Question

If \[\alpha \] and \[\beta \] are the roots of the equation \[{x^2} - p(x + 1) - c = 0\] then the numerical value of \[\dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + 2\beta + c}} = ?\]

Answer 1

Hint: In the given question, the roots of a quadratic equation is given and we are asked to find the value of \[\dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + 2\beta + c}}\]. As we know, standard form of an quadratic equation is \[a{x^2} + bx + c\], where \[a\], \[b\] and \[c\] are the coefficients and the quadratic equation in term of roots is given by: \[{x^2} - \left( {{\text{sum of the roots}}} \right)x + \left( {{\text{product of the roots}}} \right) = 0\]. Using this we will find \[c\]. Then we will substitute this \[c\] in the given expression to find its value.

Complete step by step answer:
Given, \[{x^2} - p(x + 1) - c = 0\]. \[\alpha \] and \[\beta \] are the roots of the equation. We have to find the numerical value of \[\dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + 2\beta + c}}\].
The given quadratic equation is \[{x^2} - p(x + 1) - c = 0\] i.e., \[{x^2} - px - \left( {p + c} \right) = 0 - - - (1)\].
The roots of this quadratic equation are \[\alpha \] and \[\beta \].
As we know, a quadratic equation is of form:
\[{x^2} - \left( {{\text{sum of the roots}}} \right)x + \left( {{\text{product of the roots}}} \right) = 0\]

By applying this in the given quadratic equation, we get the equation the as
\[ \Rightarrow {x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta = 0 - - - (2)\]
\[(1)\] and \[(2)\] represents the same equation, therefore on comparing we get
\[ \Rightarrow \alpha + \beta = p\] and \[\alpha \beta = - \left( {p + c} \right)\]
On solving, we get
\[ \Rightarrow c = - \alpha - \beta - \alpha \beta \]
Now putting the value of \[c\] in \[\dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + 2\beta + c}}\], we get
\[ \Rightarrow \dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + 2\beta + c}} = \dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + 2\alpha - \alpha - \beta - \alpha \beta }} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + 2\beta - \alpha - \beta - \alpha \beta }}\]

On simplifying, we get
\[ \Rightarrow \dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + 2\beta + c}} = \dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + \alpha - \beta - \alpha \beta }} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + \beta - \alpha - \alpha \beta }}\]
On taking common, we get
\[\dfrac{{{\alpha ^2} + 2\alpha + 1}}{{\alpha \left( {\alpha + 1} \right) - \beta \left( {1 + \alpha } \right)}} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{\beta \left( {\beta + 1} \right) - \alpha \left( {1 + \beta } \right)}}\]
Again, on taking common, we get
\[ \dfrac{{{\alpha ^2} + 2\alpha + 1}}{{\left( {\alpha + 1} \right)\left( {\alpha - \beta } \right)}} - \dfrac{{{\beta ^2} + 2\beta + 1}}{{\left( {\beta + 1} \right)\left( {\alpha - \beta } \right)}}\]

Using the identity \[{a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}\], we get
\[ \Rightarrow \dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + 2\beta + c}} = \dfrac{{{{\left( {\alpha + 1} \right)}^2}}}{{\left( {\alpha + 1} \right)\left( {\alpha - \beta } \right)}} - \dfrac{{{{\left( {\beta + 1} \right)}^2}}}{{\left( {\beta + 1} \right)\left( {\alpha - \beta } \right)}}\]
Cancelling the common terms from the numerator and the denominator, we get
\[ \dfrac{{\left( {\alpha + 1} \right)}}{{\left( {\alpha - \beta } \right)}} - \dfrac{{\left( {\beta + 1} \right)}}{{\left( {\alpha - \beta } \right)}}\]
\[ \Rightarrow \dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + 2\beta + c}} = \dfrac{{\left( {\alpha + 1} \right)}}{{\left( {\alpha - \beta } \right)}} - \dfrac{{\left( {\beta + 1} \right)}}{{\left( {\alpha - \beta } \right)}}\]

Taking \[\dfrac{1}{{\left( {\alpha - \beta } \right)}}\] common, we get
\[ \Rightarrow \dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + 2\beta + c}} = \dfrac{1}{{\left( {\alpha - \beta } \right)}}\left( {\alpha + 1 - \beta - 1} \right)\]
On simplifying, we get
\[ \Rightarrow \dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + 2\beta + c}} = \dfrac{1}{{\left( {\alpha - \beta } \right)}}\left( {\alpha - \beta } \right)\]
On further simplification, we get
\[ \Rightarrow \dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + 2\beta + c}} = 1\]

Therefore, the numerical value of \[\dfrac{{{\alpha ^2} + 2\alpha + 1}}{{{\alpha ^2} + 2\alpha + c}} + \dfrac{{{\beta ^2} + 2\beta + 1}}{{{\beta ^2} + 2\beta + c}}\] is \[1\].

Note: \[{x^2} - p(x + 1) - c = 0\] is a quadratic equation. A quadratic equation function may have one, two, or zero roots. Roots are also called the x-intercept or zeroes. Also, the y-coordinate of any points lying on the x-axis is zero. So, to find the roots of a quadratic function, we set \[f(x) = 0\].