Question

Solve the equation \[24{{x}^{3}}-14{{x}^{2}}-63x+45=0\] , one root being double another.

Answer 1

Hint: Let us assume p, q, and r as the roots of the cubic equation \[(24{{x}^{3}}-14{{x}^{2}}-63x+45=0)\] . We know the formula of the sum of roots, the sum of products of two roots taken at a time, and product of all roots, \[\alpha +\beta +\gamma =\dfrac{-b}{a}\] , \[\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a}\] , and \[\alpha \beta \gamma =\dfrac{-d}{a}\] where, \[\alpha ,\beta ,and\,\gamma \]are the roots of the cubic equation \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\] . Now, using these three formulas, we get three equations. Replace q by 2p in these equations. Now, solve the equations further and get the values of p. After getting the values of p we have to find the values of q and r. Then check the values of p, q, and r whether it satisfies the equations.

Complete step-by-step answer:

Assume the standard cubic equation \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\] whose roots are \[\alpha ,\beta ,and\,\gamma \] .
\[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\]……………………(1)
We know the formula of the sum of roots, sum of products of two roots taken at a time, and product of all roots.
\[\alpha +\beta +\gamma =\dfrac{-b}{a}\] …………………….(2)
\[\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a}\] ……………………(3)
\[\alpha \beta \gamma =\dfrac{-d}{a}\] …………………….(4)
According to the question, it is given that the cubic equation is,
\[24{{x}^{3}}-14{{x}^{2}}-63x+45=0\] …………………..(5)
On comparing equation (1) and equation (5), we get
\[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\]
\[24{{x}^{3}}-14{{x}^{2}}-63x+45=0\]
Here, we get a=24, b=-14, c=-63, and d=45……………..(6)
Let us assume the roots of the cubic equation \[(24{{x}^{3}}-14{{x}^{2}}-63x+45=0)\] be p, q, and r.
It is given that one root is double to another. Let us say that the root q is double of p. So,
\[q=2p\] ………………(7)
From equation (1), equation (2), and equation (3), we have the formula of the sum of roots, sum of products of two roots taken at a time, and product of all roots.
Replacing \[\alpha \] by p, \[\beta \] by q, and \[\gamma \] by r in equation (2), we get,
\[p+q+r=\dfrac{-b}{a}\]
Now, using equation(6), transforming the above equation, we get
\[p+2p+r=\dfrac{-b}{a}\]
\[\Rightarrow 3p+r=-\dfrac{b}{a}\] …………………….(8)
Replacing \[\alpha \] by p, \[\beta \] by q, and \[\gamma \] by r in equation (3), we get,
\[pq+qr+pr=\dfrac{c}{a}\] ……………………(9)
Now, using equation(6), transforming the above equation, we get
\[p(2p)+(2p)r+pr=\dfrac{c}{a}\]
\[\Rightarrow 2{{p}^{2}}+3pr=\dfrac{c}{a}\] …………………….(10)
Replacing \[\alpha \] by p, \[\beta \] by q, and \[\gamma \] by r in equation (4), we get,
\[pqr=\dfrac{-d}{a}\] …………………….(11)
Now, using equation(6), transforming the above equation, we get
\[p(2p)r=\dfrac{-d}{a}\]
\[\Rightarrow 2{{p}^{2}}r=\dfrac{-d}{a}\] ……………………..(12)
From equation (5), we have the values of a=24, b=-14, c=-63, and d=45.
Now, putting the values of a, b, c, and d in equation (8), we get
\[3p+r=-\dfrac{-14}{24}\]
\[\Rightarrow 3p+r=\dfrac{7}{12}\]
\[\Rightarrow r=\dfrac{7}{12}-3p\] …………….(13)
Now, putting the values of a, b, c, and d in equation (10), we get
\[2{{p}^{2}}+3pr=\dfrac{-63}{24}\]
\[\Rightarrow 2{{p}^{2}}+3pr=\dfrac{-21}{8}\] …………………..(14)
Now, putting the values of a, b, c, and d in equation (10), we get
\[2{{p}^{2}}r=\dfrac{-45}{24}\]
\[\Rightarrow 2{{p}^{2}}r=\dfrac{-15}{8}\] …………………..(15)
Putting the value of r from equation (13) in equation (14), we get
\[\begin{align}
  & \Rightarrow 2{{p}^{2}}+3p\left( \dfrac{7}{12}-3p \right)=\dfrac{-21}{8} \\
 & \Rightarrow \dfrac{24{{p}^{2}}+21p-108{{p}^{2}}}{12}=\dfrac{-21}{8} \\
 & \Rightarrow \dfrac{-84{{p}^{2}}+21p}{12}=\dfrac{-21}{8} \\
 & \Rightarrow \dfrac{-4{{p}^{2}}+p}{12}=\dfrac{-1}{8} \\
 & \Rightarrow -4{{p}^{2}}+p=\dfrac{-12}{8} \\
 & \Rightarrow -4{{p}^{2}}+p=\dfrac{-3}{2} \\
 & \Rightarrow 2(-4{{p}^{2}}+p)=-3 \\
 & \Rightarrow -8{{p}^{2}}+2p+3=0 \\
\end{align}\]
We have to factorize the above equation. On factorizing, we get
\[\begin{align}
  & -8{{p}^{2}}+2p+3=0 \\
 & \Rightarrow 8{{p}^{2}}-2p-3=0 \\
 & \Rightarrow 8{{p}^{2}}-6p+4p-3=0 \\
 & \Rightarrow 2p(4p-3)+1(4p-3)=0 \\
 & \Rightarrow (2p+1)(4p-3) \\
\end{align}\]
So, the value is either \[p=\dfrac{-1}{2}\] or \[p=\dfrac{3}{4}\] …………………..(16)
When \[p=\dfrac{-1}{2}\] then from equation (13), we have
\[\begin{align}
  & \Rightarrow r=\dfrac{7}{12}-3\times \dfrac{(-1)}{2} \\
 & \Rightarrow r=\dfrac{7}{12}+\dfrac{3}{2} \\
 & \Rightarrow r=\dfrac{7+18}{12} \\
\end{align}\]
\[\Rightarrow r=\dfrac{25}{12}\] …………….(17)
Putting the value of p in equation (15), we get
\[\begin{align}
  & \Rightarrow 2{{\left( \dfrac{-1}{2} \right)}^{2}}r=\dfrac{-15}{8} \\
 & \Rightarrow \dfrac{r}{2}=\dfrac{-15}{8} \\
\end{align}\]
\[\Rightarrow r=\dfrac{-15}{4}\] …………(18)
We cannot have two values of r at the same time. This contradiction arises because we have taken \[p=\dfrac{-1}{2}\] . So, p cannot be equal to \[\dfrac{-1}{2}\] .
From equation (16) we have two values of p and they are either \[p=\dfrac{-1}{2}\] or \[p=\dfrac{3}{4}\] . Since p cannot be equal to \[\dfrac{-1}{2}\] so, p is equal to \[\dfrac{3}{4}\] .
Now, putting \[p=\dfrac{3}{4}\] in equation (7), we get
\[\begin{align}
  & q=2p \\
 & \Rightarrow q=2.\dfrac{3}{4} \\
\end{align}\]
\[\Rightarrow q=\dfrac{3}{2}\]
Now, putting \[p=\dfrac{3}{4}\] in equation (15), we get
\[\begin{align}
  & \Rightarrow 2{{\left( \dfrac{3}{4} \right)}^{2}}r=\dfrac{-15}{8} \\
 & \Rightarrow \dfrac{9r}{8}=\dfrac{-15}{8} \\
 & \Rightarrow r=\dfrac{-15}{9} \\
\end{align}\]
Hence, we have the values of p, q, and r which are \[\dfrac{3}{4}\] , \[\dfrac{3}{2}\] , and \[\dfrac{-15}{9}\] .

Note: In this question, after solving the quadratic equation \[(-8{{p}^{2}}+2p+3=0)\] we have two values of p which are \[p=\dfrac{-1}{2}\] or \[p=\dfrac{3}{4}\] . Here, one can take \[p=\dfrac{-1}{2}\] and then get the value of q from the equation \[(q=2p)\] . By putting the value of p in the equation \[\left( 2{{p}^{2}}r=\dfrac{-15}{8} \right)\] , one can get the value of r and then conclude these values of p, q, and r as answer which is wrong. Therefore, we have to check the values after putting it in all the equations, and if it satisfies then conclude it as the answer.