Question

If the roots of $x^3+3x^2+4x-11=0$ 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

Answer: (B)

$23$

Solution:-

$x^3+3x^2+4x-11=0$

Here,

$A=1,B=3,C=4,D=-11$

Given that $a,b,c$ are the roots of the above equation.

Therefore,

Sum of roots $=\frac{-B}{A}$

$\Rightarrow a+b+c=\frac{-3}{1}=-3$

Sum of the product of roots $=\frac{C}{A}$

$\Rightarrow ab+bc+ca=\frac{4}{1}=4$

Product of roots $=\frac{-D}{A}$

$\Rightarrow abc=\frac{-(-11)}{1}=11$

$x^3+r{x}^2+sx+t=0$

Here,

$A=1,B=r,C=s,D=t$

Given that $(a+b),(b+c),(c+a)$ are the roots of the above equation.

Therefore,

Sum of the roots $=\frac{-B}{A}$

$\Rightarrow (a+b)+(b+c)+(c+a)=\frac{-r}{1}=-r$

Sum of the product of roots $=\frac{C}{A}$

$\Rightarrow (a+b)(b+c)+(b+c)(c+a)+(c+a)(a+b)=\frac{s}{1}=s$

Product of the roots $=\frac{-D}{A}$

$\Rightarrow (a+b)(b+c)(c+a)=\frac{-t}{1}=-t$

Now, solving for '$t$', we have

$t=-\left[(a+b)(b+c)(c+a)\right]$

$\Rightarrow t=-[ab+ac+b^2+bc)(c+a]$

$\Rightarrow t=-[a^{2}b+a^{2}c+2abc+a{b}^2+b^{2}c+a{c}^2+b{c}^2]$

$\Rightarrow t=-[a^{2}b+a^{2}c+abc+abc+a{b}^2+b^{2}c+a{c}^2+b{c}^2+abc-abc]$

$\Rightarrow t=-[a(ab+ac+bc)+b(ac+ab+bc)+c(ac+bc+ab)-abc]$

$\Rightarrow t=-[(ab+bc+ca)(a+b+c)-abc]$

From ${eq}^n\left(1\right),\left(2\right)\&\left(3\right)$, we have

$t=-[4\times (-3)-11]$

$t=-(-12-11)=23$

Hence the value of '$t$' is 23.