Sure! Let's find the quadratic polynomial given that the sum of its roots is [tex]\( \sqrt{2} \)[/tex] and the product of its roots is [tex]\( \frac{1}{3} \)[/tex].
A quadratic polynomial can be written generally as:
[tex]\[ ax^2 + bx + c = 0 \][/tex]
When the polynomial is normalized (i.e., [tex]\(a = 1\)[/tex]), this becomes:
[tex]\[ x^2 + bx + c = 0 \][/tex]
The roots of this polynomial, say [tex]\( \alpha \)[/tex] and [tex]\( \beta \)[/tex], are related to the coefficients [tex]\( b \)[/tex] and [tex]\( c \)[/tex] by Vieta's formulas:
1. The sum of the roots [tex]\( (\alpha + \beta) \)[/tex] is equal to [tex]\(-b\)[/tex].
2. The product of the roots [tex]\( (\alpha \cdot \beta) \)[/tex] is equal to [tex]\( c \)[/tex].
Given:
- The sum of the roots [tex]\( \alpha + \beta = \sqrt{2} \)[/tex]
- The product of the roots [tex]\( \alpha \cdot \beta = \frac{1}{3} \)[/tex]
From Vieta's formulas:
1. [tex]\( \alpha + \beta = -b = \sqrt{2} \)[/tex] (we will solve for b)
2. [tex]\( \alpha \cdot \beta = c = \frac{1}{3} \)[/tex]
Now, solving for the coefficients [tex]\( b \)[/tex] and [tex]\( c \)[/tex]:
1. From [tex]\( \alpha + \beta = \sqrt{2} \)[/tex]:
[tex]\[ b = -\sqrt{2} \][/tex]
2. From [tex]\( \alpha \cdot \beta = \frac{1}{3} \)[/tex]:
[tex]\[ c = \frac{1}{3} \][/tex]
Therefore, substituting these values into the general form of the quadratic polynomial:
[tex]\[
x^2 + bx + c = x^2 - (\sqrt{2})x + \left( \frac{1}{3} \right)
\][/tex]
Hence, the required quadratic polynomial is:
[tex]\[
x^2 - \sqrt{2}x + \frac{1}{3}
\][/tex]
