Question

If the roots of ${{x}^{3}}+3{{x}^{2}}+4x-11=0$ are $a$,$b$ and $c$ and the roots ${{x}^{3}}+r{{x}^{2}}+sx-t=0$ are $a+b$,$b+c$ and $c+a$, then the value of t is
A.18
B.23
C.15
D.-17

Answer 1

Hint: Here we have to find the value of $t$. We will first find the value of sum of roots , product of roots and sum of product of roots of first cubic equation. Then we will find only of product of roots of second cubic equation which is equal to $-t$ here, after simplifying the product of roots, we will get that product of roots of second cubic equation in terms of the sum of roots , product of roots and sum of product of roots of first cubic equation. We will then substitute those values to get the value of $t$.

Complete step-by-step answer:
The first cubic equation is:-
${{x}^{3}}+3{{x}^{2}}+4x-11=0$
The roots of this cubic equation are$a$,$b$ and$c$.
Now, we will find the sum of roots.
$a+b+c=-\dfrac{\text{coefficient of }{{x}^{2}}}{\text{coefficient of }{{x}^{3}}}$
Putting values, we get
$\Rightarrow$ $a+b+c=-\dfrac{3}{1}=-3............\left( 1 \right)$
Now, we will find the sum of the product of roots.
$ab+bc+ca=\dfrac{\text{coefficient of }x}{\text{coefficient of }{{x}^{3}}}$
Putting values, we get
$\Rightarrow$ $ab+bc+ca=\dfrac{4}{1}=4............\left( 2 \right)$
Now, we will find the product of roots.
$abc=-\dfrac{\text{constant term}}{\text{coefficient of }{{x}^{3}}}$
Putting values, we get
$\Rightarrow$ $abc=-\dfrac{-11}{1}=11...........\left( 3 \right)$
The first cubic equation is:-
${{x}^{3}}+r{{x}^{2}}+sx-t=0$
The roots of this cubic equation are $a+b$,$b+c$ and $c+a$.
Since, we need the value of $t$, we will only calculate the product of roots.
Now, we will find the product of roots.
$\left( a+b \right)\left( b+c \right)\left( c+a \right)=-\dfrac{\text{constant term}}{\text{coefficient of }{{x}^{3}}}$
Putting values, we get
$\Rightarrow$ $\left( a+b \right)\left( b+c \right)\left( c+a \right)=-\dfrac{t}{1}=-t$
Now, we will multiply all the terms on the left side of the equation.
$\Rightarrow$ $\left[ {{a}^{2}}b+{{a}^{2}}c+abc+abc+a{{b}^{2}}+{{b}^{2}}c+a{{c}^{2}}+b{{c}^{2}}+abc-abc \right]=-\dfrac{t}{1}=-t$
On taking common terms out, we get
$\Rightarrow$ $\left[ a\left( ab+bc+ca \right)+b\left( ab+bc+ca \right)+c\left( ab+bc+ca \right)-abc \right]=-t$
On further simplification, we get
$\Rightarrow$ $\left[ \left( a+b+c \right)\left( ab+bc+ca \right)-abc \right]=-t$
Now, we will substitute the value of $\left( a+b+c \right)$, $\left( ab+bc+ca \right)$ and $abc$ obtained in equation 1, equation 2 and equation 3 respectively.
  $\Rightarrow$ $ -3\times 4-11$=-t
 $\Rightarrow$ $ t=23$
Thus, the correct option is B.

Note: We need to remember the relation of the roots with coefficients of the equation to solve such problems. The roots of an equation are actually the values of x which we get after solving the equation.
Quadratic equation has two roots, the cubic equation has four roots and so on. Number of roots of any equation is equal to the highest power of that equation.