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

& \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{aligned}$$ We have to factorize the above equation. On factorizing, we get $$\begin{aligned} & -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{aligned}$$ So, the value is either $$p=\dfrac{-1}{2}$$ or $$p=\dfrac{3}{4}$$ …………………..(16) When $$p=\dfrac{-1}{2}$$ then from equation (13), we have $$\begin{aligned} & \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{aligned}$$ $$\Rightarrow r=\dfrac{25}{12}$$ …………….(17) Putting the value of p in equation (15), we get $$\begin{aligned} & \Rightarrow 2{{\left( \dfrac{-1}{2} \right)}^{2}}r=\dfrac{-15}{8} \\\ & \Rightarrow \dfrac{r}{2}=\dfrac{-15}{8} \\\ \end{aligned}$$ $$\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{aligned} & q=2p \\\ & \Rightarrow q=2.\dfrac{3}{4} \\\ \end{aligned}$$ $$\Rightarrow q=\dfrac{3}{2}$$ Now, putting $$p=\dfrac{3}{4}$$ in equation (15), we get $$\begin{aligned} & \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{aligned}$$ Hence, we have the values of p, q, and r which are $$\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.