \( 4x^2 - 20x + 25 \), we needed numbers that multiply to 100 (\(4 \times 25\)) and add to -20. Here, the numbers -10 and -10 satisfy both conditions.By rewriting the middle term with these numbers, grouping, and factoring out common terms, we end up with:\[ (2x - 5)(2x - 5) = 0 \], or \((2x - 5)^2 = 0\).If an expression resists factoring easily, other methods like the quadratic formula provide alternatives. Factoring, however, remains a preferred, fast technique for suitable equations.

Repeated Roots

When solving quadratic equations, discovering repeated roots can reveal unique characteristics about the graph or the physical phenomena modeled by the equation. A repeated root occurs when the quadratic can be expressed as a perfect square, like \((2x - 5)^2 = 0\).Repeated roots imply that the parabola touches the x-axis at exactly one point, instead of crossing it, illustrating a concept in calculus known as a "double root." This kind of solution informs us that not only does \( x = \frac{5}{2} \) satisfy the equation, but it's the only \( x \)-value that does. This scenario opens a wider understanding of quadratic equations, showing symmetry and how modifications to coefficients can shift or expand the graph.Identifying a repeated root helps learners see the deeper connections not just in solving problems, but in predicting behaviors in varied contexts.