Question

Find a quadratic polynomial whose zeroes are $(5 - 2\sqrt{3})$ and $(5 + 2\sqrt{3})$.

Hint:

Notice that irrational zeroes of polynomials with rational coefficients always occur in conjugate pairs ($a - \sqrt{b}$ and $a + \sqrt{b}$).
This guarantees that their sum and product are always rational numbers.

Answer 1

Step 1: Understanding the Question:
We are given two roots (zeroes) of a quadratic polynomial: $\alpha = 5 - 2\sqrt{3}$ and $\beta = 5 + 2\sqrt{3}$.
We need to construct the corresponding quadratic polynomial.

Step 2: Key Formula or Approach:
A quadratic polynomial with zeroes $\alpha$ and $\beta$ can be written in the form:
\[ p(x) = k[x^2 - S x + P] \]
where:
- $S$ is the sum of the zeroes ($S = \alpha + \beta$).
- $P$ is the product of the zeroes ($P = \alpha \times \beta$).
- $k$ is a non-zero real constant (usually taken as 1 for simplicity).

Step 3: Detailed Explanation:

• Define the two zeroes:
\[ \alpha = 5 - 2\sqrt{3} \]
\[ \beta = 5 + 2\sqrt{3} \]

• Calculate the sum of the zeroes ($S$):
\[ S = \alpha + \beta \]
\[ S = (5 - 2\sqrt{3}) + (5 + 2\sqrt{3}) \]
Combine the rational and irrational terms:
\[ S = 5 + 5 - 2\sqrt{3} + 2\sqrt{3} \]
\[ S = 10 \]

• Calculate the product of the zeroes ($P$):
\[ P = \alpha \times \beta \]
\[ P = (5 - 2\sqrt{3})(5 + 2\sqrt{3}) \]
Apply the difference of squares identity $(a - b)(a + b) = a^2 - b^2$:
\[ P = (5)^2 - (2\sqrt{3})^2 \]
\[ P = 25 - (4 \times 3) \]
\[ P = 25 - 12 = 13 \]

• Formulate the quadratic polynomial:
Substitute the sum ($S = 10$) and product ($P = 13$) into the polynomial template:
\[ p(x) = x^2 - S x + P \]
\[ p(x) = x^2 - 10x + 13 \]


Step 4: Final Answer:
The required quadratic polynomial is $x^2 - 10x + 13$.