Algebra

If a polynomial has five positive roots, by Vieta's formulas, the coefficients must satisfy specific relations. For 5x^5 - 25x^4 + ax^3 + bx^2 + cx - 5 = 0, the product of roots is 5/5 = 1. If all roots are positive, the sum of roots is 25/5 = 5. Using these, the coefficients a, b, c are determined, and 2a + 3b + 2c evaluates to 0.

Given that the roots of the equation $ ax^2+bx+c=0 $ be such that one root is $n$ times the other. 
Let one root be $\alpha$, then the other root will be $n\alpha$ by given condition.
Sum of roots $=$ $ \displaystyle S= \alpha +n\alpha = -\frac{b}{a}$ 
$  \Rightarrow  \alpha = -\dfrac{b}{a\left ( 1+n \right )}$.....(1)
Product of roots $=  n\alpha ^{2}= \dfrac{c}{a}$ 
$ \Rightarrow  \alpha ^{2}= \dfrac{c}{an}$ ....(2)
From (1) and (2), we have
$ \Rightarrow   \dfrac{c}{an}= \dfrac{b^{2}}{a^{2}\left ( 1+n \right )^{2}} $
$ \Rightarrow  \displaystyle \therefore nb^{2}= ac\left ( n+1 \right )^{2}$

General Quadratic equation is $ax^2+bx+c=0$
Given one root is three times the other.
i.e $\alpha,3\alpha$ are the roots.
Sum of the roots $=\displaystyle\frac{-b}{a}$
$\Rightarrow 4\alpha=\displaystyle\frac{-b}{a}$ ---(1)
Product of roots $=\displaystyle\frac{c}{a}$
$\Rightarrow 3\alpha^2=\displaystyle\frac{c}{a}$---(2)
From (1) and (2), we have
$3\left(\displaystyle\frac{-b}{4a}\right)^2=\displaystyle\frac{c}{a}$
$\therefore 3b^2=16ac$
Hence, option A is correct.

Rearranging the equation (x+1)(x+6) * (x+2)(x+3) = 15x^2 leads to (x^2 + 7x + 6)(x^2 + 5x + 6) = 15x^2. Dividing by x^2 gives (x + 6/x + 7)(x + 6/x + 5) = 15. Let y = x + 6/x. Then (y+7)(y+5) = 15, so y^2 + 12y + 20 = 0. Roots are y = -2, -10. Solving x + 6/x = -2 and x + 6/x = -10 yields two imaginary and two irrational roots.

Sum of two roots is zero. Using this we get the other root as $3$
Substitute $x=3$ and equating to zero we get, $k=-16$

Let $\alpha ,\beta ,\gamma ,\delta $ are roots of $x^{ 4 }-5x^{ 3 }+11x^{ 2 }-13x+6=0$
${ s } _{ 1 }=\alpha +\beta +\gamma +\delta =5\ { s } _{ 4 }=\alpha \beta \gamma \delta =6$

Let $\alpha,\beta$ are roots of ${x}^{2}-6x+q=0,$ then

=$ \because { a } _{ n }{ x }^{ 2\  }+{ a } _{ n }+{ x }^{ n-1 }+............+{ a } _{ 1 }x=\quad 0\quad \quad \quad { a } _{ 1 }\neq 0\quad n\ge 2 $

Let $\alpha, \beta$ are roots of $\displaystyle a{ x }^{ 2 }+bx+c=0$, then

Let $\alpha ,\beta ,\gamma ,\delta $ be the roots of ${ x }^{ 4 }-{ x }^{ 3 }+2{ x }^{ 2 }+kx+17=0$
Such that $\alpha +\beta =\gamma +\delta $
Then ${ s } _{ 1 }=\alpha +\beta +\gamma +\delta =1\Rightarrow \alpha +\beta =\cfrac { 1 }{ 2 } $ 
${ s } _{ 2 }=\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =2\Rightarrow { \left( \alpha +\beta  \right)  }^{ 2 }+\alpha \beta +\gamma \delta =2$
$\Rightarrow \alpha \beta +\gamma \delta =2-\cfrac { 1 }{ 4 } =\cfrac { 7 }{ 4 } $   ...(1)
${ s } _{ 3 }=\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =-k\Rightarrow \left( \alpha +\beta  \right) \left( \alpha \beta +\gamma \delta  \right) =-k$
$\Rightarrow \left( \alpha \beta +\gamma \delta  \right) =-2k$   ...(2)
From (1) and (2), we have
$-2k=\cfrac { 7 }{ 4 } \Rightarrow k=-\cfrac { 7 }{ 8 } $

let those are m, -m
now sum of three roots = p
hence third root will be p
now 
m*(-m) + m*p + (-m)*p = q
hence  –m2 = q
now m*( –m) * p = r
 –m2 p  = r
put value of  –m2 = q
hence  pq = r

Given, $x^4 - 4x^3 + ax^2 + bx + 1 = 0$

Consider the following equation
$x^{4}+ax^{2}+1=0$
$x^{3}+ax+1=0$
Subtracting equation (ii) from (i), we get 
$x^{4}-x^{3}+a(x^{2}-x)=0$
$x^{3}(x-1)+ax(x-1)=0$
$(x-1)(x^{3}+ax)=0$
$x(x-1)(x^{2}+a)=0$
Hence, we get $x=0$ $x=1$ and $x^{2}=-a$
Now out of the above two, $x=0$ is not a root of the following two equations.
We do not know the nature of '$a$'. 

$a^{2}+a-42=0\Rightarrow a=-7$ or $a=6.$
Since b is a root of $x^{2}+ax+a^{2}-37=0$
For $a=-7,$ we have $x^{2}-7x+49-37=0$
$\Rightarrow x^{2}-7x+12=0\Rightarrow x=4, 3 , so, b=4$ or $3.$
So the coordinates of P can be $(-7, 4)$ or $(-7, 3)$, 

Let $(\alpha, \beta )$ and $ (\gamma, \delta )$ be the roots of the first equation and second equation respectively.