If the roots of the equation $ \left( {{{\text{c}}^{\text{2}}} - {\text{ab}}} \right){{\text{x}}^{\text{2}}} - {\text{2}}\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right){\text{x}} + {{\text{b}}^{\text{2}}} - {\text{ac}} = {\text{0}} $ in $ {\text{x}} $ are equal, then show that either $ {\text{a}} = {\text{0}} $ or $ {{\text{a}}^{\text{3}}} + {{\text{b}}^{\text{3}}} + {{\text{c}}^{\text{3}}} = {\text{3abc}} $
Answer
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Hint: $ {\left( {{\text{x}} - {\text{y}}} \right)^{\text{2}}} = {{\text{x}}^{\text{2}}} + {{\text{y}}^{\text{2}}} - {\text{2xy}} $
General form of quadratic equation is $ {\text{a}}{{\text{x}}^{\text{2}}} + {\text{bx}} + {\text{c}} = {\text{0}} $
If $ \alpha $ and $ \beta $ are the roots of the given quadratic equation, then
Sum of the roots of the equation is $ - \dfrac{{\text{b}}}{{\text{a}}} $
Product of the roots of the equation is $ \dfrac{{\text{c}}}{{\text{a}}} $
Follow the given conditions in the question and solve accordingly.
Complete step-by-step answer:
Given: Quadratic equation is $ \left( {{{\text{c}}^{\text{2}}} - {\text{ab}}} \right){{\text{x}}^{\text{2}}} - {\text{2}}\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right){\text{x}} + {{\text{b}}^{\text{2}}} - {\text{ac}} = {\text{0}} $
Roots of the equation given are equal.
We need to prove either $ {\text{a}} = {\text{0}} $ or $ {{\text{a}}^{\text{3}}} + {{\text{b}}^{\text{3}}} + {{\text{c}}^{\text{3}}} = {\text{3abc}} $
Let the roots of the given quadratic equation be $ \alpha $ and $ \beta $ respectively.
According to question,
$ \alpha = \beta $
General form of quadratic equation is $ a{{\text{x}}^2} + b{\text{x}} + c = 0 $
Comparing the given equation $ \left( {{{\text{c}}^{\text{2}}} - {\text{ab}}} \right){{\text{x}}^{\text{2}}} - {\text{2}}\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right){\text{x}} + {{\text{b}}^{\text{2}}} - {\text{ac}} = {\text{0}} $ with the general form, we get
$
{\text{a}} = {{\text{c}}^{\text{2}}} - {\text{ab}} \\
{\text{b}} = - {\text{2}}\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right) \\
{\text{c}} = {{\text{b}}^{\text{2}}} - {\text{ac}} \;
$
Sum of the roots of the equation in general form $ {\text{a}}{{\text{x}}^{\text{2}}} + {\text{bx}} + {\text{c}} = {\text{0}} $ is $ - \dfrac{{\text{b}}}{{\text{a}}} $
$ \alpha {\text{ + }}\beta = - \dfrac{{\text{b}}}{{\text{a}}} = - \left( {\dfrac{{ - {\text{2}}\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right)}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}}} \right) $
$ \Rightarrow \alpha {\text{ + }}\beta = \dfrac{{{\text{2}}\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right)}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} $ ……………Equation(1)
From the question we know that $ \alpha = \beta $
Substituting $ \alpha = \beta $ in Equation(1), we get
$
\Rightarrow \alpha + \alpha = \dfrac{{2\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right)}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} \\
\Rightarrow 2\alpha = \dfrac{{2\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right)}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} \;
$
$ \Rightarrow \alpha = \dfrac{{{{\text{a}}^{\text{2}}} - {\text{bc}}}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} $ ……………Equation(2)
Product of the roots of the equation in general form $ {\text{a}}{{\text{x}}^{\text{2}}} + {\text{bx}} + {\text{c}} = {\text{0}} $ is $ \dfrac{{\text{c}}}{{\text{a}}} $
$ \alpha \times \beta = \dfrac{{\text{c}}}{{\text{a}}} = \dfrac{{{{\text{b}}^{\text{2}}} - {\text{ac}}}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} $ ……………Equation(3)
Substituting $ \alpha = \beta $ in Equation(3), we get
$ \alpha \times \alpha = \dfrac{{\text{c}}}{{\text{a}}} = \dfrac{{{{\text{b}}^{\text{2}}} - {\text{ac}}}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} $
$ \Rightarrow {\alpha ^2} = \dfrac{{{{\text{b}}^{\text{2}}} - {\text{ac}}}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} $ ……………Equation(4)
Substituting Equation (2) in Equation (4) , we get
\[
{\left( {\dfrac{{{{\text{a}}^{\text{2}}} - {\text{bc}}}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}}} \right)^{\text{2}}} = \dfrac{{{{\text{b}}^{\text{2}}} - {\text{ac}}}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} \\
\Rightarrow \dfrac{{{{\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right)}^{\text{2}}}}}{{{{\left( {{{\text{c}}^{\text{2}}} - {\text{ab}}} \right)}^{\text{2}}}}} = \dfrac{{{{\text{b}}^{\text{2}}} - {\text{ac}}}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} \\
\Rightarrow \dfrac{{{{\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right)}^{\text{2}}}}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} = {{\text{b}}^{\text{2}}} - {\text{ac}}
\]
\[ \Rightarrow {\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right)^{\text{2}}} = \left( {{{\text{b}}^{\text{2}}} - {\text{ac}}} \right){ \times }\left( {{{\text{c}}^{\text{2}}} - {\text{ab}}} \right)\] …………………Equation(5)
We know that $ {\left( {{\text{x}} - {\text{y}}} \right)^{\text{2}}} = {{\text{x}}^{\text{2}}} + {{\text{y}}^{\text{2}}} - {\text{2xy}} $
Therefore Equation(5) becomes
\[
{{\text{a}}^{\text{4}}} + {{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}} - {\text{2}}{{\text{a}}^{\text{2}}}{\text{bc}} = {{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}} - {\text{a}}{{\text{b}}^{\text{3}}} - {\text{a}}{{\text{c}}^{\text{3}}} + {{\text{a}}^{\text{2}}}{\text{bc}} \\
\Rightarrow {{\text{a}}^{\text{4}}} + {\text{a}}{{\text{b}}^{\text{3}}} + {\text{a}}{{\text{c}}^{\text{3}}} - {\text{3}}{{\text{a}}^{\text{2}}}{\text{bc}} = {\text{0}} \\
\Rightarrow {\text{a}}\left( {{{\text{a}}^{\text{3}}} + {{\text{b}}^{\text{3}}} + {{\text{c}}^{\text{3}}} - {\text{3abc}}} \right) = {\text{0}} \;
\]
Therefore, either $ {\text{a}} = 0 $
or
$
{{\text{a}}^{\text{3}}} + {{\text{b}}^{\text{3}}} + {{\text{c}}^{\text{3}}} - {\text{3abc}} = {\text{0}} \\
\Rightarrow {{\text{a}}^{\text{3}}} + {{\text{b}}^{\text{3}}} + {{\text{c}}^{\text{3}}} = {\text{3abc}} \;
$
Hence, proved.
Note: In this type of questions which involves the concept of quadratic equations, we need to have knowledge about the formulae related to sum of roots and product of roots of a general quadratic equation. Follow the conditions given in the question and solve them accordingly to prove what is the required.
General form of quadratic equation is $ {\text{a}}{{\text{x}}^{\text{2}}} + {\text{bx}} + {\text{c}} = {\text{0}} $
If $ \alpha $ and $ \beta $ are the roots of the given quadratic equation, then
Sum of the roots of the equation is $ - \dfrac{{\text{b}}}{{\text{a}}} $
Product of the roots of the equation is $ \dfrac{{\text{c}}}{{\text{a}}} $
Follow the given conditions in the question and solve accordingly.
Complete step-by-step answer:
Given: Quadratic equation is $ \left( {{{\text{c}}^{\text{2}}} - {\text{ab}}} \right){{\text{x}}^{\text{2}}} - {\text{2}}\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right){\text{x}} + {{\text{b}}^{\text{2}}} - {\text{ac}} = {\text{0}} $
Roots of the equation given are equal.
We need to prove either $ {\text{a}} = {\text{0}} $ or $ {{\text{a}}^{\text{3}}} + {{\text{b}}^{\text{3}}} + {{\text{c}}^{\text{3}}} = {\text{3abc}} $
Let the roots of the given quadratic equation be $ \alpha $ and $ \beta $ respectively.
According to question,
$ \alpha = \beta $
General form of quadratic equation is $ a{{\text{x}}^2} + b{\text{x}} + c = 0 $
Comparing the given equation $ \left( {{{\text{c}}^{\text{2}}} - {\text{ab}}} \right){{\text{x}}^{\text{2}}} - {\text{2}}\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right){\text{x}} + {{\text{b}}^{\text{2}}} - {\text{ac}} = {\text{0}} $ with the general form, we get
$
{\text{a}} = {{\text{c}}^{\text{2}}} - {\text{ab}} \\
{\text{b}} = - {\text{2}}\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right) \\
{\text{c}} = {{\text{b}}^{\text{2}}} - {\text{ac}} \;
$
Sum of the roots of the equation in general form $ {\text{a}}{{\text{x}}^{\text{2}}} + {\text{bx}} + {\text{c}} = {\text{0}} $ is $ - \dfrac{{\text{b}}}{{\text{a}}} $
$ \alpha {\text{ + }}\beta = - \dfrac{{\text{b}}}{{\text{a}}} = - \left( {\dfrac{{ - {\text{2}}\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right)}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}}} \right) $
$ \Rightarrow \alpha {\text{ + }}\beta = \dfrac{{{\text{2}}\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right)}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} $ ……………Equation(1)
From the question we know that $ \alpha = \beta $
Substituting $ \alpha = \beta $ in Equation(1), we get
$
\Rightarrow \alpha + \alpha = \dfrac{{2\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right)}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} \\
\Rightarrow 2\alpha = \dfrac{{2\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right)}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} \;
$
$ \Rightarrow \alpha = \dfrac{{{{\text{a}}^{\text{2}}} - {\text{bc}}}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} $ ……………Equation(2)
Product of the roots of the equation in general form $ {\text{a}}{{\text{x}}^{\text{2}}} + {\text{bx}} + {\text{c}} = {\text{0}} $ is $ \dfrac{{\text{c}}}{{\text{a}}} $
$ \alpha \times \beta = \dfrac{{\text{c}}}{{\text{a}}} = \dfrac{{{{\text{b}}^{\text{2}}} - {\text{ac}}}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} $ ……………Equation(3)
Substituting $ \alpha = \beta $ in Equation(3), we get
$ \alpha \times \alpha = \dfrac{{\text{c}}}{{\text{a}}} = \dfrac{{{{\text{b}}^{\text{2}}} - {\text{ac}}}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} $
$ \Rightarrow {\alpha ^2} = \dfrac{{{{\text{b}}^{\text{2}}} - {\text{ac}}}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} $ ……………Equation(4)
Substituting Equation (2) in Equation (4) , we get
\[
{\left( {\dfrac{{{{\text{a}}^{\text{2}}} - {\text{bc}}}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}}} \right)^{\text{2}}} = \dfrac{{{{\text{b}}^{\text{2}}} - {\text{ac}}}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} \\
\Rightarrow \dfrac{{{{\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right)}^{\text{2}}}}}{{{{\left( {{{\text{c}}^{\text{2}}} - {\text{ab}}} \right)}^{\text{2}}}}} = \dfrac{{{{\text{b}}^{\text{2}}} - {\text{ac}}}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} \\
\Rightarrow \dfrac{{{{\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right)}^{\text{2}}}}}{{{{\text{c}}^{\text{2}}} - {\text{ab}}}} = {{\text{b}}^{\text{2}}} - {\text{ac}}
\]
\[ \Rightarrow {\left( {{{\text{a}}^{\text{2}}} - {\text{bc}}} \right)^{\text{2}}} = \left( {{{\text{b}}^{\text{2}}} - {\text{ac}}} \right){ \times }\left( {{{\text{c}}^{\text{2}}} - {\text{ab}}} \right)\] …………………Equation(5)
We know that $ {\left( {{\text{x}} - {\text{y}}} \right)^{\text{2}}} = {{\text{x}}^{\text{2}}} + {{\text{y}}^{\text{2}}} - {\text{2xy}} $
Therefore Equation(5) becomes
\[
{{\text{a}}^{\text{4}}} + {{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}} - {\text{2}}{{\text{a}}^{\text{2}}}{\text{bc}} = {{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}} - {\text{a}}{{\text{b}}^{\text{3}}} - {\text{a}}{{\text{c}}^{\text{3}}} + {{\text{a}}^{\text{2}}}{\text{bc}} \\
\Rightarrow {{\text{a}}^{\text{4}}} + {\text{a}}{{\text{b}}^{\text{3}}} + {\text{a}}{{\text{c}}^{\text{3}}} - {\text{3}}{{\text{a}}^{\text{2}}}{\text{bc}} = {\text{0}} \\
\Rightarrow {\text{a}}\left( {{{\text{a}}^{\text{3}}} + {{\text{b}}^{\text{3}}} + {{\text{c}}^{\text{3}}} - {\text{3abc}}} \right) = {\text{0}} \;
\]
Therefore, either $ {\text{a}} = 0 $
or
$
{{\text{a}}^{\text{3}}} + {{\text{b}}^{\text{3}}} + {{\text{c}}^{\text{3}}} - {\text{3abc}} = {\text{0}} \\
\Rightarrow {{\text{a}}^{\text{3}}} + {{\text{b}}^{\text{3}}} + {{\text{c}}^{\text{3}}} = {\text{3abc}} \;
$
Hence, proved.
Note: In this type of questions which involves the concept of quadratic equations, we need to have knowledge about the formulae related to sum of roots and product of roots of a general quadratic equation. Follow the conditions given in the question and solve them accordingly to prove what is the required.
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