Question

\(\alpha\) and \(\beta\) are the roots of the equation \(x^{2}-3 x-16=0\). Then (a) \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\) are the roots of the equation \(16 x^{2}-41 x+16=0\) (b) \(\frac{\alpha+2}{\alpha-2}, \frac{\beta+2}{\beta-2}\) are the roots of the equation \(9 \mathrm{x}^{2}-20 \mathrm{x}+3=0\) (c) minimum value of the expression \(\left(x^{2}-3 x-16\right)\) is \(\frac{-73}{4}\) (d) The roots of the equation \((3 x+2)^{2}-3(3 x+2)-16=0\) are \(\frac{\alpha-2}{3}\) and \(\frac{\beta-2}{3}\)

Short answer

By analyzing each subproblem step-by-step, we have concluded the following: 1. The statement (a) is false. \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\) are not the roots of the equation \(16x^2 - 41x + 16= 0\). 2. The statement (b) cannot be proven nor disproven based on the given information. 3. The statement (c) is true. The minimum value of the expression \(x^2-3x-16\) is indeed \(-\frac{73}{4}\). 4. The statement (d) is true. The roots of the equation \((3x+2)^2-3(3x+2)-16=0\) are \(\frac{\alpha-2}{3}\) and \(\frac{\beta-2}{3}\).

Step-by-step solution

Find the sum and product of the roots of the first equation

The equation is \(x^2-3x-16=0\). Using Vieta's formulas, we know that \(\alpha+\beta=\frac{3}{1}=3\) and \(\alpha\beta=\frac{-16}{1}=-16\).

Determine the sum and product of \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\)

For sum: \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\). Since \(\alpha+\beta=3\) and \(\alpha\beta=-16\), then \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=3^2-2(-16)=9+32=41\). So, \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{41}{-16}=-\frac{41}{16}\). For product: \(\left(\frac{\alpha}{\beta}\right)\left(\frac{\beta}{\alpha}\right)=1\).

Compare with the given equation

Now, let's compare these with the sum and product of the roots for the equation \(16x^2-41x+16=0\). Using Vieta's formulas, we find that the sum of the roots is \(-\frac{-41}{16}=\frac{41}{16}\) and the product of the roots is \(\frac{16}{16}=1\). Therefore, the statement (a) is false. #### Subproblem (b)####

Determine the sum and product of \(\frac{\alpha+2}{\alpha-2}\) and \(\frac{\beta+2}{\beta-2}\)

For sum: \(\frac{\alpha+2}{\alpha-2}+\frac{\beta+2}{\beta-2}\). To prove or disprove this statement, we need to show whether this expression is equal to the sum of the roots of \(9x^2-20x+3=0\), which is \(-\frac{-20}{9}=\frac{20}{9}\). However, we cannot simplify this expression \(\frac{\alpha+2}{\alpha-2}+\frac{\beta+2}{\beta-2}\) further by using sum and product of the roots from the first equation only. Therefore, this statement neither can be proven nor disproven. #### Subproblem (c)####

Find the minimum value of the expression

The expression is \(x^2-3x-16\). To find its minimum value, we need to complete the square. \(x^2-3x-16=\left(x-\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2-16=\left(x-\frac{3}{2}\right)^2-\frac{73}{4}\), The minimum value occurs when \(\left(x-\frac{3}{2}\right)^2=0\), that is when \(x=\frac{3}{2}\). So, the minimum value of the expression is \(-\frac{73}{4}\), and therefore, the statement (c) is true. #### Subproblem (d)####

Identify the new equation's roots relationship

The given new equation is \((3x+2)^2-3(3x+2)-16=0\). We must show that its roots are related to \(\alpha\) and \(\beta\) in the form \(\frac{\alpha-2}{3}\) and \(\frac{\beta-2}{3}\).

Re-write the new equation in terms of x

Let \(y=3x+2\), then the new equation becomes \(y^2-3y-16=0\). Notice that this equation is same as the given equation \(x^2-3x-16=0\).

Replace x with the required expressions

The roots of the new equation are \(x_1=\frac{\alpha-2}{3}\) and \(x_2=\frac{\beta-2}{3}\). Then \(3x_1+2=\alpha\) and \(3x_2+2=\beta\). That means the sum of the new roots is \(\alpha+\beta=3\), and the product of the new roots is \(\alpha\beta=-16\). Since the equation \(y^2-3y-16=0\) has the same sum and products of the roots as \(x^2-3x-16=0\), the statement (d) is true.

Key concepts

Vieta's Formulas

Vieta's Formulas provide a way to relate the coefficients of a polynomial to sums and products of its roots. For a quadratic equation of the form \(ax^2 + bx + c = 0\), Vieta's Formulas state that:

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

These relationships are extremely helpful in solving problems related to quadratic equations as they allow us to derive properties of the roots without explicitly calculating them. For example, in the equation \(x^2 - 3x - 16 = 0\), we find \(\alpha + \beta = 3\) and \(\alpha\beta = -16\). This set of formulas not only simplifies finding roots but also aids in understanding the nature of quadratic equations and their solutions.

Roots of Quadratic Equation

The roots of a quadratic equation like \(ax^2 + bx + c = 0\) are the values of \(x\) that make the equation true. They can be found using various methods, one of which is the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(b^2 - 4ac\) is the discriminant. It determines the nature of the roots:

• If the discriminant is positive, the equation has two distinct real roots.
• If the discriminant is zero, there is exactly one real root, also known as a repeated or double root.
• If the discriminant is negative, the roots are complex and occur in conjugate pairs.

Finding the roots helps in understanding how the equation behaves graphically and forms the basis for further manipulations, such as finding the minimum or maximum values of the expression.

Completing the Square

Completing the square is a technique used to transform a quadratic expression into a perfect square trinomial, which can make solving or analyzing it easier. Consider the quadratic expression \(x^2 - 3x - 16\). Here's how you complete the square:

1. Start with the expression: \(x^2 - 3x - 16\).
2. Focus on the \(x^2 - 3x\) part. To complete the square, take half of the coefficient of \(x\), square it, and add/subtract it inside the expression:
- Half of \(-3\) is \(-\frac{3}{2}\), and squaring gives \(\left(-\frac{3}{2}\right)^2 = \frac{9}{4}\).
3. Rewrite the expression as \(\left(x - \frac{3}{2}\right)^2 - \frac{73}{4}\).

This process shows the expression's minimum value occurs, where the squared term is zero, i.e., \(x = \frac{3}{2}\). Completing the square can make finding minima or maxima more intuitive, particularly when working with parabolas.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to make equations easier to solve or analyze. It includes techniques like factoring, expanding, and simplifying fractions. Consider the expression \((3x + 2)^2 - 3(3x + 2) - 16\). By letting \(y = 3x + 2\), we convert it into a simpler quadratic form \(y^2 - 3y - 16\).

This allows us to see the parallels between the new equation and the original form \(x^2 - 3x - 16 = 0\). This transformation helps to reveal insights about the roots and facilitates solving by shifting focus to equivalent relationships. Algebraic manipulation is invaluable for exploring different perspectives on a problem and is often used in conjunction with other key concepts like Vieta’s Formulas or completing the square to solve quadratic equations.