Question

If \[\alpha ,\beta \] are the roots of \[a{{x}^{2}}+bx+c=0\], find the value of
(a) \[\left( 1+\alpha \right)\left( 1+\beta \right)\]
(b) \[{{\alpha }^{3}}\beta +\alpha {{\beta }^{3}}\]
(c) \[\dfrac{1}{\alpha }+\dfrac{1}{\beta }\]
(d) \[\dfrac{1}{a\alpha +b}+\dfrac{1}{a\beta +b}\]

Answer 1

Hint: In this question, we first need to write the formula for sum of the roots of the equation \[a{{x}^{2}}+bx+c=0\] given by \[\alpha +\beta =\dfrac{-b}{a}\] and the product of the roots given by \[\alpha \beta =\dfrac{c}{a}\]. Then convert the given expressions accordingly into the terms of \[\alpha +\beta \] and \[\alpha \beta \]to simplify further and get the result.

Complete step-by-step answer:
Relation between roots and coefficients:
If the roots of the quadratic equation \[a{{x}^{2}}+bx+c=0\] are \[\alpha ,\beta \], then
Sum of the roots is given by the formula
\[\Rightarrow \alpha +\beta =\dfrac{-b}{a}\]
Product of the roots is given by the formula
\[\Rightarrow \alpha \beta =\dfrac{c}{a}\]
(a) Now, let us first find the value of
\[\Rightarrow \left( 1+\alpha \right)\left( 1+\beta \right)\]
Let us now expand this expression first
\[\Rightarrow 1+\alpha +\beta +\alpha \beta \]
Let us now substitute the respective values to simply it further
\[\Rightarrow 1+\dfrac{-b}{a}+\dfrac{c}{a}\]
Now, on further simplification we get,
\[\Rightarrow \dfrac{a-b+c}{a}\]
Thus, the value of \[\left( 1+\alpha \right)\left( 1+\beta \right)\] is given by \[\dfrac{a-b+c}{a}\]
(b) Now, let us next find the value of the expression
\[\Rightarrow {{\alpha }^{3}}\beta +\alpha {{\beta }^{3}}\]
Now, this can be further written by taking out the common terms as
\[\Rightarrow \alpha \beta \left( {{\alpha }^{2}}+{{\beta }^{2}} \right)\]
As we already know from the factorisation of polynomials that
\[{{x}^{2}}+{{a}^{2}}={{\left( x+a \right)}^{2}}-2ax\]
Now, using the above formula the expression can be further written as
\[\Rightarrow \alpha \beta \left( {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta \right)\]
Now, on substituting the respective values we get,
\[\Rightarrow \dfrac{c}{a}\left( {{\left( \dfrac{-b}{a} \right)}^{2}}-2\dfrac{c}{a} \right)\]
Now, on further simplification we get,
\[\Rightarrow \dfrac{c}{a}\left( \dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{2c}{a} \right)\]
Now, this can be written further in the simplified form as
\[\Rightarrow \dfrac{{{b}^{2}}c-2{{c}^{2}}a}{{{a}^{3}}}\]
Thus, the value of \[{{\alpha }^{3}}\beta +\alpha {{\beta }^{3}}\] is \[\dfrac{{{b}^{2}}c-2{{c}^{2}}a}{{{a}^{3}}}\]
(c) Now, let us find the value of
\[\Rightarrow \dfrac{1}{\alpha }+\dfrac{1}{\beta }\]
Now, this can be further written as
\[\Rightarrow \dfrac{\alpha +\beta }{\alpha \beta }\]
Let us now substitute the respective values in the above expression
\[\Rightarrow \dfrac{\dfrac{-b}{a}}{\dfrac{c}{a}}\]
Now, on further simplification we get,
\[\Rightarrow \dfrac{-b}{c}\]
Thus, the value of \[\dfrac{1}{\alpha }+\dfrac{1}{\beta }\]is \[\dfrac{-b}{c}\]
(d) ) Now, let us next find the value of the expression
\[\Rightarrow \dfrac{1}{a\alpha +b}+\dfrac{1}{a\beta +b}\]
Let us first expand and simplify the given expression
\[\Rightarrow \dfrac{a\beta +b+a\alpha +b}{\left( a\alpha +b \right)\left( a\beta +b \right)}\]
Now, this can be further written in known terms as
\[\Rightarrow \dfrac{a\left( \alpha +\beta \right)+2b}{{{a}^{2}}\alpha \beta +ab\left( \alpha +\beta \right)+{{b}^{2}}}\]
Let us substitute the respective terms to simplify it further
\[\Rightarrow \dfrac{a\left( \dfrac{-b}{a} \right)+2b}{{{a}^{2}}\times \dfrac{c}{a}+ab\left( \dfrac{-b}{a} \right)+{{b}^{2}}}\]
Now, cancelling out the common terms and simplifying it further we get,
\[\Rightarrow \dfrac{-b+2b}{ac-{{b}^{2}}+{{b}^{2}}}\]
Now, again on further simplification we get,
\[\Rightarrow \dfrac{b}{ac}\]
Thus, the value of \[\dfrac{1}{a\alpha +b}+\dfrac{1}{a\beta +b}\]is \[\dfrac{b}{ac}\]

Note: Instead of using the sum of the roots and product of the roots formulae we can also solve it by finding the roots from the direct formula and then substitute them respectively. But it would be a bit lengthy.
It is important to note that while substituting and while simplifying we should not neglect any of the terms or the sign because it changes the respective expression and so the result accordingly.
It is also to be noted that we need to convert all the given expressions accordingly to the sum and product of the roots to simplify.