If α,β are the two zeros of the polynomial fy=y2-8y+a and α2+β2=40. Find the value of a.

Given that,

f (y) = y^{2} - 8 y + a

and

α

and

β

are the zeroes of the polynomial f(x),

α^{2} + β^{2} = 40

.
Comparing the given equation with

a y^{2} + by + c

a = 1

b = - 8

c = a

Now, sum of roots is,

α + β = \frac{- b}{a}

\Rightarrow α + β = \frac{- (- 8)}{1}

\Rightarrow α + β = 8

And, product of roots is,

αβ = \frac{c}{a}

\Rightarrow αβ = \frac{a}{1}

\Rightarrow αβ = a

Now, using the identity,

α^{2} + β^{2} = {(α + β)}^{2} - 2 αβ

Substituting the values,

40 = 8^{2} - 2 a

2a = 24
Thus, the value of a = 12.
Hence, option 2 is correct.

If $α, β$ are the two zeros of the polynomial $f (y) = y^{2} - 8 y + a$ and $α^{2} + β^{2} = 40$ . Find the value of a.