Question

If $\alpha ,\beta $ are the roots of $a{{x}^{2}}+bx+c=0$, then find the equation whose roots are: -
(i) $\dfrac{1}{{{\alpha }^{2}}},\dfrac{1}{{{\beta }^{2}}}$
(ii) $\dfrac{1}{a\alpha +\beta },\dfrac{1}{a\beta +\alpha }$
(iii) $\alpha +\dfrac{1}{\beta },\beta +\dfrac{1}{\alpha }$

Answer 1

Hint: Use the formula for the sum of roots of a quadratic equation given as $\alpha +\beta =\dfrac{-b}{a}$ and similarly the formula for the product of roots given as $\alpha \beta =\dfrac{c}{a}$ to obtain two relations. Now, consider each part of the question one by one and form the required quadratic equation by using the fact that the quadratic expression with the given sum of roots and product of roots is ${{x}^{2}}-\left( \text{sum of roots} \right)x+\left( \text{product of roots} \right)=0$. Use the relation obtained above to replace the values of $\alpha $ and $\beta $.

Complete step-by-step answer:
Here we have been provided with the quadratic equation $a{{x}^{2}}+bx+c=0$ with its roots as $\alpha ,\beta $. We have been asked to find the quadratic equation whose roots are provided to us as subparts (i), (ii) and (iii).
Now, the sum of roots of a quadratic equation $a{{x}^{2}}+bx+c=0$ is given as $\alpha +\beta =\dfrac{-b}{a}$ and the product of roots is given as $\alpha \beta =\dfrac{c}{a}$. Also, any quadratic equation with the given values of sum of roots and the product of roots is given by the relation ${{x}^{2}}-\left( \text{sum of roots} \right)x+\left( \text{product of roots} \right)=0$. Let us consider each subpart of the question one by one.
(i) Here we have the roots $\dfrac{1}{{{\alpha }^{2}}},\dfrac{1}{{{\beta }^{2}}}$, so we get,
$\Rightarrow $ Sum of roots = $\dfrac{1}{{{\alpha }^{2}}}+\dfrac{1}{{{\beta }^{2}}}$
$\Rightarrow $ Sum of roots = \[\dfrac{{{\alpha }^{2}}+{{\beta }^{2}}}{{{\alpha }^{2}}{{\beta }^{2}}}\]
Using the algebraic identity ${{m}^{2}}+{{n}^{2}}={{\left( m+n \right)}^{2}}-2mn$ in the numerator we get,
$\Rightarrow $ Sum of roots = \[\dfrac{{{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta }{{{\left( \alpha \beta \right)}^{2}}}\]
Substituting the values of $\alpha +\beta $ and $\alpha \beta $ we get,
$\Rightarrow $ Sum of roots = \[\dfrac{{{\left( \dfrac{-b}{a} \right)}^{2}}-2\left( \dfrac{c}{a} \right)}{{{\left( \dfrac{c}{a} \right)}^{2}}}\]
On simplifying we get,
$\Rightarrow $ Sum of roots = \[\dfrac{{{b}^{2}}-2ac}{{{c}^{2}}}\]
Now, we have product of roots = $\dfrac{1}{{{\alpha }^{2}}}\times \dfrac{1}{{{\beta }^{2}}}$
$\Rightarrow $ Product of roots = $\dfrac{1}{{{\left( \alpha \beta \right)}^{2}}}$
$\Rightarrow $ Product of roots = \[\dfrac{1}{{{\left( \dfrac{c}{a} \right)}^{2}}}\]
$\Rightarrow $ Product of roots = \[\dfrac{{{a}^{2}}}{{{c}^{2}}}\]
Therefore the required quadratic equation will be given as: -
$\begin{align}
  & \Rightarrow {{x}^{2}}-\left( \dfrac{{{b}^{2}}-2ac}{{{c}^{2}}} \right)x+\dfrac{{{a}^{2}}}{{{c}^{2}}}=0 \\
 & \therefore {{c}^{2}}{{x}^{2}}-\left( {{b}^{2}}-2ac \right)x+{{a}^{2}}=0 \\
\end{align}$
(ii) Here we have the roots $\dfrac{1}{a\alpha +\beta },\dfrac{1}{a\beta +\alpha }$, so we get,
$\Rightarrow $ Sum of roots = $\dfrac{1}{a\alpha +\beta }+\dfrac{1}{a\beta +\alpha }$
$\Rightarrow $ Sum of roots = \[\dfrac{a\beta +\alpha +a\alpha +\beta }{\left( a\alpha +\beta \right)\left( a\beta +\alpha \right)}\]
$\Rightarrow $ Sum of roots = \[\dfrac{a\left( \beta +\alpha \right)+\left( \alpha +\beta \right)}{\left( {{a}^{2}}\alpha \beta +a{{\alpha }^{2}}+a{{\beta }^{2}}+a\beta \right)}\]
$\Rightarrow $ Sum of roots = \[\dfrac{\left( a+1 \right)\left( \alpha +\beta \right)}{\left( \left( {{a}^{2}}+1 \right)\alpha \beta +a\left( {{\alpha }^{2}}+{{\beta }^{2}} \right) \right)}\]
Using the algebraic identity ${{m}^{2}}+{{n}^{2}}={{\left( m+n \right)}^{2}}-2mn$ and substituting the values of $\alpha +\beta $ and $\alpha \beta $ we get,
$\Rightarrow $ Sum of roots = \[\dfrac{\left( a+1 \right)\left( \dfrac{-b}{a} \right)}{\left( {{a}^{2}}+1 \right)\left( \dfrac{c}{a} \right)+a\left( {{\left( \dfrac{-b}{a} \right)}^{2}}-2\left( \dfrac{c}{a} \right) \right)}\]
On simplifying we get,
$\Rightarrow $ Sum of roots = \[\dfrac{-\left( a+1 \right)b}{\left( {{a}^{2}}+1 \right)c+\left( {{b}^{2}}-2ac \right)}\]
Now, we have product of roots = $\dfrac{1}{a\alpha +\beta }\times \dfrac{1}{a\beta +\alpha }$
$\Rightarrow $ Product of roots = \[\dfrac{1}{\left( a\alpha +\beta \right)\left( a\beta +\alpha \right)}\]
$\Rightarrow $ Product of roots = \[\dfrac{1}{\left( \left( {{a}^{2}}+1 \right)\alpha \beta +a\left( {{\alpha }^{2}}+{{\beta }^{2}} \right) \right)}\]
$\Rightarrow $ Product of roots = \[\dfrac{1}{\left( {{a}^{2}}+1 \right)\left( \dfrac{c}{a} \right)+a\left( {{\left( \dfrac{-b}{a} \right)}^{2}}-2\left( \dfrac{c}{a} \right) \right)}\]
On simplification we get,
$\Rightarrow $ Product of roots = \[\dfrac{a}{\left( {{a}^{2}}+1 \right)c+\left( {{b}^{2}}-2ac \right)}\]
Therefore the required quadratic equation will be given as: -
\[\begin{align}
  & \Rightarrow {{x}^{2}}-\left( \dfrac{-\left( a+1 \right)b}{\left( {{a}^{2}}+1 \right)c+\left( {{b}^{2}}-2ac \right)} \right)x+\dfrac{a}{\left( {{a}^{2}}+1 \right)c+\left( {{b}^{2}}-2ac \right)}=0 \\
 & \therefore \left[ \left( {{a}^{2}}+1 \right)c+\left( {{b}^{2}}-2ac \right) \right]{{x}^{2}}+\left( a+1 \right)bx+a=0 \\
\end{align}\]
(iii) Here we have the roots $\alpha +\dfrac{1}{\beta },\beta +\dfrac{1}{\alpha }$, so we get,
$\Rightarrow $ Sum of roots = $\alpha +\dfrac{1}{\beta }+\beta +\dfrac{1}{\alpha }$
$\Rightarrow $ Sum of roots = $\left( \alpha +\beta \right)+\dfrac{\left( \alpha +\beta \right)}{\alpha \beta }$
$\Rightarrow $ Sum of roots = $\left( \alpha +\beta \right)\left( 1+\dfrac{1}{\alpha \beta } \right)$
Substituting the values of $\alpha +\beta $ and $\alpha \beta $ and simplifying we get,
$\Rightarrow $ Sum of roots = $\left( \dfrac{-b}{a} \right)\left( 1+\dfrac{1}{\dfrac{c}{a}} \right)$
$\Rightarrow $ Sum of roots = $\left( \dfrac{-b}{a} \right)\left( 1+\dfrac{a}{c} \right)$
$\Rightarrow $ Sum of roots = $-b\left( \dfrac{a+c}{ac} \right)$
Now, we have product of roots = $\left( \alpha +\dfrac{1}{\beta } \right)\times \left( \beta +\dfrac{1}{\alpha } \right)$
$\Rightarrow $ Product of roots = $\left( \alpha \beta +1+1+\dfrac{1}{\alpha \beta } \right)$
$\Rightarrow $ Product of roots = $\left( \dfrac{c}{a}+2+\dfrac{1}{\dfrac{c}{a}} \right)$
On simplification we get,
$\Rightarrow $ Product of roots = \[\left( \dfrac{c}{a}+2+\dfrac{a}{c} \right)\]
$\Rightarrow $ Product of roots = \[\left( \dfrac{{{a}^{2}}+{{c}^{2}}+2ac}{ac} \right)\]
Using the algebraic identity ${{m}^{2}}+{{n}^{2}}+2mn={{\left( m+n \right)}^{2}}$ we get,
$\Rightarrow $ Product of roots = \[\left( \dfrac{{{\left( a+c \right)}^{2}}}{ac} \right)\]
Therefore the required quadratic equation will be given as: -
\[\begin{align}
  & \Rightarrow {{x}^{2}}-\left( \dfrac{-b\left( a+c \right)}{ac} \right)x+\dfrac{{{\left( a+c \right)}^{2}}}{ac}=0 \\
 & \therefore ac{{x}^{2}}+b\left( a+c \right)x+{{\left( a+c \right)}^{2}}=0 \\
\end{align}\]

Note: You must remember the formulas of sum and product of roots of a quadratic as well as a cubic equation as they are often used in the topic algebra. Remember all the important algebraic identities used for the conversion of expressions so that we can use the formulas. Do not write ${{m}^{2}}+{{n}^{2}}={{\left( m-n \right)}^{2}}+2mn$ as we don’t have any direct formula for the difference of roots however you can simplify it as ${{\left( m-n \right)}^{2}}={{\left( m+n \right)}^{2}}-4mn$.