Question

If ${{x}_{1}},{{x}_{2}}$ are the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ and $c\ne 0$, find the value of ${{\left( a{{x}_{1}}+b \right)}^{-2}}+{{\left( a{{x}_{2}}+b \right)}^{-2}}$in terms of $a,b,c$. \[\]

Answer 1

Hint: We use the fact that for the quadratic equation $a{{x}^{2}}+bx+c=0$ the sum and product of the roots are given by $\dfrac{-b}{a},\dfrac{c}{a}$ respectively and find ${{x}_{1}}+{{x}_{2}},{{x}_{1}}{{x}_{2}}$. W use the algebraic identity ${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$ to find ${{x}_{1}}^{2}+{{x}_{2}}^{2}$ and then use the obtained value to evaluate${{\left( a{{x}_{1}}+b \right)}^{-2}}+{{\left( a{{x}_{2}}+b \right)}^{-2}}$.\[\]

Complete step by step answer:
We know that the quadratic equation in the general form is given by $a{{x}^{2}}+bx+c=0$ where $a,b,c$ are numbers with condition$a\ne 0$. We get two roots from any quadratic equation. Let us denote them as $\alpha ,\beta $. The sum of the roots is given by
\[\alpha +\beta =\dfrac{-b}{a}\]
The product of the roots is given by
\[\alpha \beta =\dfrac{c}{a}\]
We are given in the question that ${{x}_{1}}$ and ${{x}_{2}}$ are the roots of the quadratic equation$a{{x}^{2}}+bx+c=0$. We use sum and product of the roots formula and have,
\[{{x}_{1}}+{{x}_{2}}=\dfrac{-b}{a},{{x}_{1}}{{x}_{2}}=\dfrac{c}{a}....\left( 1 \right)\]
Let us use identity ${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$ for $a={{x}_{1}},b={{x}_{2}}$ and have,
\[{{x}_{1}}^{2}+{{x}_{2}}^{2}={{\left( {{x}_{1}}+{{x}_{2}} \right)}^{2}}-2{{x}_{1}}{{x}_{2}}\]
We put the values obtained from (1) and have,
\[{{x}_{1}}^{2}+{{x}_{2}}^{2}={{\left( \dfrac{-b}{a} \right)}^{2}}-2\left( \dfrac{c}{a} \right)=\dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{2c}{a}=\dfrac{{{b}^{2}}-2ac}{{{a}^{2}}}........\left( 2 \right)\]
We are asked to find the value of value of ${{\left( a{{x}_{1}}+b \right)}^{-2}}+{{\left( a{{x}_{2}}+b \right)}^{-2}}$. Let us write the equation following the required value as
\[\begin{align}
  & a{{x}^{2}}+bx=-c \\
 & \Rightarrow ax+b=\dfrac{-c}{x} \\
\end{align}\]
As ${{x}_{1}},{{x}_{2}}$ are roots of the quadratic equation they are going to satisfy the above equation. We have,
\[\begin{align}
  & a{{x}_{1}}+b=\dfrac{-c}{{{x}_{1}}}\Rightarrow {{\left( a{{x}_{1}}+b \right)}^{-2}}={{\left( \dfrac{-c}{{{x}_{1}}} \right)}^{-2}}=\dfrac{{{x}_{1}}^{2}}{{{c}^{2}}} \\
 & a{{x}_{2}}+b=\dfrac{-c}{{{x}_{2}}}\Rightarrow {{\left( a{{x}_{2}}+b \right)}^{-2}}={{\left( \dfrac{-c}{{{x}_{2}}} \right)}^{-2}}=\dfrac{{{x}_{2}}^{2}}{{{c}^{2}}} \\
\end{align}\]
So we have
\[{{\left( a{{x}_{1}}+b \right)}^{-2}}+{{\left( a{{x}_{2}}+b \right)}^{-2}}=\dfrac{{{x}_{1}}^{2}}{{{c}^{2}}}+\dfrac{{{x}_{2}}^{2}}{{{c}^{2}}}=\dfrac{{{x}_{1}}^{2}+{{x}_{2}}^{2}}{{{c}^{2}}}\]
We put the value obtained in (2) and have,
\[{{\left( a{{x}_{1}}+b \right)}^{-2}}+{{\left( a{{x}_{2}}+b \right)}^{-2}}=\dfrac{{{x}_{1}}^{2}+{{x}_{2}}^{2}}{{{c}^{2}}}=\dfrac{{{b}^{2}}-2ac}{{{a}^{2}}{{c}^{2}}}\]

The above value in terms of $a,b,c$ is well defined because we are given the question $c\ne 0$ and from the definition of quadratic equation $a\ne 0$ and hence the denominator cannot be zero. \[\]

Note: The sum of the roots say ${{\alpha }_{1}},{{\alpha }_{2}},{{\alpha }_{3}}$of general cubic equation $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$is given by ${{\alpha }_{1}}+{{\alpha }_{2}}+{{\alpha }_{3}}=\dfrac{-b}{a}$ and the product of roots is given by ${{\alpha }_{1}}{{\alpha }_{2}}{{\alpha }_{3}}=\dfrac{-d}{a}$. The sum of the roots of general polynomial equation ${{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+...{{a}_{1}}x+{{a}_{0}}=0$ is given by $-\dfrac{{{a}_{n-1}}}{{{a}_{n}}}$ and the product of the roots is ${{\left( - \right)}^{n}}\dfrac{{{a}_{0}}}{{{a}_{n}}}$. We can find the roots of the quadratic equation by completing the square method, splitting the middle term method and the quadratic formula.