Answer:
Step-by-step explanation:
ax²+bx+c=0
Therefore α+β=-b/a
αβ=c/a
Now α²/β+β²/α
=(α³+β³)/αβ
={(α+β)³-3αβ(α+β)}/αβ
={(-b/a)³ -3×(c/a)×(-b/a)}/(c/a)
={-b³/a³+3bc/a²}/(c/a)
=(3abc-b³/a³)×(a/c)
=(3abc-b³)/a²c
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Answers & Comments
Let alpha=a
". beta=b
1/a^2+1/b^2
b^2+a^2÷a^2b^2
=(a+b)^2-2ab÷c.
=(-b)^2-2c÷c
=b^2-2c÷c
Answer:
Step-by-step explanation:
ax²+bx+c=0
Therefore α+β=-b/a
αβ=c/a
Now α²/β+β²/α
=(α³+β³)/αβ
={(α+β)³-3αβ(α+β)}/αβ
={(-b/a)³ -3×(c/a)×(-b/a)}/(c/a)
={-b³/a³+3bc/a²}/(c/a)
=(3abc-b³/a³)×(a/c)
=(3abc-b³)/a²c