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Question
If α and β are zeroes of a given polynomial f(x)=x2−p(x+2)−c then find (α+2)(β+2)
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Solution
Given that,f(x)=x2−p(x+2)−c
=x2−px−2p−c
By comparing with ax2+bx+c we have,a=1,b=−p,c=−(2p+c)
Sum of zeroes (α+β)=−ba=−(−p)1=p
Product of zeroes (αβ)=ca=−(2p+c)1=−(2p+1)
Now (α+2)(β+2)=αβ+2(α+β)+4
=−(2p+1)+2(p)+4 =−2p−1+2p+4
=3
Ans: 3
f(x)=x2−p(x+2)−c
=x2−px−2p−c
By comparing with ax2+bx+c we have,
a=1,b=−p,c=−(2p+c)
Sum of zeroes (α+β)=−ba=−(−p)1=p
Product of zeroes (αβ)=ca=−(2p+c)1=−(2p+1)
Now (α+2)(β+2)=αβ+2(α+β)+4
=−(2p+1)+2(p)+4
=−2p−1+2p+4
=3
Ans: 3
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