Solve the equation $24{{x}^{3}}-14{{x}^{2}}-63x+45=0$, one root being double the another?
Answer
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Hint: In the above problem, we have given a cubic equation $24{{x}^{3}}-14{{x}^{2}}-63x+45=0$ in which it is given that one root is equal to double of another root so let us assume if one root is $\alpha $ then the other root of the equation is $2\alpha $. Also, let us assume the third root of the equation as $\beta $. Now, we are going to write sum of the roots, sum of the roots taken two at a time and product of the roots. Then we will arrange the three equations in such a way so that we will get the values of $\alpha \And \beta $ and hence will get the roots of the equation.
Complete step by step solution:
The cubic equation given in the above problem is as follows:
$24{{x}^{3}}-14{{x}^{2}}-63x+45=0$
It is given that one root is equal to double of the other root. Let us assume one root of the equation as $\alpha $ and then the other root will be $2\alpha $ and let us assume the third root to be $\beta $.
We know that if we have a cubic equation say $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ then:
Sum of the roots is equal to:
$-\dfrac{b}{a}$
Sum of the roots taken two at a time is equal to:
$\dfrac{c}{a}$
Product of the roots is equal to:
$-\dfrac{d}{a}$
Now, adding the three roots of the equation which we have assumed above we get,
$\alpha +2\alpha +\beta $
And equating the above summation to $-\dfrac{b}{a}=-\dfrac{-14}{24}=\dfrac{14}{24}$ we get,
$\begin{align}
& \alpha +2\alpha +\beta =\dfrac{14}{24} \\
& \Rightarrow 3\alpha +\beta =\dfrac{14}{24} \\
& \Rightarrow 3\alpha +\beta =\dfrac{7}{12}........(1) \\
\end{align}$
Summing the roots taken two at a time we get,
$\alpha \left( 2\alpha \right)+2\alpha \beta +\alpha \beta $
Equating the above summation to $-\dfrac{63}{24}$ we get,
$\begin{align}
& 2{{\alpha }^{2}}+3\alpha \beta =-\dfrac{63}{24} \\
& \Rightarrow 2{{\alpha }^{2}}+3\alpha \beta =-\dfrac{21}{8}.......(2) \\
\end{align}$
Taking the product of three roots we get,
$\alpha 2\alpha \beta $
Equating the above product to $-\dfrac{45}{24}$ we get,
$\begin{align}
& 2{{\alpha }^{2}}\beta =-\dfrac{45}{24} \\
& \Rightarrow 2{{\alpha }^{2}}\beta =-\dfrac{15}{8}..........(3) \\
\end{align}$
Solving eq. (1) and eq. (2) we get,
$\begin{align}
& 3\alpha +\beta =\dfrac{7}{12} \\
& \Rightarrow \beta =\dfrac{7}{12}-3\alpha \\
\end{align}$
Substituting the above value of $\beta $ in eq. (2) we get,
$\begin{align}
& 2{{\alpha }^{2}}+3\alpha \left( \dfrac{7}{12}-3\alpha \right)=-\dfrac{21}{8} \\
& \Rightarrow 2{{\alpha }^{2}}+\dfrac{7\alpha }{4}-9{{\alpha }^{2}}=-\dfrac{21}{8} \\
\end{align}$
Taking 4 as L.C.M in the L.H.S of the above equation we get,
$\begin{align}
& \dfrac{8{{\alpha }^{2}}+7\alpha -36{{\alpha }^{2}}}{4}=-\dfrac{21}{8} \\
& \Rightarrow \dfrac{-28{{\alpha }^{2}}+7\alpha }{4}=-\dfrac{21}{8} \\
\end{align}$
Now, 4 will be cancelled out in the denominator on the both the sides of the above equation and we get,
$\dfrac{-28{{\alpha }^{2}}+7\alpha }{1}=-\dfrac{21}{2}$
Now, taking 7 as common in the numerator of the L.H.S of the above equation we get,
$7\left( -4{{\alpha }^{2}}+\alpha \right)=-\dfrac{21}{2}$
In the above equation, 7 will get cancelled out in the numerator of both the sides of the above equation and we get,
$\left( -4{{\alpha }^{2}}+\alpha \right)=-\dfrac{3}{2}$
Cross multiplying the above equation we get,
$\begin{align}
& -8{{\alpha }^{2}}+2\alpha =-3 \\
& \Rightarrow 8{{\alpha }^{2}}-2\alpha -3=0 \\
\end{align}$
Now, we are going to find the roots of the above equation by factorization method in which we multiply the coefficient of ${{\alpha }^{2}}$ and the constant which is equal to 24 then we factorize this number we get,
$\begin{align}
& 24=1\times 24 \\
& 24=2\times 12 \\
& 24=3\times 8 \\
& 24=4\times 6 \\
\end{align}$
Now, if you look towards the last factors of 24, you will find that if we subtract these two factors we will get the coefficient of $\alpha $ so doing that in the above equation we get,
$8{{\alpha }^{2}}-6\alpha +4\alpha -3=0$
Taking $2\alpha $ as common in the first two terms in the above equation we get,
$2\alpha \left( 4\alpha -3 \right)+1\left( 4\alpha -3 \right)=0$
Taking $\left( 4\alpha -3 \right)$ as common in the above equation we get,
$\left( 4\alpha -3 \right)\left( 2\alpha +1 \right)=0$
Equating each of the brackets to 0 we get,
$\begin{align}
& 4\alpha -3=0 \\
& \Rightarrow \alpha =\dfrac{3}{4} \\
& 2\alpha +1=0 \\
& \Rightarrow \alpha =-\dfrac{1}{2} \\
\end{align}$
Now, substituting these values of $\alpha $ in eq. (1) we get,
When putting $\alpha =\dfrac{3}{4}$ we get,
$\begin{align}
& 3\left( \dfrac{3}{4} \right)+\beta =\dfrac{7}{12} \\
& \Rightarrow \dfrac{9}{4}+\beta =\dfrac{7}{12} \\
& \Rightarrow \beta =\dfrac{7}{12}-\dfrac{9}{4} \\
\end{align}$
Taking 12 as L.C.M in the above we get,
$\begin{align}
& \beta =\dfrac{7-27}{12} \\
& \Rightarrow \beta =-\dfrac{20}{12} \\
& \Rightarrow \beta =-\dfrac{5}{3} \\
\end{align}$
When putting $\alpha =-\dfrac{1}{2}$ we get,
$\begin{align}
& 3\left( -\dfrac{1}{2} \right)+\beta =\dfrac{7}{12} \\
& \Rightarrow -\dfrac{3}{2}+\beta =\dfrac{7}{12} \\
& \Rightarrow \beta =\dfrac{7}{12}+\dfrac{3}{2} \\
\end{align}$
Taking 12 as L.C.M in the above we get,
$\begin{align}
& \beta =\dfrac{7+18}{12} \\
& \Rightarrow \beta =\dfrac{25}{12} \\
\end{align}$
Now, checking the values of $\left( \alpha ,\beta \right)$ which we have found above by substituting in eq. (3) we get,
Checking $\left( \dfrac{3}{4},-\dfrac{5}{3} \right)$ we get,
$\begin{align}
& 2{{\left( \dfrac{3}{4} \right)}^{2}}\left( -\dfrac{5}{3} \right)=-\dfrac{15}{8} \\
& \Rightarrow 2\left( \dfrac{9}{16} \right)\left( -\dfrac{5}{3} \right)=-\dfrac{15}{8} \\
\end{align}$
Now, in the L.H.S of the above equation, 16 will be divided by 2 and 9 will be divided by 3 and we get,
$-\dfrac{15}{8}=-\dfrac{15}{8}$
This value of $\left( \alpha ,\beta \right)$ is acceptable.
Checking $\left( -\dfrac{1}{2},\dfrac{25}{12} \right)$ we get,
$\begin{align}
& 2{{\left( -\dfrac{1}{2} \right)}^{2}}\left( \dfrac{25}{12} \right)=-\dfrac{15}{8} \\
& \Rightarrow 2\left( \dfrac{1}{4} \right)\left( \dfrac{25}{12} \right)=-\dfrac{15}{8} \\
& \Rightarrow \dfrac{25}{24}=-\dfrac{15}{8} \\
\end{align}$
As you can see that L.H.S is not equal to R.H.S so this value of $\left( \alpha ,\beta \right)$ is not acceptable.
Hence, the solutions of the given cubic equation are:
$\dfrac{3}{4},\dfrac{6}{4},-\dfrac{5}{3}$
Note: The mistake that could be possible in the above problem is that you might forget to check the values of $\left( \alpha ,\beta \right)$ by substituting these values in the third equation in the above solution so make sure you won’t make this mistake in the above solution.
Complete step by step solution:
The cubic equation given in the above problem is as follows:
$24{{x}^{3}}-14{{x}^{2}}-63x+45=0$
It is given that one root is equal to double of the other root. Let us assume one root of the equation as $\alpha $ and then the other root will be $2\alpha $ and let us assume the third root to be $\beta $.
We know that if we have a cubic equation say $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ then:
Sum of the roots is equal to:
$-\dfrac{b}{a}$
Sum of the roots taken two at a time is equal to:
$\dfrac{c}{a}$
Product of the roots is equal to:
$-\dfrac{d}{a}$
Now, adding the three roots of the equation which we have assumed above we get,
$\alpha +2\alpha +\beta $
And equating the above summation to $-\dfrac{b}{a}=-\dfrac{-14}{24}=\dfrac{14}{24}$ we get,
$\begin{align}
& \alpha +2\alpha +\beta =\dfrac{14}{24} \\
& \Rightarrow 3\alpha +\beta =\dfrac{14}{24} \\
& \Rightarrow 3\alpha +\beta =\dfrac{7}{12}........(1) \\
\end{align}$
Summing the roots taken two at a time we get,
$\alpha \left( 2\alpha \right)+2\alpha \beta +\alpha \beta $
Equating the above summation to $-\dfrac{63}{24}$ we get,
$\begin{align}
& 2{{\alpha }^{2}}+3\alpha \beta =-\dfrac{63}{24} \\
& \Rightarrow 2{{\alpha }^{2}}+3\alpha \beta =-\dfrac{21}{8}.......(2) \\
\end{align}$
Taking the product of three roots we get,
$\alpha 2\alpha \beta $
Equating the above product to $-\dfrac{45}{24}$ we get,
$\begin{align}
& 2{{\alpha }^{2}}\beta =-\dfrac{45}{24} \\
& \Rightarrow 2{{\alpha }^{2}}\beta =-\dfrac{15}{8}..........(3) \\
\end{align}$
Solving eq. (1) and eq. (2) we get,
$\begin{align}
& 3\alpha +\beta =\dfrac{7}{12} \\
& \Rightarrow \beta =\dfrac{7}{12}-3\alpha \\
\end{align}$
Substituting the above value of $\beta $ in eq. (2) we get,
$\begin{align}
& 2{{\alpha }^{2}}+3\alpha \left( \dfrac{7}{12}-3\alpha \right)=-\dfrac{21}{8} \\
& \Rightarrow 2{{\alpha }^{2}}+\dfrac{7\alpha }{4}-9{{\alpha }^{2}}=-\dfrac{21}{8} \\
\end{align}$
Taking 4 as L.C.M in the L.H.S of the above equation we get,
$\begin{align}
& \dfrac{8{{\alpha }^{2}}+7\alpha -36{{\alpha }^{2}}}{4}=-\dfrac{21}{8} \\
& \Rightarrow \dfrac{-28{{\alpha }^{2}}+7\alpha }{4}=-\dfrac{21}{8} \\
\end{align}$
Now, 4 will be cancelled out in the denominator on the both the sides of the above equation and we get,
$\dfrac{-28{{\alpha }^{2}}+7\alpha }{1}=-\dfrac{21}{2}$
Now, taking 7 as common in the numerator of the L.H.S of the above equation we get,
$7\left( -4{{\alpha }^{2}}+\alpha \right)=-\dfrac{21}{2}$
In the above equation, 7 will get cancelled out in the numerator of both the sides of the above equation and we get,
$\left( -4{{\alpha }^{2}}+\alpha \right)=-\dfrac{3}{2}$
Cross multiplying the above equation we get,
$\begin{align}
& -8{{\alpha }^{2}}+2\alpha =-3 \\
& \Rightarrow 8{{\alpha }^{2}}-2\alpha -3=0 \\
\end{align}$
Now, we are going to find the roots of the above equation by factorization method in which we multiply the coefficient of ${{\alpha }^{2}}$ and the constant which is equal to 24 then we factorize this number we get,
$\begin{align}
& 24=1\times 24 \\
& 24=2\times 12 \\
& 24=3\times 8 \\
& 24=4\times 6 \\
\end{align}$
Now, if you look towards the last factors of 24, you will find that if we subtract these two factors we will get the coefficient of $\alpha $ so doing that in the above equation we get,
$8{{\alpha }^{2}}-6\alpha +4\alpha -3=0$
Taking $2\alpha $ as common in the first two terms in the above equation we get,
$2\alpha \left( 4\alpha -3 \right)+1\left( 4\alpha -3 \right)=0$
Taking $\left( 4\alpha -3 \right)$ as common in the above equation we get,
$\left( 4\alpha -3 \right)\left( 2\alpha +1 \right)=0$
Equating each of the brackets to 0 we get,
$\begin{align}
& 4\alpha -3=0 \\
& \Rightarrow \alpha =\dfrac{3}{4} \\
& 2\alpha +1=0 \\
& \Rightarrow \alpha =-\dfrac{1}{2} \\
\end{align}$
Now, substituting these values of $\alpha $ in eq. (1) we get,
When putting $\alpha =\dfrac{3}{4}$ we get,
$\begin{align}
& 3\left( \dfrac{3}{4} \right)+\beta =\dfrac{7}{12} \\
& \Rightarrow \dfrac{9}{4}+\beta =\dfrac{7}{12} \\
& \Rightarrow \beta =\dfrac{7}{12}-\dfrac{9}{4} \\
\end{align}$
Taking 12 as L.C.M in the above we get,
$\begin{align}
& \beta =\dfrac{7-27}{12} \\
& \Rightarrow \beta =-\dfrac{20}{12} \\
& \Rightarrow \beta =-\dfrac{5}{3} \\
\end{align}$
When putting $\alpha =-\dfrac{1}{2}$ we get,
$\begin{align}
& 3\left( -\dfrac{1}{2} \right)+\beta =\dfrac{7}{12} \\
& \Rightarrow -\dfrac{3}{2}+\beta =\dfrac{7}{12} \\
& \Rightarrow \beta =\dfrac{7}{12}+\dfrac{3}{2} \\
\end{align}$
Taking 12 as L.C.M in the above we get,
$\begin{align}
& \beta =\dfrac{7+18}{12} \\
& \Rightarrow \beta =\dfrac{25}{12} \\
\end{align}$
Now, checking the values of $\left( \alpha ,\beta \right)$ which we have found above by substituting in eq. (3) we get,
Checking $\left( \dfrac{3}{4},-\dfrac{5}{3} \right)$ we get,
$\begin{align}
& 2{{\left( \dfrac{3}{4} \right)}^{2}}\left( -\dfrac{5}{3} \right)=-\dfrac{15}{8} \\
& \Rightarrow 2\left( \dfrac{9}{16} \right)\left( -\dfrac{5}{3} \right)=-\dfrac{15}{8} \\
\end{align}$
Now, in the L.H.S of the above equation, 16 will be divided by 2 and 9 will be divided by 3 and we get,
$-\dfrac{15}{8}=-\dfrac{15}{8}$
This value of $\left( \alpha ,\beta \right)$ is acceptable.
Checking $\left( -\dfrac{1}{2},\dfrac{25}{12} \right)$ we get,
$\begin{align}
& 2{{\left( -\dfrac{1}{2} \right)}^{2}}\left( \dfrac{25}{12} \right)=-\dfrac{15}{8} \\
& \Rightarrow 2\left( \dfrac{1}{4} \right)\left( \dfrac{25}{12} \right)=-\dfrac{15}{8} \\
& \Rightarrow \dfrac{25}{24}=-\dfrac{15}{8} \\
\end{align}$
As you can see that L.H.S is not equal to R.H.S so this value of $\left( \alpha ,\beta \right)$ is not acceptable.
Hence, the solutions of the given cubic equation are:
$\dfrac{3}{4},\dfrac{6}{4},-\dfrac{5}{3}$
Note: The mistake that could be possible in the above problem is that you might forget to check the values of $\left( \alpha ,\beta \right)$ by substituting these values in the third equation in the above solution so make sure you won’t make this mistake in the above solution.
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