Solve the math problems in the attached image.

Steps to Solve

Here's how to solve each of the three questions:

Question 1: If $\alpha$ and $\beta$ are the zeros of the polynomial $f(x) = x^2 + px + q$, then the polynomial having $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ as its zeros is

1. Find the sum and product of the new zeros

The new zeros are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$. We need to find their sum and product:

Sum: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}$ Product: $\frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha \beta}$

2. Relate the original zeros to the coefficients of the polynomial

For the polynomial $f(x) = x^2 + px + q$, we know that the sum of the zeros is $-\frac{p}{1} = -p$ and the product of the zeros is $\frac{q}{1} = q$. So, $\alpha + \beta = -p$ and $\alpha \beta = q$.

3. Substitute the relationships into the sum and product of new roots

Sum of new zeros: $\frac{\alpha + \beta}{\alpha \beta} = \frac{-p}{q}$ Product of new zeros: $\frac{1}{\alpha \beta} = \frac{1}{q}$

4. Form the new polynomial

A polynomial with zeros $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ can be written as: $x^2 - (\text{sum of zeros})x + (\text{product of zeros}) = 0$ $x^2 - (\frac{-p}{q})x + \frac{1}{q} = 0$ $x^2 + \frac{p}{q}x + \frac{1}{q} = 0$

Multiplying by $q$ to eliminate fractions, we get: $qx^2 + px + 1 = 0$

Thus, the required polynomial is $qx^2 + px + 1$.

Question 2: If $\alpha, \beta$ are the zeros of the polynomial $f(x) = x^2 - p(x+1) - c$, then $(\alpha + 1)(\beta + 1) = $

1. Rewrite the polynomial in standard form

First, rewrite $f(x)$ as $f(x) = x^2 - px - p - c = x^2 - px - (p+c)$.

2. Relate the zeros to the coefficients

From the polynomial $f(x)$, we have: $\alpha + \beta = p$ $\alpha \beta = -(p+c)$

3. Expand the expression and substitute

Now, let's find $(\alpha + 1)(\beta + 1)$: $(\alpha + 1)(\beta + 1) = \alpha \beta + \alpha + \beta + 1$ Substitute the values we found: $= -(p+c) + p + 1$ $= -p - c + p + 1$ $= 1 - c$ So, $(\alpha + 1)(\beta + 1) = 1 - c$.

Question 3: If $f(x)=ax^2 + bx + c$ has no real zeros and $a + b + c < 0$, then

1. Consider the condition of no real zeros

If the quadratic has no real zeros, then its discriminant must be negative. Discriminant, $D = b^2 - 4ac < 0$.

2. Analyze $a + b + c < 0$

Notice that $f(1) = a(1)^2 + b(1) + c = a + b + c$. Since $a + b + c < 0$, we have $f(1) < 0$.

3. Consider the sign of $a$

Since $f(x)$ has no real zeros, the parabola does not intersect the x-axis. Also, $f(1)<0$. This means that the parabola must open downwards, because if it opened upwards, $f(x)$ would have to cross the x-axis to have $f(1) < 0 $ which violates the problem condition of no real zeros i.e. no intersection with x-axis at all. Therefore, $a < 0$.

4. Determine the sign of $c$

Since $a < 0$ and $b^2 - 4ac < 0$, we have $b^2 < 4ac$. Since $a < 0$, for $4ac$ to be greater than $b^2$, $c$ must also be negative. If $c$ was positive then $b^2 < 4ac$ would not hold because the sign of the expression is negative. If $c$ was zero then $b^2 < 4ac$ would not hold because $b^2$ is always positive or zero.

Thus, $c < 0$.