Question

Find a quadratic polynomial, the sum and product of whose zeroes are 3 and 2, respectively.

Answer 1

Let's denote the two zeroes (or roots) of the quadratic polynomial as x1 and x2. We know that their sum is 3 and their product is 2. Therefore:

x1 + x2 = 3 (Equation 1)

x1 * x2 = 2 (Equation 2)

We can use Equation 1 to find x2 in terms of x1:

x2 = 3 - x1

We can then substitute this expression for x2 into Equation 2:

x1 * (3 - x1) = 2

Expanding the left side, we get:

3x1 - x1^2 = 2

Rearranging and simplifying, we get:

x1^2 - 3x1 + 2 = 0

This is a quadratic polynomial in x1. We can factor it as:

(x1 - 1)(x1 - 2) = 0

Therefore, the two zeroes of the polynomial are x1 = 1 and x1 = 2 (which also means that x2 = 2 and x2 = 1, respectively).

Thus, the quadratic polynomial with the given properties is:

f(x) = (x - 1)(x - 2) = x^2 - 3x + 2

Answer 2

Let the quadratic polynomial be ax^2 + bx + c.

The sum of the zeroes of a quadratic equation is given by -b/a, and the product of the zeroes is given by c/a.

We are given that the sum of the zeroes is 3 and the product is 2.

So, we have:

-b/a = 3
c/a = 2

Multiplying the two equations, we get:

-bc/a^2 = 6

Rearranging, we get:

bc = -6a^2

We can choose any value for 'a', and solve for 'b' and 'c' using the given conditions.

Let's choose a = 1 for simplicity.

Using the equation bc = -6a^2, we get:

bc = -6

Now, using the condition that the sum of the zeroes is 3, we get:

-b/a = 3
-b = 3a
-b = 3

Solving for 'b', we get:

b = -3

Finally, using the condition that the product of the zeroes is 2, we get:

c/a = 2
c = 2a
c = 2

Therefore, the quadratic polynomial is:

x^2 - 3x + 2.