Question

$If \: \alpha \:and \: \beta \: are \: the \: zeros \: of \: the \: quadratic \: polynomial \: f(x) = = t² - p(t + 1)- c \: show \: that \: ( \alpha + 1)( \beta + 1) = 1 - c$
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Answer 1

Answer:

$</p><p>if \alpha and \beta are zeros of polyn. t^2 -p(t+1)-c</p><p>then</p><p>(t-\alpha)(t-\beta)=t^2-p(t+1)-c</p><p>or t^2 -(\alpha +\beta)t +\alpha\beta=t^2-pt-(p+c)</p><p>\implies \alpha+\beta = p ; \alpha\beta=-(p+c)</p><p>so</p><p>(\alpha +1)(\beta +1)</p><p>=\alpha\beta +(\alpha+\beta)+1</p><p>=-p-c+p+1</p><p>=1-c</p><p>$

Answer 2

Answer:

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

Step-by-step explanation:

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