Question

Hey mate!! Hope you are doing well!! Here's another question for you...... If alpha and ß are zeroes of the quadratic polynomial f(x) = x² - p(x+1) - c , then show that (alpha + 1) (ß+1) = 1 - c​

Answer 1

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If α and β are the zeroes of the polynomial $f(x) = {x}^{2} - p(x + 1) - c \: \: then \: show \: \: that \: \: (\alpha + 1)( \beta + 1) = (1 - c)$

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We know, for any polynomial of the form $a{x}^{2} \: + \: bx \: + \: c$

Sum of zeroes = $\frac{ - b}{a}$

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Product of zeroes = $\frac{c}{a}$

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Given polynomial :

${x}^{2} + p(x + 1) - c \\ = {x}^{2} + px + p - c$

Now according to sum,

Sum of zeroes ( α + β ) = $\frac{ - (p)}{1}$

Product of zeroes (α×β) = $\frac{p \: - \: c}{1}$

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

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

$= \frac{ - p}{1} + \frac{(p \: - \: c)}{1} + 1$

$= - p + p - c + 1$

$= 1 - c$

--------- ( Proved )