To solve this problem, we start by using the properties of the roots of the polynomial [tex]\( f(y) = y^2 - 8y + a \)[/tex].
Given:
- [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] are the zeros/roots of the polynomial, satisfying [tex]\( \alpha^2 + \beta^2 = 40 \)[/tex].
### Step 1: Use Vieta's formulas for the sums and products of the roots
For any polynomial of the form [tex]\( ay^2 + by + c \)[/tex], the following properties hold true:
- The sum of the roots, [tex]\( \alpha + \beta \)[/tex], is given by [tex]\( -\frac{b}{a} \)[/tex].
- The product of the roots, [tex]\( \alpha\beta \)[/tex], is given by [tex]\( \frac{c}{a} \)[/tex].
In our specific polynomial [tex]\( f(y) = y^2 - 8y + a \)[/tex]:
- [tex]\( \alpha + \beta = -(-8)/1 = 8 \)[/tex].
- [tex]\( \alpha\beta = a \)[/tex].
### Step 2: Use the given condition
We are given that:
[tex]\[ \alpha^2 + \beta^2 = 40 \][/tex].
### Step 3: Express [tex]\(\alpha^2 + \beta^2\)[/tex] in a different form
We recall that:
[tex]\[ (\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 \][/tex]
Substituting the known sum of the roots:
[tex]\[ 8^2 = \alpha^2 + 2\alpha\beta + \beta^2 \][/tex]
[tex]\[ 64 = \alpha^2 + \beta^2 + 2\alpha\beta \][/tex]
### Step 4: Substitute the value for [tex]\(\alpha^2 + \beta^2\)[/tex]
Given that:
[tex]\[ \alpha^2 + \beta^2 = 40 \][/tex]
We can substitute this into our expanded sum of roots equation:
[tex]\[ 64 = 40 + 2\alpha\beta \][/tex]
### Step 5: Solve for [tex]\(\alpha\beta\)[/tex] (which is [tex]\( a \)[/tex])
[tex]\[ 64 - 40 = 2\alpha\beta \][/tex]
[tex]\[ 24 = 2\alpha\beta \][/tex]
[tex]\[ \alpha\beta = 12 \][/tex]
### Final result
Therefore, the value of [tex]\( a \)[/tex] is:
[tex]\[ a = 12 \][/tex]
So, the value of [tex]\( a \)[/tex] is [tex]\( \boxed{12} \)[/tex].
