If alpha and Beeta are zeros of the polynomial x2-p(x+1)-c.show that (alpha+1)(Beeta+1)=1-c
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Answered by
536
given
x² - p(x+1)-c = x² -px-p -c
compare it with ax²+bx+c =0
a= 1, b= -p , c= -p-c
α and β are two zeroes
i) sum of the zeros= -b/a
α+β = - (-p)/1= p -----(1)
ii) product of the zeroes = c/a
αβ = (-p-c) /1 = -p-c -----(2)
now take
lhs = (α+1)(β+1)
= α(β+1) +1(β+1)
=αβ +α +β +1
=-p-c+p+1 [from (2) and (1) ]
= -c+1
=1 -c
=RHS
x² - p(x+1)-c = x² -px-p -c
compare it with ax²+bx+c =0
a= 1, b= -p , c= -p-c
α and β are two zeroes
i) sum of the zeros= -b/a
α+β = - (-p)/1= p -----(1)
ii) product of the zeroes = c/a
αβ = (-p-c) /1 = -p-c -----(2)
now take
lhs = (α+1)(β+1)
= α(β+1) +1(β+1)
=αβ +α +β +1
=-p-c+p+1 [from (2) and (1) ]
= -c+1
=1 -c
=RHS
mysticd:
7+q= 18
q= 18-7
q= 11
therefore p=7, q= 11
required polynomial is in the form x^2 - (p+q)x+pq
=x^2-18x+77
i hope it will be useful to u
Tnks a loot
U r a life saver
best of luck
Answered by
127
Answer:
Step-by-step explanation:
given
x² - p(x+1)-c = x² -px-p -c
compare it with ax²+bx+c =0
a= 1, b= -p , c= -p-c
α and β are two zeroes
i) sum of the zeros= -b/a
α+β = - (-p)/1= p -----(1)
ii) product of the zeroes = c/a
αβ = (-p-c) /1 = -p-c -----(2)
now take
lhs = (α+1)(β+1)
= α(β+1) +1(β+1)
=αβ +α +β +1
=-p-c+p+1 [from (2) and (1) ]
= -c+1
=1 -c
=RHS
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