Question

If α and β are the zeroes of the quadratic polynomial f(t) = t²-p(t+1) -c , then show that (α+1)(β+1) = 1 - c. ​

Answer 1

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Answer 2

f(t)= $t^{2} - p(t+1) - c$

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opening the bracket we get:-

f(t)=  $t^{2} -pt- p - c$

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now we know that the expression of a quadratic polynomial is as follows:-

p(x)= $ax^{2} +bx+c$

Thus,

Making the given polynomial in that form:-

f(t)=  $t^{2} -pt- (p + c)$

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now,

here we have :-

a=1

b=(-p)

c= -(p+c)

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we know that:-

sum of zeroes =$\boxed{\dfrac{-b}{a}}$

product of zeroes=$\boxed{\dfrac{c}{a}}$

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we have :-

$(\alpha +1)(\beta +1) \\ \\ \\= \alpha \beta + \alpha + \beta +1$

now putting the formula of sum and zeroes:-

$(\alpha +1)(\beta +1) \\ \\ \\= \alpha \beta + \alpha + \beta +1 \\ \\ \\ = \dfrac{-(p+c)}{1} + \dfrac{p}{1} +1 \\ \\ \\= (-p)+(-c)+p+1 \\ \\ \\=1-c (RHS)$

$\blue{\bold{\sf{HENCE\:PROVED}}}$

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