If one root of the quadratic equation \[{{a}_{1}}{{x}^{2}}+{{b}_{1}}x+{{c}_{1}}=0\] is numerically equal but opposite in sign to one root of \[{{a}_{2}}{{x}^{2}}+{{b}_{2}}x+{{c}_{2}}=0\], then prove that the quadratic equation whose roots are the other roots of the both of these equations is $\dfrac{{{x}^{2}}}{\dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}}}+x+\dfrac{1}{\dfrac{{{b}_{1}}}{{{c}_{1}}}+\dfrac{{{b}_{2}}}{{{c}_{2}}}}=0$.
Answer
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Hint: We start solving the problem by assigning the variables for the roots of both given quadratic equations. We then find the sum and product of the roots of both the given quadratic equations. We then make use of addition and division operations for the sum and product of the roots we just obtained to get the values of sum and product of the new quadratic equation. Once, we find the sum and product of the roots of the new equation, we substitute them to complete the required proof.
Complete step by step answer:
According to the problem, we have given one root of the quadratic equation \[{{a}_{1}}{{x}^{2}}+{{b}_{1}}x+{{c}_{1}}=0\] is numerically equal but opposite in sign to one root of \[{{a}_{2}}{{x}^{2}}+{{b}_{2}}x+{{c}_{2}}=0\]. We need to prove that the quadratic equation whose roots are other roots of the both of these equations is $\dfrac{{{x}^{2}}}{\dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}}}+x+\dfrac{1}{\dfrac{{{b}_{1}}}{{{c}_{1}}}+\dfrac{{{b}_{2}}}{{{c}_{2}}}}=0$.
Let us assume the magnitude of the common root be $\alpha $.
Let us assume the roots of the equation \[{{a}_{1}}{{x}^{2}}+{{b}_{1}}x+{{c}_{1}}=0\] be $\alpha $ and $\beta $.
We know that the sum and product of the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b}{a}$ and $\dfrac{c}{a}$.
So, we get \[\alpha +\beta =\dfrac{-{{b}_{1}}}{{{a}_{1}}}\] ---(1) and $\alpha \beta =\dfrac{{{c}_{1}}}{{{a}_{1}}}$ ---(2).
Let us divide equations (1) and (2)
$\Rightarrow \dfrac{\alpha +\beta }{\alpha \beta }=\dfrac{\dfrac{-{{b}_{1}}}{{{a}_{1}}}}{\dfrac{{{c}_{1}}}{{{a}_{1}}}}$.
$\Rightarrow \dfrac{1}{\alpha }+\dfrac{1}{\beta }=\dfrac{-{{b}_{1}}}{{{c}_{1}}}$ ---(3).
Let us assume the roots of the equation \[{{a}_{2}}{{x}^{2}}+{{b}_{2}}x+{{c}_{2}}=0\] be $-\alpha $ and ${{\beta }_{1}}$.
we get \[-\alpha +{{\beta }_{1}}=\dfrac{-{{b}_{2}}}{{{a}_{2}}}\] ---(4) and $-\alpha {{\beta }_{1}}=\dfrac{{{c}_{2}}}{{{a}_{2}}}$ ---(5).
Let us divide equations (4) and (5)
$\Rightarrow \dfrac{-\alpha +{{\beta }_{1}}}{-\alpha {{\beta }_{1}}}=\dfrac{\dfrac{-{{b}_{2}}}{{{a}_{2}}}}{\dfrac{{{c}_{2}}}{{{a}_{2}}}}$.
$\Rightarrow -\dfrac{1}{\alpha }+\dfrac{1}{{{\beta }_{1}}}=\dfrac{-{{b}_{2}}}{{{c}_{2}}}$---(6).
Now, we assume the new quadratic equations whose roots are $\beta $ and ${{\beta }_{1}}$ be ${{x}^{2}}+{{b}_{3}}x+{{c}_{3}}=0$.
So, we have $\beta +{{\beta }_{1}}=-{{b}_{3}}$ ---(7) and $\beta {{\beta }_{1}}={{c}_{3}}$ ---(8).
Let us add equations (1) and (4).
We get \[\alpha +\beta -\alpha +{{\beta }_{1}}=\dfrac{-{{b}_{1}}}{{{a}_{1}}}-\dfrac{{{b}_{2}}}{{{a}_{2}}}\].
$\Rightarrow \beta +{{\beta }_{1}}=\dfrac{-{{b}_{1}}}{{{a}_{1}}}-\dfrac{{{b}_{2}}}{{{a}_{2}}}$. We substitute this in equation (7).
$\Rightarrow -{{b}_{3}}=\dfrac{-{{b}_{1}}}{{{a}_{1}}}-\dfrac{{{b}_{2}}}{{{a}_{2}}}$.
\[\Rightarrow {{b}_{3}}=\dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}}\] ---(9).
Let us add equations (3) and (6).
We get $\dfrac{1}{\alpha }+\dfrac{1}{\beta }-\dfrac{1}{\alpha }+\dfrac{1}{{{\beta }_{1}}}=\dfrac{-{{b}_{1}}}{{{c}_{1}}}-\dfrac{{{b}_{2}}}{{{c}_{2}}}$.
$\Rightarrow \dfrac{1}{\beta }+\dfrac{1}{{{\beta }_{1}}}=\dfrac{-{{b}_{1}}}{{{c}_{1}}}-\dfrac{{{b}_{2}}}{{{c}_{2}}}$.
$\Rightarrow \dfrac{\beta +{{\beta }_{1}}}{\beta {{\beta }_{1}}}=\dfrac{-{{b}_{1}}}{{{c}_{1}}}-\dfrac{{{b}_{2}}}{{{c}_{2}}}$.
$\Rightarrow \dfrac{\dfrac{-{{b}_{1}}}{{{a}_{1}}}-\dfrac{{{b}_{2}}}{{{a}_{2}}}}{\beta {{\beta }_{1}}}=\dfrac{-{{b}_{1}}}{{{c}_{1}}}-\dfrac{{{b}_{2}}}{{{c}_{2}}}$.
$\Rightarrow \beta {{\beta }_{1}}=\dfrac{\dfrac{-{{b}_{1}}}{{{a}_{1}}}-\dfrac{{{b}_{2}}}{{{a}_{2}}}}{\dfrac{-{{b}_{1}}}{{{c}_{1}}}-\dfrac{{{b}_{2}}}{{{c}_{2}}}}$.
$\Rightarrow \beta {{\beta }_{1}}=\dfrac{\dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}}}{\dfrac{{{b}_{1}}}{{{c}_{1}}}+\dfrac{{{b}_{2}}}{{{c}_{2}}}}$.
From equation (8), we get ${{c}_{3}}=\dfrac{\dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}}}{\dfrac{{{b}_{1}}}{{{c}_{1}}}+\dfrac{{{b}_{2}}}{{{c}_{2}}}}$.
Using these values, we get the quadratic equation as ${{x}^{2}}+\left( \dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}} \right)x+\left( \dfrac{\dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}}}{\dfrac{{{b}_{1}}}{{{c}_{1}}}+\dfrac{{{b}_{2}}}{{{c}_{2}}}} \right)=0$.
$\Rightarrow \dfrac{{{x}^{2}}}{\dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}}}+x+\dfrac{1}{\dfrac{{{b}_{1}}}{{{c}_{1}}}+\dfrac{{{b}_{2}}}{{{c}_{2}}}}=0$.
We have proved the required quadratic equation as $\dfrac{{{x}^{2}}}{\dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}}}+x+\dfrac{1}{\dfrac{{{b}_{1}}}{{{c}_{1}}}+\dfrac{{{b}_{2}}}{{{c}_{2}}}}=0$.
Note:
We can also take the new quadratic equation as ${{a}_{3}}{{x}^{2}}+{{b}_{3}}x+{{c}_{3}}=0$, which will also give the same result but with an extra step. We should not confuse with the sum and product of the roots of the quadratic equations. We should not make calculation mistakes while solving this problem. We can also find the roots of both the equations using$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and compare them to solve the given problem.
Complete step by step answer:
According to the problem, we have given one root of the quadratic equation \[{{a}_{1}}{{x}^{2}}+{{b}_{1}}x+{{c}_{1}}=0\] is numerically equal but opposite in sign to one root of \[{{a}_{2}}{{x}^{2}}+{{b}_{2}}x+{{c}_{2}}=0\]. We need to prove that the quadratic equation whose roots are other roots of the both of these equations is $\dfrac{{{x}^{2}}}{\dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}}}+x+\dfrac{1}{\dfrac{{{b}_{1}}}{{{c}_{1}}}+\dfrac{{{b}_{2}}}{{{c}_{2}}}}=0$.
Let us assume the magnitude of the common root be $\alpha $.
Let us assume the roots of the equation \[{{a}_{1}}{{x}^{2}}+{{b}_{1}}x+{{c}_{1}}=0\] be $\alpha $ and $\beta $.
We know that the sum and product of the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b}{a}$ and $\dfrac{c}{a}$.
So, we get \[\alpha +\beta =\dfrac{-{{b}_{1}}}{{{a}_{1}}}\] ---(1) and $\alpha \beta =\dfrac{{{c}_{1}}}{{{a}_{1}}}$ ---(2).
Let us divide equations (1) and (2)
$\Rightarrow \dfrac{\alpha +\beta }{\alpha \beta }=\dfrac{\dfrac{-{{b}_{1}}}{{{a}_{1}}}}{\dfrac{{{c}_{1}}}{{{a}_{1}}}}$.
$\Rightarrow \dfrac{1}{\alpha }+\dfrac{1}{\beta }=\dfrac{-{{b}_{1}}}{{{c}_{1}}}$ ---(3).
Let us assume the roots of the equation \[{{a}_{2}}{{x}^{2}}+{{b}_{2}}x+{{c}_{2}}=0\] be $-\alpha $ and ${{\beta }_{1}}$.
we get \[-\alpha +{{\beta }_{1}}=\dfrac{-{{b}_{2}}}{{{a}_{2}}}\] ---(4) and $-\alpha {{\beta }_{1}}=\dfrac{{{c}_{2}}}{{{a}_{2}}}$ ---(5).
Let us divide equations (4) and (5)
$\Rightarrow \dfrac{-\alpha +{{\beta }_{1}}}{-\alpha {{\beta }_{1}}}=\dfrac{\dfrac{-{{b}_{2}}}{{{a}_{2}}}}{\dfrac{{{c}_{2}}}{{{a}_{2}}}}$.
$\Rightarrow -\dfrac{1}{\alpha }+\dfrac{1}{{{\beta }_{1}}}=\dfrac{-{{b}_{2}}}{{{c}_{2}}}$---(6).
Now, we assume the new quadratic equations whose roots are $\beta $ and ${{\beta }_{1}}$ be ${{x}^{2}}+{{b}_{3}}x+{{c}_{3}}=0$.
So, we have $\beta +{{\beta }_{1}}=-{{b}_{3}}$ ---(7) and $\beta {{\beta }_{1}}={{c}_{3}}$ ---(8).
Let us add equations (1) and (4).
We get \[\alpha +\beta -\alpha +{{\beta }_{1}}=\dfrac{-{{b}_{1}}}{{{a}_{1}}}-\dfrac{{{b}_{2}}}{{{a}_{2}}}\].
$\Rightarrow \beta +{{\beta }_{1}}=\dfrac{-{{b}_{1}}}{{{a}_{1}}}-\dfrac{{{b}_{2}}}{{{a}_{2}}}$. We substitute this in equation (7).
$\Rightarrow -{{b}_{3}}=\dfrac{-{{b}_{1}}}{{{a}_{1}}}-\dfrac{{{b}_{2}}}{{{a}_{2}}}$.
\[\Rightarrow {{b}_{3}}=\dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}}\] ---(9).
Let us add equations (3) and (6).
We get $\dfrac{1}{\alpha }+\dfrac{1}{\beta }-\dfrac{1}{\alpha }+\dfrac{1}{{{\beta }_{1}}}=\dfrac{-{{b}_{1}}}{{{c}_{1}}}-\dfrac{{{b}_{2}}}{{{c}_{2}}}$.
$\Rightarrow \dfrac{1}{\beta }+\dfrac{1}{{{\beta }_{1}}}=\dfrac{-{{b}_{1}}}{{{c}_{1}}}-\dfrac{{{b}_{2}}}{{{c}_{2}}}$.
$\Rightarrow \dfrac{\beta +{{\beta }_{1}}}{\beta {{\beta }_{1}}}=\dfrac{-{{b}_{1}}}{{{c}_{1}}}-\dfrac{{{b}_{2}}}{{{c}_{2}}}$.
$\Rightarrow \dfrac{\dfrac{-{{b}_{1}}}{{{a}_{1}}}-\dfrac{{{b}_{2}}}{{{a}_{2}}}}{\beta {{\beta }_{1}}}=\dfrac{-{{b}_{1}}}{{{c}_{1}}}-\dfrac{{{b}_{2}}}{{{c}_{2}}}$.
$\Rightarrow \beta {{\beta }_{1}}=\dfrac{\dfrac{-{{b}_{1}}}{{{a}_{1}}}-\dfrac{{{b}_{2}}}{{{a}_{2}}}}{\dfrac{-{{b}_{1}}}{{{c}_{1}}}-\dfrac{{{b}_{2}}}{{{c}_{2}}}}$.
$\Rightarrow \beta {{\beta }_{1}}=\dfrac{\dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}}}{\dfrac{{{b}_{1}}}{{{c}_{1}}}+\dfrac{{{b}_{2}}}{{{c}_{2}}}}$.
From equation (8), we get ${{c}_{3}}=\dfrac{\dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}}}{\dfrac{{{b}_{1}}}{{{c}_{1}}}+\dfrac{{{b}_{2}}}{{{c}_{2}}}}$.
Using these values, we get the quadratic equation as ${{x}^{2}}+\left( \dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}} \right)x+\left( \dfrac{\dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}}}{\dfrac{{{b}_{1}}}{{{c}_{1}}}+\dfrac{{{b}_{2}}}{{{c}_{2}}}} \right)=0$.
$\Rightarrow \dfrac{{{x}^{2}}}{\dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}}}+x+\dfrac{1}{\dfrac{{{b}_{1}}}{{{c}_{1}}}+\dfrac{{{b}_{2}}}{{{c}_{2}}}}=0$.
We have proved the required quadratic equation as $\dfrac{{{x}^{2}}}{\dfrac{{{b}_{1}}}{{{a}_{1}}}+\dfrac{{{b}_{2}}}{{{a}_{2}}}}+x+\dfrac{1}{\dfrac{{{b}_{1}}}{{{c}_{1}}}+\dfrac{{{b}_{2}}}{{{c}_{2}}}}=0$.
Note:
We can also take the new quadratic equation as ${{a}_{3}}{{x}^{2}}+{{b}_{3}}x+{{c}_{3}}=0$, which will also give the same result but with an extra step. We should not confuse with the sum and product of the roots of the quadratic equations. We should not make calculation mistakes while solving this problem. We can also find the roots of both the equations using$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and compare them to solve the given problem.
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