Suppose that 𝑎+1𝑏a+ b1​  and 𝑏+1𝑎b+ a1​  are the roots of the equation 𝑥2−𝑝𝑥+𝑞=0x 2 −px+q=0. If 𝑎𝑏=1ab=1, what is the value of 𝑞q?

1. Break Down the Problem

We need to determine the value of $q$ given that $\frac{a+1}{b}$ and $\frac{b+1}{a}$ are the roots of the quadratic equation $x^2 - px + q = 0$ and $ab = 1$.

2. Relevant Concepts

For a quadratic equation of the form $x^2 - px + q = 0$:

• The sum of the roots $r_1 + r_2 = p$
• The product of the roots $r_1 \cdot r_2 = q$

Where $r_1 = \frac{a+1}{b}$ and $r_2 = \frac{b+1}{a}$.

3. Analysis and Detail

First, calculate the sum of the roots: $r_1 + r_2 = \frac{a+1}{b} + \frac{b+1}{a}$

Finding a common denominator: $r_1 + r_2 = \frac{(a+1)a + (b+1)b}{ab} = \frac{a^2 + a + b^2 + b}{ab}$

Since $ab = 1$: $r_1 + r_2 = a^2 + a + b^2 + b$

Next, calculate the product of the roots: $r_1 \cdot r_2 = \left(\frac{a+1}{b}\right) \left(\frac{b+1}{a}\right) = \frac{(a+1)(b+1)}{ab} = (a+1)(b+1)$

Again using $ab = 1$: $r_1 \cdot r_2 = ab + a + b = 1 + a + b$

4. Verify and Summarize

Using the results for the sum and product of the roots:

• From $r_1 + r_2 = p$ and $r_1 \cdot r_2 = q$:

For $q$: $q = 1 + a + b$

Since $ab = 1$ implies $b = \frac{1}{a}$, substitute $b$ into the equation: $q = 1 + a + \frac{1}{a}$

To simplify $1 + a + \frac{1}{a}$, we find a common expression: Let $x = a + \frac{1}{a}$, which achieves a minimum value of $2$ (by AM-GM inequality) when $a = 1$.

Thus:

• Minimum value occurs at $a = 1$, hence $q = 1 + 1 + 1 = 3$.

Final Answer

The value of $q$ is $3$.