Question

The solution of the equation \[2{x^3} - {x^2} - 22x - 24 = 0\] when two of the roots are in ratio \[3:4\] , is
A) \[3,4,\dfrac{1}{2}\]
B) \[ - \dfrac{3}{2}, - 2,4\]
C) \[ - \dfrac{1}{2},\dfrac{3}{2},2\]
D) \[\dfrac{3}{2},2,\dfrac{5}{2}\]

Answer 1

Hint: First, we will assume the three roots. Then, we make three equations by the knowledge of the cubic equation and the relationship between the coefficients and roots. We should know that if \[a{x^3} + b{x^2} + cx + d = 0\] is the general equation, then, the sum of the roots is \[ - \dfrac{b}{a}\] , where b is the coefficient of \[{x^2}\] and a is the coefficient of \[{x^3}\] , product of roots is \[ - \dfrac{d}{a}\] , where d is the constant term and a is the coefficient of \[{x^3}\] , and the sum of product of roots taken two at a time is, \[\dfrac{c}{a}\] , where c is the coefficient of \[x\] and a is the coefficient of \[{x^3}\] .

Complete step-by-step answer:
We know that the equation \[2{x^3} - {x^2} - 22x - 24 = 0\] is cubic so it will have three roots. We are given that two of the roots are in ratio \[3:4\] , so let the three roots be 3m, 4m, n.
Comparing the given equation with general equation, \[a{x^3} + b{x^2} + cx + d = 0\] we get,
 \[a = 2\]
 \[b = - 1\]
 \[c = - 22\]
 \[d = - 24\]
Now, we know that, the sum of the roots is \[ - \dfrac{b}{a}\] , so we get,
 \[3m + 4m + n = - \dfrac{{\left( { - 1} \right)}}{2}\]
Simplifying the equation,
 \[ \Rightarrow 7m + n = \dfrac{1}{2} - - - - \left( 1 \right)\]
Now, the product of roots is \[ - \dfrac{d}{a}\] . So, we get,
 \[\left( {3m} \right)\left( {4m} \right)\left( n \right) = - \dfrac{{\left( { - 24} \right)}}{2}\]
 \[ \Rightarrow 12{m^2}n = 12\]
 \[ \Rightarrow {m^2}n = 1 - - - - \left( 2 \right)\]
Now, the sum of product of roots taken two at a time is \[\dfrac{c}{a}\] ,
 \[\left( {3m} \right)\left( {4m} \right) + \left( {4m} \right)\left( n \right) + \left( n \right)\left( {3m} \right) = \dfrac{{\left( { - 22} \right)}}{2}\]
Simplifying the equation, we get,
 \[ \Rightarrow 12{m^2} + 4mn + 3mn = - 11\]
 \[ \Rightarrow 12{m^2} + 7mn = - 11 - - - - \left( 3 \right)\]
Substituting value of n from equation $\left( 1 \right)$, that is \[n = \dfrac{1}{2} - 7m\] , in equation $\left( 2 \right)$, we get,
 \[12{m^2} + 7m\left( {\dfrac{1}{2} - 7m} \right) = - 11\]
Now, on solving we get,
 \[ \Rightarrow 12{m^2} + \dfrac{7}{2}m - 49{m^2} = - 11\]
 \[ \Rightarrow - 37{m^2} + \dfrac{7}{2}m + 11 = 0\]
Multiplying the whole equation by two, we get,
 \[ \Rightarrow - 74{m^2} + 7m + 22 = 0\]
Multiplying the whole equation by $ - 1$, we get,
 \[ \Rightarrow 74{m^2} - 7m - 22 = 0\]
Now, by splitting the middle term method, in this method, we split the middle terms in two parts such that if we multiply them we get the product of the coefficient of \[{x^2}\] and constant term and on its addition, we get the coefficient of x.
So, this equation changes to
 \[ \Rightarrow 74{m^2} - 44m + 37m - 22 = 0\]
We split the middle term $ - 7m$ into two terms $ - 44m$ and $37m$ since the product of these terms, $ - 1628{m^2}$ is equal to the product of the constant term and coefficient of ${x^2}$ and sum of these terms gives us the original middle term, $ - 7m$. So, we get,
 \[ \Rightarrow 2m\left( {37m - 22} \right) + 1\left( {37m - 22} \right) = 0\]
Factoring out the common terms, we get,
 \[ \Rightarrow \left( {2m + 1} \right)\left( {37m - 22} \right) = 0\]
Now, since product of two terms is zero. So, either of the two terms have to be zero.
Here, either \[2m + 1 = 0\] or \[37m - 22 = 0\]
 \[ \Rightarrow m = - \dfrac{1}{2}\] or \[m = \dfrac{{22}}{{37}}\]
If \[m = - \dfrac{1}{2}\] , then we get from equation $\left( 1 \right)$,
 \[n = \dfrac{1}{2} - 7\left( { - \dfrac{1}{2}} \right)\]
 \[ \Rightarrow n = \dfrac{{1 + 7}}{2} = \dfrac{8}{2} = 4\]
And if \[m = \dfrac{{22}}{{37}}\] , then from equation $\left( 1 \right)$,
 \[n = \dfrac{1}{2} - 7\left( {\dfrac{{22}}{{37}}} \right)\]
 \[ \Rightarrow n = \dfrac{{37 - 308}}{{74}} = \dfrac{{271}}{{74}}\]
To check the roots of the given equation, \[2{x^3} - {x^2} - 22x - 24 = 0\] we substitute values of m and in in equation $\left( 3 \right)$.
If \[m = \dfrac{{22}}{{37}}\] and \[n = \dfrac{{217}}{{74}}\] .
Then,
 \[{m^2}n = {\left( {\dfrac{{22}}{{37}}} \right)^2}\left( {\dfrac{{271}}{{74}}} \right) \ne 1\]
So this cannot be the solution of the given cubic equation.
Now, check for \[m = - \dfrac{1}{2}\] and \[n = 4\] .
 \[{m^2}n = {\left( { - \dfrac{1}{2}} \right)^2}\left( 4 \right) = \dfrac{1}{4} \times 4 = 1\]
Hence, \[m = - \dfrac{1}{2}\] and \[n = 4\] these are solutions.
Substituting value of m, we get the roots of cubic equation as, \[ - \dfrac{3}{2}, - 2,4\]
Hence, option (B) is the correct answer.
So, the correct answer is “Option B”.

Note: The main thing to keep in mind while doing these questions is that , the sum of the roots is \[ - \dfrac{b}{a}\] , where b is the coefficient of \[{x^2}\] and a is the coefficient of \[{x^3}\] , product of roots is \[ - \dfrac{d}{a}\] , where d is the constant term and a is the coefficient of \[{x^3}\] , and the sum of product of roots taken two at a time is, \[\dfrac{c}{a}\] , where c is the coefficient of \[x\] and a is the coefficient of \[{x^3}\] , where, \[a{x^3} + b{x^2} + cx + d = 0\] is the general equation. We must know the splitting the middle terms method to solve the equation formed in the problem.