$\left( \cos6x+1 \right) { \tan }^{ 2 }x$

A cubic Equation can have at max 3 distinct roots, so it will cut the $x$ axis at $3$ points.

$x+y=-1\x=-1-y\x-y=15\-1-y-y=15\-2y=15+1\y=\dfrac{16}{-2}\y=-8\x-8=-1\x=-1+8\x=7\(x.y)=(7,-8)$

Descartes rule counts the number of times the sign changes from either $+$ to $-$ or from $-$ to $+$

Cubic Equations can have at max $3$ roots. Also, imaginary roots always occur in a pair of conjugates.
Since here one root is real and the other is imaginary, the third one must be imaginary and it will be the conjugate of the second root.

$x^3+6x^2+11x+6=0$

${x}^{4}-2a{x}^{2}+{a}^{2}-a=0$

Given equation is $x^{3}+ax^{2}+bx+c= 0$

Let $\alpha ,-\alpha ,\beta $ be the roots of $x^{ 3 }-mx^{ 2 }-3x+2=0$
Then
${ s } _{ 1 }=\alpha -\alpha +\beta =m\ \Rightarrow \beta =m$
Substituting $x=m$ in equation, we get
$m^{ 3 }-m.m^{ 2 }-3.m+2=0\ \Rightarrow m=\cfrac { 2 }{ 3 } $
Hence, option 'B' is correct.

Let $f(x)=x^4-x^3+1$

For equation $x-\dfrac { 2 }{ x-1 } =1-\dfrac { 2 }{ x-1 } $, the term $'x-1'$ is in the denominator. Hence the solution isn't defined. For $x=1$ $\Rightarrow $ $x\neq 1$

We have our equation as $x-\dfrac { 2 }{ x-1 } =1-\dfrac { 2 }{ x-1 } $
cancelling the common term on both sides,we get $x=1$. 
But for well defined solution $x\neq 1$. Hence,this equation has no solution.

$x-\dfrac { 5 }{ x-2 } =2-\dfrac { 5 }{ x-2 } $       ...(1)
Equation (1) is valid when $x\neq 2$
Rewriting eq. (1), we get $x=2$
But $x\neq 2$
Therefore, number of roots satisfying eq. (1) are zero.

 $2x^{99}+3x^{98}+2x^{97}+3x^{96}+........+2x+3=0$ .... $(i)$

$x^{10}-x^9-2=x^{10}-2x^9+1x^9-2=(x^9+1)(x-2)$

$P(x)=2x^{98}+3x^{97}+2x^{96}+.....+2x+3=0$

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.

Let the roots be $\alpha,\beta,\gamma$
Then
$\alpha=\beta+\gamma$.
Hence
$\alpha+\beta+\gamma=-a$
$2(\beta+\gamma)=-a$
$\beta+\gamma=\alpha=\dfrac{-a}{2}$ ...(i)
$\alpha.\beta+\beta.\gamma+\gamma.\alpha=b$
$\alpha(\beta+\gamma)+\beta.\gamma=b$
$\alpha^{2}+\beta.\gamma=b$
Or 
$\dfrac{a^{2}}{4}+\beta.\gamma=b$
$a^{2}+4\beta.\gamma=4b$ ...(ii)
And 
$\alpha.\beta.\gamma=-c$
Or 
$\dfrac{-a}{2}.\beta.\gamma=-c$
Or 
$\beta.\gamma=\dfrac{2c}{a}$.
Then
$a^{2}+4\beta.\gamma=4b$
$a^{2}+4\dfrac{2c}{a}=4b$
$a^{3}+8c=4ab$
Or 
$a^{3}=4(ab-2c)$.

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

$x^{ 3 }+x^{ 2 }-2x-1=0$
$f\left( 0 \right) =-1\ f\left( -1 \right) =1\ f\left( -2 \right) =-1\ f\left( -3 \right) =-13\ f\left( -4 \right) =-71$
As $f\left( -1 \right) >0$ and $f\left( -2 \right) <0$
Therefore one roots lies between $-2$ and $-1$

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 $

$let\quad \alpha ,-\alpha ,\beta \quad be\quad the\quad roots\ Given\quad sum\quad of\quad the\quad roots\quad is\quad zero\ \alpha -\alpha +\beta =0\ \beta =0\ Therefore\quad product\quad of\quad the\quad roots\quad is\quad zero,\quad i.e.,\quad b=0$

$ Let\quad \alpha \quad and\quad \alpha ^{ 2 }\quad be\quad the\quad roots\quad as\quad per\quad the\quad given\quad condition.\ \therefore \quad \alpha +\alpha ^{ 2 }=1\quad and\quad { \alpha  }^{ 3 }=-k\ Now\quad (\alpha +\alpha ^{ 2 })^{ 3 }=1\ \Rightarrow { \alpha  }^{ 3 }+({ \alpha  }^{ 2 })^{ 3 }+3{ \alpha  }^{ 3 }(\alpha +\alpha ^{ 2 })=1\ Replacing\quad { \alpha  }^{ 3 }\quad by\quad -k\quad we\quad get\quad \ -k+k^{ 2 }-3k-1=0\ \Rightarrow { k }^{ 2 }-4k-1=0\ \Rightarrow k=\frac { 4+\sqrt { 20 }  }{ 2 } =2+\sqrt { 5 } \ or\quad k=\frac { 4-\sqrt { 20 }  }{ 2 } =2-\sqrt { 5 } <0\ But\quad k>0\quad therefore\quad we\quad reject\quad this\quad value.\ \therefore \quad k=2+\sqrt { 5 } \ Answer-\quad Option\quad B. $

Let  $f(x) =x^{7}+14x^{5}+16x^{3}+30\mathrm{x} -560$

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

Let $\alpha,\beta$ are roots of ${x}^{2}+px-3=0,$ 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 roots are $\pm a,b,c\ b+c=2\ -{ a }^{ 2 }bc=-10\ { a }^{ 2 }bc=10\ { -a }^{ 2 }+ab+ac+bc-ab-ac=-3\ bc-{ a }^{ 2 }=-3\ { a }^{ 2 }-bc=3\ $

In the functional relation replace $x$ with $x+1$.

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

$\begin{array}{l} \frac { 1 }{ { x-\sin  \alpha  } } +\frac { 1 }{ { x-\sin  \beta  } } +\frac { 1 }{ { x-\sin  \gamma  } } =0 \ 0<\alpha <\beta <\gamma <\frac { \pi  }{ 2 }  \ let\, \alpha ={ 30^{ 0 } },\beta ={ 45^{ 0 } },\gamma ={ 60^{ 0 } } \ \Rightarrow \frac { 1 }{ { x-\frac { 1 }{ 2 }  } } =\frac { 1 }{ { x-2\sqrt { 2 }  } } =\frac { 1 }{ { x-\sqrt { \frac { 3 }{ 2 }  }  } } =0 \end{array}$

For 
$ax^{2}+bx+c=0$
$b^{2}-4ac\geq 0$ for real roots ...(i)
and for 
$ax^{2}-dx-c=0$
$d^{2}+4ac\geq 0$ for real roots ...(ii)
Now, 
$ac\neq 0$
Hence, 
Case I
If $ac>0$
Hence, 
$ax^{2}-dx-x=0$ has positive roots.
Case II
If $ac<0$
Then,
$ax^{2}+bx+c=0$
has Real roots.
Hence, the above polynomial has atleast two real roots.

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. 

Given, $64(81^{ x })-84(144^{ x })+27(256^{ x })=0$

Let the roots of the equation be $a,b,c,d$. 
As per the question,
$a+b = c+d$ 
From the theory of polynomials, 
$a+b+c+d = -p$
$ \Rightarrow a+b=c+d= \displaystyle \frac{-p}{2} $

Also,
$ab+ac+ad+bd+bc+cd  = q $
$ \Rightarrow (a+b)(c+d) +ab+cd  =q $
$ \Rightarrow ab+cd = q - \displaystyle \frac{p^2}{4} $

Also, 
$abc+abd+bcd+adc = -r $
$ \Rightarrow ab(c+d) +cd(a+b) = -r $
$ \Rightarrow \displaystyle \frac {-p}{2} (ab+cd) = -r $
$ \Rightarrow \displaystyle \frac {-p}{2} ( q - \displaystyle \frac{p^2}{4} ) = -r $
$ \Rightarrow -4pq + p^3 = -8r $
$ \Rightarrow p^3 + 8r = 4pq $

Given equation $a{ x }^{ 2 }+bx+c$
Given that, one root of the equation is square of another.
So, lets assume $\alpha$ , ${ \alpha  }^{ 2 }$ are roots of the given equation
We know that,
Sum of roots $=$ $\alpha +{ \alpha  }^{ 2 }=\dfrac { -b }{ a }$ 
Product of roots $=$ $ \alpha \times { \alpha  }^{ 2 }=\dfrac { c }{ a }$
$\alpha (1+\alpha )=\dfrac { -b }{ a } \longrightarrow 1  $
${ \alpha  }^{ 3 }=\dfrac { c }{ a } \longrightarrow 2 $
Cubing equation (1) on both sides and substitute the value from equation (2).
${ \alpha  }^{ 3 }{ (1+\alpha ) }^{ 3 }=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ { \alpha  }^{ 3 }({ \alpha  }^{ 3 }+1+3{ \alpha  }(1+\alpha ))=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ \dfrac { c }{ a } \left (\dfrac { c }{ a } +1+3\left (\dfrac { -b }{ a } \right)\right)=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ \dfrac { ({ c }^{ 2 }+ac-3bc) }{ { a }^{ 2 } } =\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ a({ c }^{ 2 }+ac-3bc)=-{ b }^{ 3 }\ { b }^{ 3 }+a{ c }^{ 2 }+{ a }^{ 2 }c=3abc $