Problem 12 - Algebra for College Students

Question

Solve each quadratic equation by using (a) the factoring method and (b) the method of completing the square. $$2 n^{2}+7 n-4=0$$

Step-by-Step Solution

Answer

Solutions: \(n = \frac{1}{2}\) and \(n = -4\).

Step 1: Factoring - Rearrange the Equation

First, ensure the quadratic equation is written in standard form: \(2n^2 + 7n - 4 = 0\). The equation is already arranged this way.

Step 2: Factoring - Factor the Quadratic Expression

To factor the expression \(2n^2 + 7n - 4\), we seek two numbers that multiply to \(2 \times -4 = -8\) and add to \(7\). These numbers are \(8\) and \(-1\). Start by rewriting the middle term as a sum: \(2n^2 + 8n - n - 4 = 0\).

Step 3: Factoring - Group Terms

Group the terms into pairs: \((2n^2 + 8n) + (-n - 4) = 0\).

Step 4: Factoring - Factor by Grouping

Factor out the greatest common factor from each group: \(2n(n + 4) - 1(n + 4) = 0\).

Step 5: Factoring - Solve the Factored Equation

Factor out \(n + 4\): \((2n - 1)(n + 4) = 0\). Solve each factor for zero: \(2n - 1 = 0\) or \(n + 4 = 0\). This gives the solutions \(n = \frac{1}{2}\) and \(n = -4\).

Step 6: Completing the Square - Start with the Original Equation

Return to the original equation: \(2n^2 + 7n - 4 = 0\). First, divide all terms by 2 to ease the process of completing the square: \(n^2 + \frac{7}{2}n = 2\).

Step 7: Completing the Square - Make the Left Side a Perfect Square

Add \(\left(\frac{7}{4}\right)^2 = \frac{49}{16}\) to both sides: \(n^2 + \frac{7}{2}n + \frac{49}{16} = 2 + \frac{49}{16}\). This rewrites as \((n + \frac{7}{4})^2 = \frac{81}{16}\).

Step 8: Completing the Square - Solve for n

Take the square root of both sides: \(n + \frac{7}{4} = \pm \frac{9}{4}\). Solve for \(n\): \(n = \frac{9}{4} - \frac{7}{4} = \frac{1}{2}\) or \(n = -\frac{9}{4} - \frac{7}{4} = -4\).

Step 9: Final Conclusion

The solutions to the equation \(2n^2 + 7n - 4 = 0\) are \(n = \frac{1}{2}\) and \(n = -4\).

Key Concepts

Factoring Method

The factoring method is a straightforward approach used to solve quadratic equations like \(2n^2 + 7n - 4 = 0\). Initially, ensure the equation is in standard form, which it is in this example. The next step involves breaking it into simpler expressions multiplied together, allowing you to solve for the variable \(n\).

To factor \(2n^2 + 7n - 4\), we need numbers that multiply to \(-8\) (the product of the leading coefficient and the constant term) and add up to \(7\) (the coefficient of the middle term). Here, those numbers are \(8\) and \(-1\).

We then rewrite the quadratic equation: substitute the middle term \(7n\) with \(8n - n\):
\[(2n^2 + 8n) + (-n - 4) = 0\]
Next, group the terms and factor them:
\[2n(n + 4) - 1(n + 4) = 0\]
Finally, factor by grouping:
\[(2n - 1)(n + 4) = 0\]
Solving these factors leads to the solutions:\(n = \frac{1}{2}\) and \(n = -4\).

This method is practical when the quadratic expression can be easily factored into a product of binomials.

Completing the Square

Completing the square is another method to solve quadratic equations by transforming the expression into a perfect square trinomial. Starting with the equation \(2n^2 + 7n - 4 = 0\), the first step is dividing all terms by 2 to make it simpler: \(n^2 + \frac{7}{2}n = 2\).

To complete the square, add \(\left(\frac{7}{4}\right)^2\) to both sides, which balances the equation while transforming the left side into a perfect square trinomial.

Here's the breakdown:

• Add \(\frac{49}{16}\) to both sides:
• Equates to \[(n + \frac{7}{4})^2 = 2 + \frac{49}{16}\]
• Simplifies to \[(n + \frac{7}{4})^2 = \frac{81}{16}\]

With the equation now in perfect square form, take the square root of both sides:
\[n + \frac{7}{4} = \pm \frac{9}{4}\]
Finally, solve for \(n\):

• \(n = \frac{9}{4} - \frac{7}{4} = \frac{1}{2}\)
• or \(n = -\frac{9}{4} - \frac{7}{4} = -4\)

Thus, the solutions to the equation are \(n = \frac{1}{2}\) and \(n = -4\).

Completing the square is especially helpful when the quadratic does not easily factor, providing precise results.

Solving Quadratic Equations

Solving quadratic equations involves finding the values of the variable \(n\) that make the equation true. Quadratic equations generally take the form \(ax^2 + bx + c = 0\). Understanding different techniques like the factoring method and completing the square is essential.

Both methods aim to simplify the quadratic equation, making it easier to identify the solutions where the function equals zero. When using these methods:

• Factoring relies on breaking down the quadratic into products of binomials.
• Completing the square modifies the equation to identify its vertex form, then finds the roots via simplification.

Each technique has its strengths; factoring is often quick if the numbers are manageable, while completing the square is useful for more complicated expressions where factoring isn't straightforward.

Ultimately, learning both enables you to tackle a wide range of quadratic problems effectively, equipping you with robust mathematical tools.

Factoring by Grouping

Factoring by grouping is a specific strategy used within the factoring method, especially when dealing with quadratic expressions that are hard to factor at first glance, like \(2n^2 + 7n - 4\).

To apply this method, the key is to rearrange the quadratic in such a way that common factors in grouped terms become evident. Here's how it works for \(2n^2 + 7n - 4\):

• Rewrite \(7n\) as \(8n - n\).
• Group the expression into \((2n^2 + 8n) + (-n - 4)\).
• Factor out common factors from each group:
• \[2n(n + 4) - 1(n + 4)\]

After grouping, notice \(n + 4\) appears in both sub-expressions, allowing further factoring into:
\[(2n - 1)(n + 4) = 0\]
This reveals the roots upon solving the factors: \(n = \frac{1}{2}\) and \(n = -4\).

Factoring by grouping is efficient for quadratics where simple factoring is not initially apparent, offering a step-by-step decomposition to achieve the solution.