Question

If α and β are the roots of the equation$a{{x}^{2}}+bx+c$, then find the values of
(a). $\dfrac{1}{{{\alpha }^{2}}}+\dfrac{1}{{{\beta }^{2}}}$
(b). ${{\alpha }^{4}}{{\beta }^{7}}+{{\alpha }^{7}}{{\beta }^{4}}$
(c). ${{\left( \dfrac{\alpha }{\beta }-\dfrac{\beta }{\alpha } \right)}^{2}}$

Answer 1

Hint: For a given quadratic equation of form $a{{x}^{2}}+bx+c$whose roots are α and β,
the sum of the roots = $\left( \alpha +\beta \right)=\dfrac{-b}{a}$and the product of the roots =$\alpha \beta =\dfrac{c}{a}$. We have to substitute these equations in the solution to get our desired answer.

Complete step-by-step answer:

So here we are an equation $a{{x}^{2}}+bx+c$whose roots are α and β.
We have to find
        $\dfrac{1}{{{\alpha }^{2}}}+\dfrac{1}{{{\beta }^{2}}}$
So taking LCM of the equation we get;
$\dfrac{{{\alpha }^{2}}\text{+}{{\beta }^{2}}}{{{\alpha }^{2}}{{\beta }^{2}}}$
=$\dfrac{{{(\alpha \text{+}\beta )}^{2}}-2.\alpha .\beta }{{{\left( \alpha \beta \right)}^{2}}}$ (Because${{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2.a.b$)
= \[\dfrac{{{\left( \dfrac{-b}{a} \right)}^{2}}-2.\left( \dfrac{c}{a} \right)}{{{\left( \dfrac{c}{a} \right)}^{2}}}\] (Because α + β = $\dfrac{-b}{a}$ and αβ =$\dfrac{c}{a}$)
=$\dfrac{\dfrac{{{b}^{2}}}{{{a}^{2}}}-2.\dfrac{c}{a}}{\dfrac{{{c}^{2}}}{{{a}^{2}}}}$ =$\dfrac{\dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{2c}{a}}{\dfrac{{{c}^{2}}}{{{a}^{2}}}}$
Taking ${{a}^{2}}$as LCM in the numerator, we get;
= $\dfrac{\dfrac{{{b}^{2}}-2ac}{{{a}^{2}}}}{\dfrac{{{c}^{2}}}{{{a}^{2}}}}$
= $\dfrac{{{b}^{2}}-2ac}{{{a}^{2}}}*\dfrac{{{a}^{2}}}{{{c}^{2}}}$ (By reciprocal method)
Cancelling common term${{a}^{2}}$, we get;
= $\dfrac{{{b}^{2}}-2ac}{{{c}^{2}}}$
Therefore the required solution is $\dfrac{{{b}^{2}}-2ac}{{{c}^{2}}}$
We have to find
               ${{\alpha }^{4}}{{\beta }^{7}}+{{\alpha }^{7}}{{\beta }^{4}}$
Taking ${{\alpha }^{4}}{{\beta }^{4}}$as common, we get;
${{\alpha }^{4}}{{\beta }^{4}}({{\beta }^{3}}+{{\alpha }^{3}})$
= ${{\left( \alpha \beta \right)}^{4}}\left( \alpha +\beta \right)\left( {{\alpha }^{2}}+{{\beta }^{2}}-\alpha \beta \right)$ (Because${{a}^{3}}+{{b}^{3}}=(a+b)({{a}^{2}}-ab+{{b}^{2}})$)
= ${{\left( \alpha \beta \right)}^{4}}\left( \alpha +\beta \right)\left( {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta -\alpha \beta \right)$ (Because${{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2.a.b$)
= ${{\left( \alpha \beta \right)}^{4}}\left( \alpha +\beta \right)\left( {{\left( \alpha +\beta \right)}^{2}}-3\alpha \beta \right)$
We know \[\alpha \text{ }+\text{ }\beta =\dfrac{-b}{a}\] and \[\alpha \beta =\dfrac{c}{a}\]
Applying them in the equation, we get;
=${{\left( \dfrac{c}{a} \right)}^{4}}\left( \dfrac{-b}{a} \right)\left( {{\left( \dfrac{-b}{a} \right)}^{2}}-\dfrac{3c}{a} \right)$
= ${{\left( \dfrac{c}{a} \right)}^{4}}\left( \dfrac{-b}{a} \right)\left( {{\left( \dfrac{-b}{a} \right)}^{2}}-\dfrac{3c}{a} \right)$
=${{\left( \dfrac{c}{a} \right)}^{4}}\left( \dfrac{-b}{a} \right)\left( \dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{3c}{a} \right)$
Taking ${{a}^{2}}$ as LCM, we get;
= ${{\left( \dfrac{c}{a} \right)}^{4}}\left( \dfrac{-b}{a} \right)\left( \dfrac{{{b}^{2}}-3ac}{{{a}^{2}}} \right)$
=${{\left( \dfrac{c}{a} \right)}^{4}}\left( \dfrac{-{{b}^{3}}+3abc}{{{a}^{3}}} \right)$
=\[\left( \dfrac{-{{b}^{3}}{{c}^{4}}+3ab{{c}^{5}}}{{{a}^{7}}} \right)\]
Therefore the required solution is \[\left( \dfrac{-{{b}^{3}}{{c}^{4}}+3ab{{c}^{5}}}{{{a}^{7}}} \right)\].
We have to find
${{\left( \dfrac{\alpha }{\beta }-\dfrac{\beta }{\alpha } \right)}^{2}}$
 Taking LCM we get;
${{\left( \dfrac{{{\alpha }^{2}}-{{\beta }^{2}}}{\alpha \beta } \right)}^{2}}$
We know${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$. Applying this in the equation we get,
=\[{{\left( \dfrac{(\alpha -\beta )(\alpha +\beta )}{\alpha \beta } \right)}^{2}}\]
= \[\dfrac{{{(\alpha \text{ - }\beta )}^{2}}{{(\alpha \text{ + }\beta )}^{2}}}{{{(\alpha \beta )}^{2}}}\] …… (i)
We know α + β = $\dfrac{-b}{a}$ and αβ = $\dfrac{c}{a}$
And ${{(\alpha \text{ - }\beta )}^{2}}={{(\alpha \text{ + }\beta )}^{2}}-4\alpha \beta $
So , ${{(\alpha \text{ - }\beta )}^{2}}={{\left( \dfrac{-b}{a} \right)}^{2}}-4\left( \dfrac{c}{a} \right)$
                       = $\dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{4c}{a}$
Taking ${{a}^{2}}$as LCM, we get;
${{(\alpha \text{ - }\beta )}^{2}}$ =$\dfrac{{{b}^{2}}-4ac}{{{a}^{2}}}$
Putting the value of ${{(\alpha \text{ - }\beta )}^{2}}$ in equation (i), we get;
${{\dfrac{\dfrac{\left( {{b}^{2}}-4ac \right)}{{{a}^{2}}}.\left( \dfrac{-b}{a} \right)}{{{\left( \dfrac{c}{a} \right)}^{2}}}}^{2}}$
= $\dfrac{\dfrac{\left( {{b}^{2}}-4ac \right)}{{{a}^{2}}}.\left( \dfrac{{{b}^{2}}}{{{a}^{2}}} \right)}{\dfrac{{{c}^{2}}}{{{a}^{2}}}}$
Multiplying and taking reciprocal we get;
$\dfrac{{{b}^{4}}-4ac{{b}^{2}}}{{{a}^{4}}}.\dfrac{{{a}^{2}}}{{{c}^{2}}}$
Cancelling the common term ${{a}^{2}}$we get;
$\dfrac{{{b}^{4}}-4ac{{b}^{2}}}{{{a}^{2}}{{c}^{2}}}$
So, the required answer is$\dfrac{{{b}^{4}}-4ac{{b}^{2}}}{{{a}^{2}}{{c}^{2}}}$.


Note: It is very important to remember the formulas of
$\alpha \text{+}\beta =\dfrac{-b}{a}$ , $\alpha \beta =\dfrac{c}{a}$ and ${{\left( \alpha -\beta \right)}^{2}}=\dfrac{{{b}^{2}}-4ac}{{{a}^{2}}}$ many times they will be used in quadratic equation problems. We can also write $\alpha -\beta =\dfrac{|\sqrt{D}|}{a}$ where D is the discriminant where Discriminant $D={{b}^{2}}-4ac$.