Question

If \[\alpha ,\beta \] are the roots of ${x^2} - p(x + 1) - c = 0$ . Show that
(i) $(\alpha + 1)(\beta + 1) = 1 - c$
(ii) $\dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + 2\beta + c}} = 1$

Answer 1

Hint: First apply that if the quadratic equation is $a{x^2} + bx + d = 0$ then , Sum of roots is equal to $\dfrac{{ - b}}{a}$ and product of roots is $\dfrac{d}{a}$ then for part (i) $(\alpha + 1)(\beta + 1) = 1 - c$ expand the LHS term and put the value of $\alpha + \beta = \dfrac{{ - ( - p)}}{1}$ and $\alpha \times \beta = - (p + c)$ .For part (ii) Convert it into $\dfrac{{{{(\alpha + 1)}^2}}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{{(\beta + 1)}^2}}}{{{\beta ^2} + 2\beta + c}}$ and then put the value of c and $\alpha + \beta = \dfrac{{ - ( - p)}}{1}$ and $\alpha \times \beta = - (p + c)$ for proving this .

Complete step-by-step answer:
It is given that \[\alpha ,\beta \] are the roots of ${x^2} - p(x + 1) - c = 0$ ,
So we know that if the quadratic equation is $a{x^2} + bx + d = 0$ then , Sum of roots is equal to $\dfrac{{ - b}}{a}$ and product of roots is $\dfrac{d}{a}$
So on comparing with the real equation we get $a = 1,b = - p,d = - (p + c)$
then we can write it as , $\alpha + \beta = \dfrac{{ - ( - p)}}{1}$ and $\alpha \times \beta = - (p + c)$
So now from the part (i) $(\alpha + 1)(\beta + 1) = 1 - c$
From LHS , $(\alpha + 1)(\beta + 1) = (\alpha \times \beta ) + \alpha + \beta + 1$
Hence we know that the value of $\alpha + \beta = \dfrac{{ - ( - p)}}{1}$ and $\alpha \times \beta = - (p + c)$
on putting in the above equation we get ,
$ - (p + c) + p + 1$
On solving we get , $1 - c$ = RHS
hence proved ,
Now for the part (ii) that is $\dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + 2\beta + c}} = 1$
From LHS , $\dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + 2\beta + c}}$ we know that ,
${\alpha ^2} + 2\alpha + 1 = {(\alpha + 1)^2}$ and ${\beta ^2} + 2\beta + 1 = {(\beta + 1)^2}$
$\dfrac{{{{(\alpha + 1)}^2}}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{{(\beta + 1)}^2}}}{{{\beta ^2} + 2\beta + c}}$
Now from above we prove that the $(\alpha + 1)(\beta + 1) = 1 - c$ from this the value of c we get it as
$c = - \alpha - \beta - \alpha \beta $ now put it this value in the equation $\dfrac{{{{(\alpha + 1)}^2}}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{{(\beta + 1)}^2}}}{{{\beta ^2} + 2\beta + c}}$ we get as
$\Rightarrow$ $\dfrac{{{{(\alpha + 1)}^2}}}{{{\alpha ^2} + 2\alpha - \alpha - \beta - \alpha \beta }} + \dfrac{{{{(\beta + 1)}^2}}}{{{\beta ^2} + 2\beta - \alpha - \beta - \alpha \beta }}$ ,
On solving ,
$\Rightarrow$ \[\dfrac{{{{(\alpha + 1)}^2}}}{{{\alpha ^2} + \alpha - \beta - \alpha \beta }} + \dfrac{{{{(\beta + 1)}^2}}}{{{\beta ^2} + \beta - \alpha - \alpha \beta }}\]
$\Rightarrow$ \[\dfrac{{{{(\alpha + 1)}^2}}}{{\alpha (\alpha + 1) - \beta (\alpha + 1)}} + \dfrac{{{{(\beta + 1)}^2}}}{{\beta (\beta + 1) - \alpha (\beta + 1)}}\]
$\Rightarrow$ \[\dfrac{{{{(\alpha + 1)}^2}}}{{(\alpha - \beta )(\alpha + 1)}} + \dfrac{{{{(\beta + 1)}^2}}}{{(\beta - \alpha )(\beta + 1)}}\]
As $\alpha + 1,\beta + 1$ is common in both numerator and denominator so it will cancel out and we get ,
\[\dfrac{{\alpha + 1}}{{\alpha - \beta }} - \dfrac{{\beta + 1}}{{\alpha - \beta }}\]
that is = $1$
RHS hence proved ,

Note: We can also solve this equation by making a quadratic equation of having roots $\alpha + 1,\beta + 1$ for this let us suppose $\alpha + 1 = y$ then $\alpha = y - 1$ hence we know that \[\alpha ,\beta \] are the roots of ${x^2} - p(x + 1) - c = 0$ now put $\alpha = y - 1$ in this equation ${(y - 1)^2} - p((y - 1) + 1) - c = 0$ hence this is given quadratic equation whose roots is $\alpha + 1,\beta + 1$ now we can find the value that product of roots or $(\alpha + 1)(\beta + 1) = 1 - c$ .