Solve the equation $24{{x}^{3}}-14{{x}^{2}}-63x+\lambda =0$ , one root being double of another. Hence find the value(s) of $\lambda $ .
Answer
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Hint: To solve the equation $24{{x}^{3}}-14{{x}^{2}}-63x+\lambda =0$ and find the value of $\lambda $ , we will consider the roots to be $\alpha ,\beta $ and $\gamma $ . From the given condition, we will assume $\beta =2\alpha $ . Using the properties of roots of cubic equation, we will get three equations $\alpha +\beta +\gamma =\dfrac{14}{24}$ , $\alpha \beta \gamma =\dfrac{-\lambda }{24}$ and $\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{-63}{24}$ . We have to substitute the given condition in these equations and solve to get the values of the roots and $\lambda $ .
Complete step by step solution:
We have to solve the equation $24{{x}^{3}}-14{{x}^{2}}-63x+\lambda =0$ and find the value of $\lambda $ . Let us consider the roots to be $\alpha ,\beta $ and $\gamma $ . We know that the sum of roots in a cubic equation of the form $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ is given as
$\alpha +\beta +\gamma =\dfrac{-b}{a}$
Here, we can see that $a=24,b=-14,c=-63$ and $d=\lambda $ .
$\begin{align}
& \Rightarrow \alpha +\beta +\gamma =\dfrac{-\left( -14 \right)}{24} \\
& \Rightarrow \alpha +\beta +\gamma =\dfrac{14}{24} \\
\end{align}$
Let us cancel the common factor of 2 from the RHS.
$\Rightarrow \alpha +\beta +\gamma =\dfrac{7}{12}...\left( i \right)$
We are given that one root is double of another. Let us consider $\beta $ to be the double of $\alpha $
$\Rightarrow \beta =2\alpha ...\left( ii \right)$
Let us substitute the above value in the equation (i).
$\begin{align}
& \Rightarrow \alpha +2\alpha +\gamma =\dfrac{7}{12} \\
& \Rightarrow 3\alpha +\gamma =\dfrac{7}{12}...\left( iii \right) \\
\end{align}$
We know that the product of the roots of cubic equation is given as
$\alpha \beta \gamma =\dfrac{-d}{a}$
Let us substitute the values.
$\Rightarrow \alpha \beta \gamma =\dfrac{-\lambda }{24}$
Let us substitute (ii) in the above equation.
$\begin{align}
& \Rightarrow \alpha \cdot 2\alpha \cdot \gamma =\dfrac{-\lambda }{24} \\
& \Rightarrow 2{{\alpha }^{2}}\gamma =\dfrac{-\lambda }{24} \\
\end{align}$
Let us take 2 from LHS to the RHS.
$\begin{align}
& \Rightarrow {{\alpha }^{2}}\gamma =\dfrac{-\lambda }{24\times 2} \\
& \Rightarrow {{\alpha }^{2}}\gamma =\dfrac{-\lambda }{48}...\left( iv \right) \\
\end{align}$
We know that the sum of the products of the roots of cubic equation is given by
$\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a}$ .
Let us substitute the values in the above equation.
$\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{-63}{24}$
Let us cancel the common factor 3 from the RHS.
$\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{-21}{8}$
Now, we have to substitute (ii) in the above equation.
$\begin{align}
& \Rightarrow 2{{\alpha }^{2}}+2\alpha \gamma +\gamma \alpha =\dfrac{-21}{8} \\
& \Rightarrow 2{{\alpha }^{2}}+3\alpha \gamma =\dfrac{-21}{8}...\left( v \right) \\
\end{align}$
Let us solve equation (iii) and (v). Let us use a substitution method. From equation (iii), we can find the value of $\gamma $ as follows.
$\Rightarrow 3\alpha +\gamma =\dfrac{7}{12}$
Let us take $3\alpha $ to the RHS.
$\Rightarrow \gamma =\dfrac{7}{12}-3\alpha ...\left( vi \right)$
Let us substitute equation (vi) in equation (v).
$\Rightarrow 2{{\alpha }^{2}}+3\alpha \left( \dfrac{7}{12}-3\alpha \right)=\dfrac{-21}{8}$
We have to apply distributive property on the second term of the LHS.
\[\Rightarrow 2{{\alpha }^{2}}+\dfrac{21\alpha }{12}-9{{\alpha }^{2}}=\dfrac{-21}{8}\]
We have to solve the LHS.
\[\begin{align}
& \Rightarrow -7{{\alpha }^{2}}+\dfrac{21\alpha }{12}=\dfrac{-21}{8} \\
& \Rightarrow -7{{\alpha }^{2}}+\dfrac{7\alpha }{4}=\dfrac{-21}{8} \\
\end{align}\]
Let us take the common term 7 outside from the LHS.
\[\Rightarrow 7\left( -{{\alpha }^{2}}+\dfrac{\alpha }{4} \right)=\dfrac{-21}{8}\]
We can take 7 to the RHS.
\[\Rightarrow -{{\alpha }^{2}}+\dfrac{\alpha }{4}=\dfrac{-21}{8\times 7}\]
Let us cancel the common factor 7 from the RHS.
\[\Rightarrow -{{\alpha }^{2}}+\dfrac{\alpha }{4}=\dfrac{-3}{8}\]
We have to simplify the LHS by taking the LCM and solving.
\[\Rightarrow \dfrac{-4{{\alpha }^{2}}+\alpha }{4}=\dfrac{-3}{8}\]
Let us take 4 from the denominator of the LHS to the RHS.
\[\begin{align}
& \Rightarrow -4{{\alpha }^{2}}+\alpha =\dfrac{-3\times 4}{8} \\
& \Rightarrow -4{{\alpha }^{2}}+\alpha =\dfrac{-3}{2} \\
\end{align}\]
Let us take the RHS to the LHS.
\[\Rightarrow -4{{\alpha }^{2}}+\alpha +\dfrac{3}{2}=0\]
Now, we have to solve this equation. The above equation is of the form $a{{x}^{2}}+bx+c=0$ . We have to use the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
$\begin{align}
& \Rightarrow \alpha =\dfrac{-1\pm \sqrt{{{1}^{2}}-4\times -4\times \dfrac{3}{2}}}{2\times -4} \\
& \Rightarrow \alpha =\dfrac{-1\pm \sqrt{1-4\times -2\times 3}}{-8} \\
& \Rightarrow \alpha =\dfrac{-1\pm \sqrt{1+24}}{-8} \\
& \Rightarrow \alpha =\dfrac{-1\pm \sqrt{25}}{-8} \\
& \Rightarrow \alpha =\dfrac{-1\pm 5}{-8} \\
\end{align}$
Let us split the signs.
$\Rightarrow \alpha =\dfrac{-1+5}{-8},\alpha =\dfrac{-1-5}{-8}$
Now, let us simplify the above expression.
$\begin{align}
& \Rightarrow \alpha =\dfrac{4}{-8},\alpha =\dfrac{-6}{-8} \\
& \Rightarrow \alpha =\dfrac{-1}{2},\alpha =\dfrac{3}{4} \\
\end{align}$
Now, we have obtained two values of $\alpha $ . Let us substitute each of these values in equation (vi).
Let us consider $\alpha =\dfrac{-1}{2}$ .
$\begin{align}
& \Rightarrow \gamma =\dfrac{7}{12}-3\times \dfrac{-1}{2} \\
& \Rightarrow \gamma =\dfrac{7}{12}+\dfrac{3}{2} \\
\end{align}$
We have to take the LCM and simplify.
$\Rightarrow \gamma =\dfrac{7}{12}+\dfrac{18}{12}=\dfrac{7+18}{12}=\dfrac{25}{12}$
Let us find $\beta $ from equation (ii).
$\Rightarrow \beta =2\times \dfrac{-1}{2}=-1$
Now, we have to find $\lambda $ using equation (iv).
$\begin{align}
& \Rightarrow {{\left( \dfrac{-1}{2} \right)}^{2}}\left( \dfrac{25}{12} \right)=\dfrac{-\lambda }{48} \\
& \Rightarrow \dfrac{1}{4}\times \left( \dfrac{25}{12} \right)=\dfrac{-\lambda }{48} \\
\end{align}$
Let us take 48 to the LHS.
$\Rightarrow -\lambda =\dfrac{1}{4}\times \dfrac{25}{12}\times 48$
Let us cancel the common factor 4.
$\Rightarrow -\lambda =\dfrac{25}{12}\times 12$
We have to cancel the common factor 12.
$\begin{align}
& \Rightarrow -\lambda =25 \\
& \Rightarrow \lambda =-25 \\
\end{align}$
Therefore, we found that $\alpha =\dfrac{-1}{2}$ , $\beta =-1,\gamma =\dfrac{25}{12},\lambda =-25$ .
Now, let us consider $\alpha =\dfrac{3}{4}$ .
Using equation (ii), we will get
$\beta =2\times \dfrac{3}{4}=\dfrac{3}{2}$
Let us substitute this value of $\alpha $ in equation (iv).
$\begin{align}
& \Rightarrow \gamma =\dfrac{7}{12}-3\times \dfrac{3}{4} \\
& \Rightarrow \gamma =\dfrac{7}{12}-\dfrac{9}{4} \\
\end{align}$
Let us take the LCM.
$\begin{align}
& \Rightarrow \gamma =\dfrac{7}{12}-\dfrac{27}{12}=\dfrac{7-27}{12} \\
& \Rightarrow \gamma =\dfrac{-20}{12} \\
\end{align}$
Let us cancel the common factor 4.
$\Rightarrow \gamma =\dfrac{-5}{3}$
Now, we have to substitute these values in equation (iv).
\[\begin{align}
& \Rightarrow {{\left( \dfrac{3}{4} \right)}^{2}}\left( \dfrac{-5}{3} \right)=\dfrac{-\lambda }{48} \\
& \Rightarrow \dfrac{9}{16}\times \dfrac{-5}{3}=\dfrac{-\lambda }{48} \\
\end{align}\]
Let us cancel the common factor 3.
\[\Rightarrow \dfrac{3}{16}\times -5=\dfrac{-\lambda }{48}\]
We have to take 48 from RHS to the LHS.
\[\Rightarrow \dfrac{3}{16}\times -5\times 48=-\lambda \]
Let us cancel the common factor 16.
\[\begin{align}
& \Rightarrow 3\times -5\times 3=-\lambda \\
& \Rightarrow -\lambda =-45 \\
& \Rightarrow \lambda =45 \\
\end{align}\]
Hence, we found that $\alpha =\dfrac{3}{4},\beta =\dfrac{3}{2},\gamma =\dfrac{-5}{3},\lambda =45$ .
Therefore, the solutions of $24{{x}^{3}}-14{{x}^{2}}-63x+\lambda =0$ are \[x=\dfrac{-1}{2},-1,\dfrac{25}{12}\] when \[\lambda =-25\] and $x=\dfrac{3}{4},\dfrac{3}{2},\dfrac{-5}{3}$ when $\lambda =45$ .
Note: Students can also consider $\alpha =2\beta $ , $\alpha =2\gamma $ , $\gamma =2\alpha ,\gamma =2\beta $ or $\beta =2\gamma $ . In either case the value of the roots and the $\lambda $ will be similar to the above solution. Students can also use other methods like elimination or cross-multiplication to solve the equations.
Complete step by step solution:
We have to solve the equation $24{{x}^{3}}-14{{x}^{2}}-63x+\lambda =0$ and find the value of $\lambda $ . Let us consider the roots to be $\alpha ,\beta $ and $\gamma $ . We know that the sum of roots in a cubic equation of the form $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ is given as
$\alpha +\beta +\gamma =\dfrac{-b}{a}$
Here, we can see that $a=24,b=-14,c=-63$ and $d=\lambda $ .
$\begin{align}
& \Rightarrow \alpha +\beta +\gamma =\dfrac{-\left( -14 \right)}{24} \\
& \Rightarrow \alpha +\beta +\gamma =\dfrac{14}{24} \\
\end{align}$
Let us cancel the common factor of 2 from the RHS.
$\Rightarrow \alpha +\beta +\gamma =\dfrac{7}{12}...\left( i \right)$
We are given that one root is double of another. Let us consider $\beta $ to be the double of $\alpha $
$\Rightarrow \beta =2\alpha ...\left( ii \right)$
Let us substitute the above value in the equation (i).
$\begin{align}
& \Rightarrow \alpha +2\alpha +\gamma =\dfrac{7}{12} \\
& \Rightarrow 3\alpha +\gamma =\dfrac{7}{12}...\left( iii \right) \\
\end{align}$
We know that the product of the roots of cubic equation is given as
$\alpha \beta \gamma =\dfrac{-d}{a}$
Let us substitute the values.
$\Rightarrow \alpha \beta \gamma =\dfrac{-\lambda }{24}$
Let us substitute (ii) in the above equation.
$\begin{align}
& \Rightarrow \alpha \cdot 2\alpha \cdot \gamma =\dfrac{-\lambda }{24} \\
& \Rightarrow 2{{\alpha }^{2}}\gamma =\dfrac{-\lambda }{24} \\
\end{align}$
Let us take 2 from LHS to the RHS.
$\begin{align}
& \Rightarrow {{\alpha }^{2}}\gamma =\dfrac{-\lambda }{24\times 2} \\
& \Rightarrow {{\alpha }^{2}}\gamma =\dfrac{-\lambda }{48}...\left( iv \right) \\
\end{align}$
We know that the sum of the products of the roots of cubic equation is given by
$\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a}$ .
Let us substitute the values in the above equation.
$\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{-63}{24}$
Let us cancel the common factor 3 from the RHS.
$\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{-21}{8}$
Now, we have to substitute (ii) in the above equation.
$\begin{align}
& \Rightarrow 2{{\alpha }^{2}}+2\alpha \gamma +\gamma \alpha =\dfrac{-21}{8} \\
& \Rightarrow 2{{\alpha }^{2}}+3\alpha \gamma =\dfrac{-21}{8}...\left( v \right) \\
\end{align}$
Let us solve equation (iii) and (v). Let us use a substitution method. From equation (iii), we can find the value of $\gamma $ as follows.
$\Rightarrow 3\alpha +\gamma =\dfrac{7}{12}$
Let us take $3\alpha $ to the RHS.
$\Rightarrow \gamma =\dfrac{7}{12}-3\alpha ...\left( vi \right)$
Let us substitute equation (vi) in equation (v).
$\Rightarrow 2{{\alpha }^{2}}+3\alpha \left( \dfrac{7}{12}-3\alpha \right)=\dfrac{-21}{8}$
We have to apply distributive property on the second term of the LHS.
\[\Rightarrow 2{{\alpha }^{2}}+\dfrac{21\alpha }{12}-9{{\alpha }^{2}}=\dfrac{-21}{8}\]
We have to solve the LHS.
\[\begin{align}
& \Rightarrow -7{{\alpha }^{2}}+\dfrac{21\alpha }{12}=\dfrac{-21}{8} \\
& \Rightarrow -7{{\alpha }^{2}}+\dfrac{7\alpha }{4}=\dfrac{-21}{8} \\
\end{align}\]
Let us take the common term 7 outside from the LHS.
\[\Rightarrow 7\left( -{{\alpha }^{2}}+\dfrac{\alpha }{4} \right)=\dfrac{-21}{8}\]
We can take 7 to the RHS.
\[\Rightarrow -{{\alpha }^{2}}+\dfrac{\alpha }{4}=\dfrac{-21}{8\times 7}\]
Let us cancel the common factor 7 from the RHS.
\[\Rightarrow -{{\alpha }^{2}}+\dfrac{\alpha }{4}=\dfrac{-3}{8}\]
We have to simplify the LHS by taking the LCM and solving.
\[\Rightarrow \dfrac{-4{{\alpha }^{2}}+\alpha }{4}=\dfrac{-3}{8}\]
Let us take 4 from the denominator of the LHS to the RHS.
\[\begin{align}
& \Rightarrow -4{{\alpha }^{2}}+\alpha =\dfrac{-3\times 4}{8} \\
& \Rightarrow -4{{\alpha }^{2}}+\alpha =\dfrac{-3}{2} \\
\end{align}\]
Let us take the RHS to the LHS.
\[\Rightarrow -4{{\alpha }^{2}}+\alpha +\dfrac{3}{2}=0\]
Now, we have to solve this equation. The above equation is of the form $a{{x}^{2}}+bx+c=0$ . We have to use the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
$\begin{align}
& \Rightarrow \alpha =\dfrac{-1\pm \sqrt{{{1}^{2}}-4\times -4\times \dfrac{3}{2}}}{2\times -4} \\
& \Rightarrow \alpha =\dfrac{-1\pm \sqrt{1-4\times -2\times 3}}{-8} \\
& \Rightarrow \alpha =\dfrac{-1\pm \sqrt{1+24}}{-8} \\
& \Rightarrow \alpha =\dfrac{-1\pm \sqrt{25}}{-8} \\
& \Rightarrow \alpha =\dfrac{-1\pm 5}{-8} \\
\end{align}$
Let us split the signs.
$\Rightarrow \alpha =\dfrac{-1+5}{-8},\alpha =\dfrac{-1-5}{-8}$
Now, let us simplify the above expression.
$\begin{align}
& \Rightarrow \alpha =\dfrac{4}{-8},\alpha =\dfrac{-6}{-8} \\
& \Rightarrow \alpha =\dfrac{-1}{2},\alpha =\dfrac{3}{4} \\
\end{align}$
Now, we have obtained two values of $\alpha $ . Let us substitute each of these values in equation (vi).
Let us consider $\alpha =\dfrac{-1}{2}$ .
$\begin{align}
& \Rightarrow \gamma =\dfrac{7}{12}-3\times \dfrac{-1}{2} \\
& \Rightarrow \gamma =\dfrac{7}{12}+\dfrac{3}{2} \\
\end{align}$
We have to take the LCM and simplify.
$\Rightarrow \gamma =\dfrac{7}{12}+\dfrac{18}{12}=\dfrac{7+18}{12}=\dfrac{25}{12}$
Let us find $\beta $ from equation (ii).
$\Rightarrow \beta =2\times \dfrac{-1}{2}=-1$
Now, we have to find $\lambda $ using equation (iv).
$\begin{align}
& \Rightarrow {{\left( \dfrac{-1}{2} \right)}^{2}}\left( \dfrac{25}{12} \right)=\dfrac{-\lambda }{48} \\
& \Rightarrow \dfrac{1}{4}\times \left( \dfrac{25}{12} \right)=\dfrac{-\lambda }{48} \\
\end{align}$
Let us take 48 to the LHS.
$\Rightarrow -\lambda =\dfrac{1}{4}\times \dfrac{25}{12}\times 48$
Let us cancel the common factor 4.
$\Rightarrow -\lambda =\dfrac{25}{12}\times 12$
We have to cancel the common factor 12.
$\begin{align}
& \Rightarrow -\lambda =25 \\
& \Rightarrow \lambda =-25 \\
\end{align}$
Therefore, we found that $\alpha =\dfrac{-1}{2}$ , $\beta =-1,\gamma =\dfrac{25}{12},\lambda =-25$ .
Now, let us consider $\alpha =\dfrac{3}{4}$ .
Using equation (ii), we will get
$\beta =2\times \dfrac{3}{4}=\dfrac{3}{2}$
Let us substitute this value of $\alpha $ in equation (iv).
$\begin{align}
& \Rightarrow \gamma =\dfrac{7}{12}-3\times \dfrac{3}{4} \\
& \Rightarrow \gamma =\dfrac{7}{12}-\dfrac{9}{4} \\
\end{align}$
Let us take the LCM.
$\begin{align}
& \Rightarrow \gamma =\dfrac{7}{12}-\dfrac{27}{12}=\dfrac{7-27}{12} \\
& \Rightarrow \gamma =\dfrac{-20}{12} \\
\end{align}$
Let us cancel the common factor 4.
$\Rightarrow \gamma =\dfrac{-5}{3}$
Now, we have to substitute these values in equation (iv).
\[\begin{align}
& \Rightarrow {{\left( \dfrac{3}{4} \right)}^{2}}\left( \dfrac{-5}{3} \right)=\dfrac{-\lambda }{48} \\
& \Rightarrow \dfrac{9}{16}\times \dfrac{-5}{3}=\dfrac{-\lambda }{48} \\
\end{align}\]
Let us cancel the common factor 3.
\[\Rightarrow \dfrac{3}{16}\times -5=\dfrac{-\lambda }{48}\]
We have to take 48 from RHS to the LHS.
\[\Rightarrow \dfrac{3}{16}\times -5\times 48=-\lambda \]
Let us cancel the common factor 16.
\[\begin{align}
& \Rightarrow 3\times -5\times 3=-\lambda \\
& \Rightarrow -\lambda =-45 \\
& \Rightarrow \lambda =45 \\
\end{align}\]
Hence, we found that $\alpha =\dfrac{3}{4},\beta =\dfrac{3}{2},\gamma =\dfrac{-5}{3},\lambda =45$ .
Therefore, the solutions of $24{{x}^{3}}-14{{x}^{2}}-63x+\lambda =0$ are \[x=\dfrac{-1}{2},-1,\dfrac{25}{12}\] when \[\lambda =-25\] and $x=\dfrac{3}{4},\dfrac{3}{2},\dfrac{-5}{3}$ when $\lambda =45$ .
Note: Students can also consider $\alpha =2\beta $ , $\alpha =2\gamma $ , $\gamma =2\alpha ,\gamma =2\beta $ or $\beta =2\gamma $ . In either case the value of the roots and the $\lambda $ will be similar to the above solution. Students can also use other methods like elimination or cross-multiplication to solve the equations.
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