Problem 25 - Algebra 2 and Trigonometry

Question

One of the roots is given. Find the other root. \(z^{2}+2 z+k=0 ; \frac{3}{4}\)

Step-by-Step Solution

Answer

The other root is \(-\frac{11}{4}\) and the value of \(k\) is \(-\frac{33}{16}\).

Step 1: Understand the Roots' Relationship

When a quadratic equation is given by \(ax^2 + bx + c = 0\), if \(r_1\) and \(r_2\) are its roots, the sum and product of the roots can be determined by Vieta's formulas: \(r_1 + r_2 = -\frac{b}{a}\) and \(r_1 \cdot r_2 = \frac{c}{a}\). In this question, \(r_1 = \frac{3}{4}\) and we need to find \(r_2\).

Step 2: Apply Vieta's Formula for Sum of Roots

Use the formula for the sum of roots \(r_1 + r_2 = -\frac{b}{a}\). Here, \(a = 1\), \(b = 2\), so the equation becomes \(r_1 + r_2 = -2\). Substitute \(r_1 = \frac{3}{4}\) to find \(r_2\): \[\frac{3}{4} + r_2 = -2\]

Step 3: Solve for the Other Root

To isolate \(r_2\), subtract \(\frac{3}{4}\) from both sides of the equation:\[r_2 = -2 - \frac{3}{4}\]Convert \(-2\) into quarters to combine these:\[-2 = -\frac{8}{4}\]Thus, \[r_2 = -\frac{8}{4} - \frac{3}{4} = -\frac{11}{4}\]

Step 4: Verify the Product of the Roots

Use Vieta's product formula \(r_1 \cdot r_2 = \frac{c}{a}\). Substituting for \(r_1 = \frac{3}{4}\) and \(r_2 = -\frac{11}{4}\), we find:\[\frac{3}{4} \times -\frac{11}{4} = -\frac{33}{16}\]Since the product should equal \(\frac{k}{1}\), it follows \(k = -\frac{33}{16}\).

Key Concepts

Vieta's formulas

Vieta's formulas offer a simple yet powerful way to relate the coefficients of a quadratic equation to its roots. Given a quadratic equation in the form of \(ax^2 + bx + c = 0\), Vieta's formulas tell us:

• The sum of the roots \((r_1 + r_2)\) is equal to \(-\frac{b}{a}\).
• The product of the roots \((r_1 \cdot r_2)\) is \(\frac{c}{a}\).

These formulas provide an effective shortcut when working with quadratic equations because they allow you to find relationships between the roots without solving the equation directly.
In the given problem, we see Vieta’s formulas come into play when determining that for \(z^2 + 2z + k = 0\), the sum of the roots \(\frac{3}{4}\) and \(r_2\) is \(-2\), and their product is \(\frac{k}{1}\). This framework simplifies the process of finding the unknown coefficients and missing roots.

Sum and Product of Roots

Understanding the sum and product of the roots delivers a focused perspective when analyzing quadratic equations. The sum of the roots result \(-\frac{b}{a}\) highlights how root values can create a simple linear relationship with the coefficients of the quadratic.
For the equation \(z^2 + 2z + k = 0\), the root \(r_1 = \frac{3}{4}\) is given.

• Applying the sum formula \(r_1 + r_2 = -2\), we identify \(r_2\) by substituting \(r_1\).
• Through subtraction and conversion of fractions, we determine that \(r_2 = -\frac{11}{4}\).

Simultaneously, using the product of the roots \(\frac{k}{1}\), we check our work: the product \(\frac{3}{4} \times -\frac{11}{4} = -\frac{33}{16}\) confirms \(k\). These checks ensure that we have correctly understood the roots' interactions with the equation’s coefficients.

Solving Quadratic Equations

Solving quadratic equations typically involves finding where the quadratic expression equals zero—its roots. This can be done through factoring, completing the square, or using the quadratic formula.
In this problem, however, the sum and product knowledge from Vieta's formulas streamline the solving process substantially.

• First, identify the given coefficients and note the given root \(\frac{3}{4}\).
• Next, apply Vieta’s sum formula to find the other root, no traditional algebraic manipulation is necessary beyond basic arithmetic.
• Finally, confirm by using the product of roots form, which closes the loop by ensuring that both the derived sum and product align with the original equation’s coefficients.

This strategic approach reduces complexity and highlights how understanding relationships within a quadratic enhances solving ability and comprehension, making solving quadratic equations less daunting and more intuitive.