if a and b are zeros of x^2+7x + 121 . find a^2 + b^2
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Answer:
Given that,
α,β are the zeroes of the polynomial p(x)=2x
2
−7x+3
Sum of zeroes (α+β)=
a
−b
=
2
−(−7)
=
2
7
Product of zeroes αβ=
a
c
=
2
3
∴α
2
+β
2
=(α+β)
2
−2αβ
=(
2
7
)
2
−2(
2
3
)
=
4
49
−
4
12
=
4
37
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