Question

Form a quadratic equation, one of whose zero is \( 2 + \sqrt{5} \) and the sum of zeros is 4.

Hint:

For quadratic equations with real coefficients, if one zero is \( p + \sqrt{q} \), the other zero must be \( p - \sqrt{q} \).

Answer 1

Step 1: Use the sum and product of zeros.}
For a quadratic equation of the form \( ax^2 + bx + c = 0 \), the sum and product of its zeros are related to the coefficients as follows: \[ \text{Sum of zeros} = -\frac{b}{a}, \quad \text{Product of zeros} = \frac{c}{a} \] We are given: - One zero is \( 2 + \sqrt{5} \). - The sum of zeros is 4. Since the zeros of a quadratic equation with real coefficients must come in conjugate pairs, the other zero must be \( 2 - \sqrt{5} \).
Step 2: Calculate the product of the zeros.}
The product of the zeros is: \[ (2 + \sqrt{5})(2 - \sqrt{5}) = 2^2 - (\sqrt{5})^2 = 4 - 5 = -1 \]
Step 3: Write the quadratic equation.}
We now know: - Sum of zeros = \( 4 \) - Product of zeros = \( -1 \) Thus, the quadratic equation is: \[ x^2 - (\text{Sum of zeros})x + (\text{Product of zeros}) = 0 \] \[ x^2 - 4x - 1 = 0 \] % Final Answer Final Answer:
The required quadratic equation is \( x^2 - 4x - 1 = 0 \).