Problem 29 - Algebra 2 and Trigonometry

Question

One root of the equation \(-x^{2}-11 x+c=0\) is \(\sqrt{3}\) a. Find the other root. b. Find the value of c. c. Explain why the roots of this equation are not conjugates.

Step-by-Step Solution

Answer

The other root is \(11 - \sqrt{3}\). The value of \(c\) is \(3 - 11\sqrt{3}\). The roots are not conjugates because they do not fit the form \(a \pm b\sqrt{d}\).

Step 1: Understanding the Roots

Given that one root of the equation \[-x^{2} - 11x + c = 0\] is \(\sqrt{3}\). By Vieta's formulas, the sum and product of the roots of a quadratic equation \(ax^2 + bx + c = 0\) are given by \(-\frac{b}{a}\) and \(\frac{c}{a}\) respectively.

Step 2: Using Vieta's Formulas

From Vieta’s formulas, we know the sum of the roots \(r_1 + r_2 = -\frac{-11}{-1} = 11\). So,\[\sqrt{3} + r_2 = 11\] Solving for \(r_2\), we find that \(r_2 = 11 - \sqrt{3}\).

Step 3: Finding the Other Root

If one root is \(\sqrt{3}\), the other root \(r_2\) is given by \[r_2 = 11 - \sqrt{3}\].

Step 4: Finding the Value of c

Using the product of the roots condition from Vieta’s formulas:\(\frac{c}{-1} = \sqrt{3}(11-\sqrt{3})\).Thus, \[c = - (11\sqrt{3} - 3) = 3 - 11\sqrt{3}\].

Step 5: Explanation of Non-Conjugate Roots

The roots of a quadratic equation are conjugates if they have the form \(a \pm b\sqrt{d}\). Here, the roots \(\sqrt{3}\) and \(11-\sqrt{3}\) do not fit this form, as they do not have the same coefficient in front of \(\sqrt{d}\) and do not have opposite sign.

Key Concepts

Vieta's Formulas

Vieta's Formulas offer a relation between the coefficients of a quadratic equation and its roots. This is especially useful when solving equations like \(-x^{2} - 11x + c = 0\).

The general quadratic equation is expressed as \(ax^2 + bx + c = 0\).The roots of this equation, often represented by \(r_1\) and \(r_2\), relate to the coefficients \(a, b,\) and \(c\) in specific ways:

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

Considering our specific equation \(-x^{2} - 11x + c = 0\), translated into the standard form as \(-1x^{2} - 11x + c = 0\),Vieta's Formulas become:

• Sum of roots: \(r_1 + r_2 = -\frac{-11}{-1} = 11\).
• Product of roots: \(r_1 \cdot r_2 = \frac{c}{-1}\).

Using these relationships, we can easily derive the other root, given that one root is \(\sqrt{3}\). This makes solving quadratic equations more systematic and simple.

Sum and Product of Roots

The relationship between the sum and product of the roots gives insightful shortcuts when solving quadratics.Given one root as \(\sqrt{3}\),we apply the sum formula:\(r_1 + r_2 = 11\).

In this problem:

• One root is known to be \(\sqrt{3}\).
• Thus, the equation becomes\(\sqrt{3} + r_2 = 11\).

Solving for \(r_2\), we find that \(r_2 = 11 - \sqrt{3}\).
For the product of the roots, given by \(r_1 \cdot r_2 = \frac{c}{-1}\),substitute the roots to find:\(\sqrt{3} \cdot (11 - \sqrt{3}) = \frac{c}{-1}\).Simplify to obtain \(c = 3 - 11\sqrt{3}\).This confirms the connections Vieta's Formulas offer between the coefficients of the equation and its roots.

Conjugate Roots

The idea of conjugate roots is tied to the characteristics of certain quadratic equations and their solutions.Roots are conjugates when they can be expressed as \(a \pm b\sqrt{d}\).

For an equation to have conjugate roots, the roots should

• Have the same whole number part (\(a\)).
• Possess opposite signs in their irrational part \(b\sqrt{d}\).

In our problem, the roots are \(\sqrt{3}\) and \(11 - \sqrt{3}\).These do not match the conjugate form because they don't share a common number in the form of a whole number or an opposite irrational component.
Hence, since \(\sqrt{3}\) is derived from the irrational number’s square root and the second root includes a different whole number (11),they are not conjugate.