Problem 44 - Algebra A Combined Function

Question

Solve. See Examples 1 through 5. $$ (2 n-3)^{2}-7(2 n-3)+12=0 $$

Step-by-Step Solution

Answer

The solutions are \(n = \frac{7}{2}\) and \(n = 3\).

Step 1: Recognize the Quadratic Form

We recognize that the given equation \((2n-3)^2 - 7(2n-3) + 12 = 0\) is similar to a quadratic equation in the standard form \(ax^2 + bx + c = 0\). Let \(u = 2n-3\). The equation becomes a simpler quadratic \(u^2 - 7u + 12 = 0\).

Step 2: Apply the Quadratic Formula

For the equation \(u^2 - 7u + 12 = 0\), compare with \(au^2 + bu + c = 0\) to find \(a = 1\), \(b = -7\), and \(c = 12\). Use the quadratic formula: \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute the values:\[ u = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \times 1 \times 12}}{2 \times 1} \] This simplifies to:\[ u = \frac{7 \pm \sqrt{49 - 48}}{2} \]\[ u = \frac{7 \pm \sqrt{1}}{2} \] Thus, \[ u = \frac{7 + 1}{2} \] or \[ u = \frac{7 - 1}{2} \] which gives \(u = 4\) or \(u = 3\).

Step 3: Substitute Back to Find n

We substitute values of \(u\) back to find \(n\). For \(u = 4\):\[ 2n - 3 = 4 \] \[ 2n = 7 \] \[ n = \frac{7}{2} \]For \(u = 3\):\[ 2n - 3 = 3 \] \[ 2n = 6 \] \[ n = 3 \]

Step 4: Verify the Solutions

Substituting \(n = \frac{7}{2}\) and \(n = 3\) back into the original equation verifies the correctness:- For \(n = \frac{7}{2}\), \((2 \times \frac{7}{2} - 3)^2 = 4\) satisfies the equation.- For \(n = 3\), \((2 \times 3 - 3)^2 = 9\) satisfies the equation. Both are correct solutions.

Key Concepts

Quadratic Formula

The quadratic formula is a universal tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). It provides a method to find solutions even when an equation cannot be easily factored. The formula is written as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where:

• \( a \), \( b \), and \( c \) are coefficients from the equation.
• \( b^2 - 4ac \) is called the discriminant, which helps determine the nature of the roots.

To use the quadratic formula, simply substitute the values of \( a \), \( b \), and \( c \) into the formula and solve for \( x \). In our example, substituting \( a = 1 \), \( b = -7 \), and \( c = 12 \) gives us:\[ u = \frac{7 \pm \sqrt{1}}{2} \]resulting in the solutions \( u = 4 \) or \( u = 3 \). The values of \( u \) must be further processed to find \( n \).

Completing the Square

Completing the square is an algebraic method used to solve quadratic equations by turning them into a perfect square trinomial. Although the quadratic formula is more commonly used, understanding completing the square gives insight into the derivation of the quadratic formula itself.Here's a quick overview of the method:

• Start with a quadratic equation in standard form \( ax^2 + bx + c = 0 \).
• Ensure \( a \) (the coefficient of \( x^2 \)) is 1. If not, divide through by \( a \).
• Move the constant term, \( c \), to the other side of the equation.
• Add \( \left(\frac{b}{2}\right)^2 \) to both sides to form a perfect square on one side.
• Factor the perfect square trinomial and solve for \( x \).

For example, given the quadratic \( u^2 - 7u + 12 = 0 \), it could be rearranged and manipulated:\[ u^2 - 7u = -12 \]Then make it a complete square by:\[ u^2 - 7u + \left(\frac{7}{2}\right)^2 = \frac{49}{4} -12 \]

Solving Quadratics by Factoring

Factoring is a straightforward method to solve quadratic equations, particularly when the equation can be neatly expressed as a product of binomials. However, not all quadratics can be easily factored, which is why other methods like the quadratic formula and completing the square are often employed.To solve a quadratic equation by factoring:

• Rearrange the equation into the standard form \( ax^2 + bx + c = 0 \).
• Identify two numbers that multiply to \( ac \) and add to \( b \).
• Rewrite the middle term using these two numbers and factor by grouping.
• Set each factor equal to zero and solve for \( x \).

For instance, for \( u^2 - 7u + 12 = 0 \), it can be factored as:\[ (u - 4)(u - 3) = 0 \]This directly gives \( u = 4 \) or \( u = 3 \), demonstrating how factoring provides a quick resolution when applicable.