1. To find a quadratic polynomial with the given zeroes 5-√3 and 5+√3, we use the fact that if α and β are the zeroes of a quadratic polynomial ax² + bx + c, then the polynomial can be written as a(x-α)(x-β).
Given that the zeroes are 5-√3 and 5+√3, the factors of the polynomial are (x - (5-√3))(x - (5+√3)). Expanding this expression gives us the quadratic polynomial in the standard form.
(x - (5-√3))(x - (5+√3)) = (x - 5 + √3)(x - 5 - √3)
= x² - 5x - √3x - 5x + 25 + 5√3 + √3x - 5√3 - 3
= x² - 10x + 22
Therefore, the quadratic polynomial with zeroes 5-√3 and 5+√3 is x² - 10x + 22.
2. Given that the zeroes of the polynomial x² + 4x + 2a are a and 1/a, and the sum and product of the zeroes are 21/8, we can use Vieta's formulas. Vieta's formulas state that for a quadratic polynomial ax² + bx + c, the sum of the zeroes is -b/a and the product of the zeroes is c/a.
In this case, the sum of the zeroes is a + 1/a = -4, and the product of the zeroes is a * (1/a) = 1 = 2a. Therefore, we have two equations:
1. a + 1/a = -4
2. 2a = 1
From equation 2, we find that a = 1/2. Substituting this value back into equation 1, we get:
1/2 + 2 = -4
1/2 - 8/2 = -4
-7/2 = -4
7/2 = 4
Therefore, the value of a is 1/2.
