If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ -\ p(x\ +\ 1)\ –\ c$, show that $(α\ +\ 1)(β\ +\ 1)\ =\ 1\ -\ c$.

Given:

$α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ -\ p(x\ +\ 1)\ –\ c$.

To do:

Here, we have to prove that $(α\ +\ 1)(β\ +\ 1)\ =\ 1\ -\ c$.

Solution: 

$f(x)\ =\ x^2\ -\ p(x\ +\ 1)\ –\ c$ can be written as $f(x)=x^2-px-(p+c)$.

 We know that, 

The standard form of a quadratic equation is $ax^2+bx+c=0$, where a, b and c are

constants and $a≠0$

Comparing the given equation with the standard form of a quadratic equation, 

$a=1$, $b=-p$ and $c=-(p+c)$

Sum of the roots $= α+β = \frac{-b}{a} = \frac{-(-p)}{1}=p$.

Product of the roots $= αβ = \frac{c}{a} = \frac{-(p+c)}{1}=-(p+c)$.

LHS

$ \begin{array}{l}
( \alpha +1)( \beta +1) =\alpha \beta +\alpha +\beta +1\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\alpha \beta +( \alpha +\beta ) +1\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-( p+c) +p+1\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-p-c+p+1\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1-c
\end{array}$

RHS$=1-c$

LHS$=$RHS

Hence proved.

Tutorialspoint

Updated on: 10-Oct-2022

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