Quadratic Equations

$ \begin{aligned} & \alpha+\beta+\gamma=0 \\ & \quad \alpha \beta+\beta \gamma+\alpha \gamma=+\frac{a}{2} \\ & \quad \alpha \beta \gamma=-b \end{aligned} $

Let $y+a=z$

$\Rightarrow z^3+K z^2+L=0$ has roots.

$z_1, z_2$ and $z_3$

$ \begin{array}{lc} \Rightarrow & z_1=3 \beta \gamma, z_2=3 \alpha \gamma, z_3=3 \alpha \beta \\ \Rightarrow & 3(\alpha \beta+\beta \gamma+\alpha \gamma)=-K \\ \Rightarrow & 27(\alpha \beta \gamma)^2=-L \Rightarrow-27 b^2=L \\ \Rightarrow & K=\frac{-3 a}{2} \\ \Rightarrow & \frac{L}{K}=\frac{27 b^2 \times 2}{3 a}=\frac{18 b^2}{a} \end{array} $

$ \begin{aligned} &\text { } f \text { is symmetric about } x=1\\ &\begin{aligned} \Rightarrow & & f^{\prime}(1) & =0 \\ \Rightarrow & & 2+b & =0 \\ \Rightarrow & & b & =-2 \end{aligned} \end{aligned} $

$ \begin{aligned} f(x) & =x^2-2 x+c \\ f(0) & =c \\ f(1) & =c-1 \\ f(-1) & =c+3 \\ \Rightarrow \quad f(-1) & >f(0)>f(1) \end{aligned} $

$ \begin{gathered} \alpha+\beta=-3 \\ \alpha \beta=k \end{gathered} $

$ \begin{aligned} & \Rightarrow \quad\left(\alpha+\frac{1}{\alpha}\right)\left(\beta+\frac{1}{\beta}\right)=\frac{18}{4} \\ & \Rightarrow \quad \alpha \beta+\frac{\alpha^2+\beta^2}{\alpha \beta}+\frac{1}{\alpha \beta}=\frac{9}{2} \\ & \Rightarrow \quad k+\frac{9-2 k}{k}+\frac{1}{k}=\frac{9}{2} \\ & \Rightarrow \quad \frac{k^2-2 k+10}{k}=\frac{9}{2} \\ & \Rightarrow \quad 2 k^2-4 k+20=9 k \\ & \Rightarrow \quad 2 k^2-13 k+20=0 \\ & \Rightarrow k \text { satisfies } 2 x^2-13 x+20=0 \end{aligned} $

$ \begin{array}{ll} \text { Let } & f(x)=a x^2+b x+c \\ & f(-3)=0 \\ \Rightarrow & f(x)=a(x+3)^2 \\ \Rightarrow & f(0)=9 a \Rightarrow a=2 \\ \text { So, } & f(x)=2(x+3)^2 \\ & f(3)=2 \times 6^2=72 \end{array} $

$ \begin{aligned} &\text { Let the integral root be } \gamma \text {. }\\ &\begin{aligned} & \Rightarrow \quad 6 \gamma^3-25 \gamma^2+2 \gamma+8=0 \\ & \Rightarrow \quad 6 \gamma^3-24 \gamma^2-\gamma^2+2 \gamma+8=0 \\ & \Rightarrow \quad 6 \gamma^2(\gamma-4)-\left(\gamma^2-2 \gamma-8\right)=0 \\ & \Rightarrow \quad 6 \gamma^2(\gamma-4)-(\gamma-4)(\gamma+2)=0 \\ & \Rightarrow \quad(\gamma-4)\left(6 \gamma^2-\gamma-2\right)=0 \\ & \Rightarrow \quad(\gamma-4)\left(6 \gamma^2+3 \gamma-4 \gamma-2\right)=0 \\ & \Rightarrow \quad(\gamma-4)(3 \gamma(2 \gamma+1))-2(2 \gamma+1))=0 \\ & \Rightarrow \quad \gamma=4, \beta=\frac{-1}{2}, \quad \alpha=\frac{2}{3} . \\ & \therefore \quad \frac{4}{\alpha}+\frac{1}{\beta}=\frac{4 \times 3}{2}-2=4 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow x^5+x^4-13 x^3-13 x^2+36 x+36=0 \\ & \Rightarrow x^4(x+1)-13 x^2(x+1)+36(x+1)=0 \\ & \Rightarrow(x+1)\left(x^4-4 x^2-9 x^2+36\right)=0 \\ & \Rightarrow(x+1)\left(x^2-4\right)\left(x^2-9\right)=0 \\ & x=-1, \pm 2, \pm 3 \\ & \text { So, } \varepsilon=3, \delta=2, \gamma=-1 \\ & \beta=-2 \alpha=-3 \\ & \text { So, } \frac{\varepsilon}{\alpha}+\frac{\delta}{\beta}+\frac{1}{\gamma}=\frac{3}{-3}+\frac{2}{-2}+\frac{1}{-1} \\ & \quad=-3 \end{aligned} $

The equation $ax^2+bx+c=0$ has the roots $\tan \theta$ and $\cot \theta$.

For a quadratic equation $ax^2+bx+c=0$, if $p$ and $q$ are roots, then:

Sum of roots: $p+q=-\frac{b}{a}$
Product of roots: $pq=\frac{c}{a}$

Here, the roots are $\tan \theta$ and $\cot \theta$. So, $\tan \theta + \cot \theta = -\frac{b}{a}$ and $\tan \theta \cdot \cot \theta = \frac{c}{a}$.

$\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}$

This simplifies to: $\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}$,
because $\sin^2 \theta+\cos^2 \theta = 1$

Now, $\tan \theta + \cot \theta = \frac{1}{\sin \theta \cos \theta}$
But from earlier, $\tan \theta + \cot \theta = -\frac{b}{a}$.
So, $\frac{1}{\sin \theta \cos \theta} = -\frac{b}{a}$.

For $\tan \theta \cdot \cot \theta = \frac{c}{a}$:
$\tan \theta \cdot \cot \theta = 1$, so $\frac{c}{a} = 1 \implies c = a$.

Now, take the result $\frac{1}{\sin \theta \cos \theta} = -\frac{b}{a}$.
Flip both sides: $\sin \theta \cos \theta = -\frac{a}{b}$

$2 \sin \theta \cos \theta = 2 \times \left(-\frac{a}{b}\right) = -\frac{2a}{b}$

But $2 \sin \theta \cos \theta = \sin 2\theta$.
So, $\sin 2\theta = -\frac{2a}{b}$.

Since $a = c$, we can also write $\sin 2\theta = -\frac{2c}{b}$.

We have,

$ \begin{array}{ll} & ||2 x-3|-4|=2 \\ \Rightarrow & |2 x-3|-4= \pm 2 \\ \Rightarrow & |2 x-3|= \pm 2+4 \\ \Rightarrow & |2 x-3|=6 \text { or }|2 x-3|=2 \\ \Rightarrow & 2 x-3= \pm 6 \text { or } 2 x-3= \pm 2 \\ \Rightarrow & 2 x= \pm 6+3 \text { or } 2 x= \pm 2+3 \\ \Rightarrow & 2 x=9,-3 \text { or } 2 x=5,1 \\ \Rightarrow & x=\frac{9}{2}, \frac{-3}{2} \text { or } x=\frac{5}{2}, \frac{1}{2} \end{array} $

∴ Sum of roots

$ \begin{aligned} & =\frac{9}{2}-\frac{3}{2}+\frac{5}{2}+\frac{1}{2} \\ & =\frac{9-3+5+1}{2}=\frac{12}{2}=6 \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & 3 x^5-6 x^4+2 x^3+4 x^2-5 x+8 \\ \Rightarrow \quad & 3 x^3\left(x^2-2 x+3\right)-7 x\left(x^2-2 x+3\right) -10\left(x^2-2 x+3\right)-4 x+38 \\ \Rightarrow \quad & \left(x^2-2 x+3\right)\left(3 x^3-7 x-10\right)-4 x+38 \\ \therefore \quad & a=3, b=0, c=-7, d=-10 \\ \quad & p=-4, q=38 \\ \therefore \quad & a b+c d=70 \\ \quad & p+2 q-2=-4+76-2=70 \\ \therefore \quad & a b+c d=p+2 q-2 \end{aligned} \end{aligned} $

$ \begin{aligned} & \alpha \beta=\alpha \delta=1 \\ & 12\left(x^2+\frac{1}{x^2}\right)-56\left(x+\frac{1}{x}\right)+89=0 \\ & 12\left(x+\frac{1}{x}\right)^2-24-56\left(x+\frac{1}{x}\right)+89=0 \\ & 12\left(x+\frac{1}{x}\right)^2-56\left(x+\frac{1}{x}\right)+65=0 \\ & 12 t^2-56 t+65=0, \quad x+\frac{1}{x}=t \\ & t=\frac{56 \pm \sqrt{(56)^2-4(12)(65)}}{2 \times 12} \end{aligned} $

$ \begin{aligned} & x+\frac{1}{x}=\frac{5}{2}, x+\frac{1}{x}=\frac{13}{6} \\ & x=2, \frac{1}{2}, x=\frac{2}{3}, \frac{3}{2} \\ \therefore & \frac{\alpha+\beta}{\gamma+\delta}=\frac{\frac{5}{2}}{\frac{13}{6}}=\frac{15}{13} \end{aligned} $

$ \begin{aligned} & \text { } x^2+p x+2=0 \\ & x^2+x+2 p=0 \end{aligned} $

Let ' $\alpha$ ' be the common root

$ \begin{array}{ll} \therefore & \alpha=2 \\ & 4+2 p+2=0 \\ & 2 p=-6 \\ & p=-3 \end{array} $

∴ Sum of roots of $x^2+2 p x+8=0$ (i.e., $x^2-6 x+8=0$ ) is 6 .

$x^2-5 a x+6 a=0$

∵ Both roots are greater than 2 .

$ \begin{aligned} & D \geq 0, \frac{-b}{2 a}>1, f(0)>0 \\ & (5 a)^2-4(1)(6 a) \geq 0, \frac{5 a}{2}>1,6 a>0 \\ & \Rightarrow a(25 a-24) \geq 0, a>\frac{2}{5}, a>0 \\ & \therefore a \in\left[\frac{24}{25}, \infty\right) \end{aligned} $

$ \begin{aligned} & \Rightarrow x^3(x-1)-3 x^2(x-1)+2(x-1)=0 \\ & \Rightarrow(x-1)\left(x^3-3 x^2+2\right)=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow(x-1)\left[x^2(x-1)-2 x(x-1)-2(x-1)\right]=0 \\ & \Rightarrow(x-1)(x-1)\left(x^2-2 x-2\right)=0 \\ & \therefore \quad \alpha=1, \gamma+\delta=2 \\ & \quad \beta=1, \gamma \delta=-2 \\ & \therefore \alpha+2 \beta+\gamma^2+\delta^2 \end{aligned} $

$ \begin{aligned} & =1+2+(\gamma+\delta)^2-2 \gamma \delta \\ & =3+4+4=11 \end{aligned} $

We have, $x^4+4 x^3-16 x-16=0$

$ \begin{aligned} & \Rightarrow x^3(x-2)+6 x^2(x-2)+12 x(x-2) +8(x-2)=0 \\ & \Rightarrow(x-2)\left(x^3+6 x^2+12 x+8\right)=0 \\ & \Rightarrow(x-2)\left[x^2(x+2)+4 x(x+2)+4(x+2)\right]=0 \\ & \Rightarrow(x-2)(x+2)\left(x^2+4 x+4\right)=0 \\ & \Rightarrow(x-2)(x+2)^3=0 \end{aligned} $

∴ Multiply roots is -2 .

(a) $x^2+2 x-3=0 \Rightarrow(x+3)(x-1)=0$

(b) $x^2-3 x+2=0 \Rightarrow(x-2)(x-1)=0$

(c) $x^2+x-2=0 \Rightarrow(x+2)(x-1)=0$

(d) $x^2-4 x+3=0 \Rightarrow(x-3)(x-1)=0$

Given equation is

$ x^2-3 a x+a^2-2 a-k=0 $

For distinct real roots, $D>0$

$ \begin{aligned} & =(-3 a)^2-4 \times 1 \times\left(a^2-2 a-k\right)>0 \\ & =9 a^2-4 a^2+8 a+4 k>0 \\ & =5 a^2+8 a+4 k>0 \end{aligned} $

Since, leading coefficient of $a^2$ is $5>0$

So, $(8)^2-4 \times(5)(4 k)<0=64-80 k<0$

$ \begin{aligned} \Rightarrow \quad 64 & <80 k \\ & =k>\frac{4}{5} \Rightarrow k \in\left(\frac{4}{5}, \infty\right) \end{aligned} $

Let

$ \begin{aligned} & f(x)=x^4-10 x^3+37 x^2-60 x+36=0 \\ & \text { At } x=1, f(1)=1-10+37-60+36 \neq 0 \\ & \text { At } x=2, f(2)=\left(2^4-10\left(2^3+37\left(2^2\right.\right.\right. \\ & -60(2+36=0 \\ & \quad=16-80+148-120+36 \\ & =200-200=0 \end{aligned} $

So, $x=2$ is a factor of $f(x)$

Similarly $x=3$ is a factor of $f(x)$

$ \therefore f(x)=(x-2)^2(x-3)^2 $

$\therefore$ Original roots are $2,2,3,3$

The distinct roots are 2 and 3

There are three possible pairs of distinct roots

(i) Increase the two $2 s$ by 1

(ii) Increase the two $3 s$ by 1

(iii) Increase one 2 and one 3 by 1

Let us assume we choose one root 2 and one root 3 to increase by 1

Since, these distinct roots.

Increase one ' 2 ' by $1: 2+1=3$

Increase one ' 3 ' by $1: 3+1=4$

Other two roots ( 2 and 3 ) are kept fixed

$\therefore$ New sets of roots of he transformed equation is $3,4,2,3$

Let the transformed equation be $g(x)$. its roots are $2,3,3,4$

$ \begin{aligned} \therefore g(x) & =(x-2)(x-3)(x-3)(x-4) \\ & =(x-2)(x-3)^2(x-4) \end{aligned} $

The common roots are value of $x$ for which $f(x)=0$ and $g(x)=0$

$\therefore$ Common roots are $2,3,3$

$\therefore$ Number of common roots $=3$

Given, $\alpha, \beta, \gamma$ are the roots of the equations $x^3-P x^2+Q x-R=0$

$ \therefore \quad \begin{array}{r} \alpha+\beta+\gamma=P \\ \alpha \beta+\beta \gamma+\gamma \alpha=Q \end{array} $

And $\alpha \beta \gamma=R$

Also, $(\alpha-2)^2,(\beta-2)^2,(\gamma-2)^2$ are the roots of equations, $x^3-5 x^2+4 x=0$

Now, $x^2-5 x^2+4 x=0$

$ \begin{array}{lc} \Rightarrow & x\left(x^2-5 x+4\right)=0 \\ \Rightarrow & x(x-1)(x-4)=0 \\ \Rightarrow & x=0,1,4 \\ \therefore & (\alpha-2)^2=0 \Rightarrow \alpha=2 \\ & (\beta-2)^2=1 \Rightarrow \beta-2= \pm 1 \Rightarrow \beta=1,3 \end{array} $

And $(\gamma-2)^2=4 \Rightarrow \gamma-2= \pm 2 \Rightarrow \gamma=0,4$

Now,

$ \begin{aligned} P+Q+R=(\alpha+\beta+\gamma)+(\alpha \beta+\beta \gamma & +\gamma \alpha) +\alpha \beta \gamma \end{aligned} $

Case I when $\alpha=2, \beta=1, \gamma=0$

then,

$ \begin{aligned} P+Q+R & =(2+1+0)+(2+0+0) \\ & =5 \end{aligned} $

Case II when $\alpha=2, \beta=1, \gamma=4$

then,

$ \begin{aligned} P+Q+R=(2+1+4)+(2+4+8) & + (2 \times 1 \times 4)=29 \end{aligned} $

Case III when $\alpha=2, \beta=3, \gamma=0$ then

$ \begin{aligned} P+Q+R & =(2+3+0)+(6+0+0)+0 \\ & =11 \end{aligned} $

Case IV when $\alpha=2, \beta=3, \gamma=4$

then

$ \begin{aligned} P+Q+R & =(2+3+4)+(6+12+8)+24 \\ & =59 \end{aligned} $

Clearly, least value of $P+Q+R$ is 5 .

Given, quadratic equation is

$ a x^2+a x+5=0 $

For a quadratic equation

$ \begin{gathered} A x^2+B x+C=0 \text { to have no real roots, } \\ \Delta=B^2-4 A C<0 \end{gathered} $

So, $\quad \Delta=a^2-4(a)(5)<0$

$ \begin{aligned} & \Rightarrow \quad a^2-20 a<0 \\ & \Rightarrow \quad a(a-20)<0 \end{aligned} $

So, $\quad 0

Thus, $a$ must be an integer.

So, the integral values of $a$ in this range are $1,2,3, \ldots, 19$.

The number of such value is

$ 19-1+1=19 $

Given equation is

$ 32 x^3-48 x^2+22 x-3=0 $

Since, the roots of this equation are in AP .

So, sum of roots $=a-d+a+a+d=3 a$

$ =\frac{-(-48)}{32}=\frac{3}{2} \Rightarrow a=\frac{1}{2} $

Now, sum of the products of roots $=\frac{22}{32}$

$ \Rightarrow(a-d) a+a(a+d)+(a-d)(a+d)=\frac{11}{16} $

$ \begin{aligned} & \Rightarrow a^2-a d+a^2+a d+a^2-d^2=\frac{11}{16} \\ & \Rightarrow 3 a^2-d^2=\frac{11}{16} \Rightarrow 3\left(\frac{1}{2}\right)^2-d^2=\frac{11}{16} \\ & \Rightarrow d^2=\frac{3}{4}-\frac{11}{16}=\frac{12-11}{16}=\frac{1}{16} \\ & \Rightarrow d^2=\frac{1}{16} \end{aligned} $

So, the square of the common difference of the root is $\frac{1}{16}$.

Given, equation is

$ x^4-2 x^3+x^2+4 x-6=0 $

Let the roots of this equation be $\alpha, \beta, \gamma$ and $\delta$.

Sum of the roots

$ =\alpha+\beta+\gamma+\delta=\frac{-(-2)}{1}=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Sum of the products of roots

$ \begin{aligned} & =\alpha \beta+\alpha \gamma+\alpha \delta+\beta \gamma+\beta \delta+\gamma \delta \\ & =\frac{1}{1}=1 \end{aligned} $

Sum of products of roots taken three at a time

$ \begin{aligned} & =\alpha \beta \gamma+\alpha \beta \delta+\alpha \gamma \delta+\beta \gamma \delta \\ & =\frac{-4}{1}=-4 \end{aligned} $

Product of roots $=\alpha \beta \gamma \delta=\frac{-6}{1}=-6$

Since, the sum of two roots be zero, i.e.,

$ \begin{aligned} & & \alpha+\beta & =0 \\ \Rightarrow & & \beta & =-\alpha \end{aligned} $

So, $\quad \gamma+\delta=2$        [from Eq. (i)]

And,

$ \begin{aligned} & \alpha(-\alpha)+\alpha \gamma+\alpha \delta+(-\alpha \gamma)+(-\alpha) \delta+\gamma \delta=1 \\ & \Rightarrow \quad-\alpha^2+\alpha \gamma+\alpha \delta-\alpha \gamma-\alpha \delta+\gamma \delta=1 \\ & \Rightarrow \quad-\alpha^2+\gamma \delta=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

And $\quad \alpha(-\alpha) \gamma \delta=-6$

$ \Rightarrow \quad-\alpha^2 \gamma \delta=-6 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

From Eqs. (ii) and (iii), we get

$ \begin{aligned} & \gamma \delta=1+\alpha^2 \\ \Rightarrow \quad & \alpha^2\left(1+\alpha^2\right)=6 \\ \Rightarrow \quad & \alpha^4+\alpha^2=6 \end{aligned} $

Let $y=\alpha^2$, then we get

$ \begin{array}{ll} & y^2+y-6=0 \\ \Rightarrow & (y+3)(y-2)=0 \\ \Rightarrow & y=-3 \text { and } y=2 \end{array} $

Since, $\alpha^2$ must be non-negative,

$ y=\alpha^2=2 $

Then, $\gamma \delta=1+\alpha^2=1+2=3$

Now, $(\gamma+\delta)^2=\gamma^2+\delta^2+2 \gamma \delta$

$ \begin{aligned} \Rightarrow \quad & \gamma^2+\delta^2=(\gamma+\delta)^2-2 \gamma \delta \\ & =2^2-2(3)=4-6=-2 \end{aligned} $

$x^2-4 a x+5+a>0$

so, $D<0$

$ \begin{aligned} & \Rightarrow \quad(-4 a)^2-4(1)(5+a)<0 \\ & \Rightarrow \quad 16 a^2-20-4 a<0 \\ & \Rightarrow \quad 4 a^2-a-5<0 \\ & \Rightarrow \quad 4 a^2-5 a+4 a-5<0 \\ & \Rightarrow \quad a(4 a-5)+1(4 a-5)<0 \\ & \Rightarrow \quad(4 a-5)(a+1)<0 \end{aligned} $

So, $-1 < a < \frac{3}{4}$ that means, $\alpha=-1, \beta=5 / 4$

$ 4 \beta+\alpha=4 \times \frac{5}{4}-1=4 $

$\because x^3-12 x^2+k x-18=0$

So, $\quad \alpha+\beta+\gamma=-(-12) / 1=12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \alpha \beta+\beta \gamma+\gamma \alpha=k \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

and $\alpha \beta \gamma=-(-18) / 1=18\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Let, $\alpha=3(\beta+\gamma)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

Using Eq. (i), we get

$ \begin{aligned} & 3(\beta+\gamma)+(\beta+\gamma)=12 \\ \Rightarrow \quad & \beta+\gamma=3 \end{aligned} $

So, $\alpha=3 \times 3=9\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)$

and $9 \beta \gamma=18$

$ \Rightarrow \quad \beta \gamma=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi)$

From Eqs. (v) and (vi), we get

$ \beta=1, \gamma=2 $

$ \begin{aligned}Therefore, k & =9 \times 1+1 \times 2+2 \times 9 \\ & =9+2+18=29 \end{aligned} $

Hence, $\alpha^2+\beta^2+\gamma^2-k$

$ \begin{aligned} & =9^2+1^2+2^2-29 \\ & =81+1+4-29=57 \end{aligned} $

Let

$ P(x)=x^5-3 x^4-x^3+11 x^2-12 x+4 $

When each roots of the equation increased by 2 , then

$ P(x) \rightarrow P(x-2) $

So, $P(x-2)=(x-2)^5-3(x-2)^4$

$ \begin{gathered} -(x-2)^3+11(x-2)^2-12(x-2)+4 \\ =x^5-10 x^4+40 x^3-80 x^2+80 x-32 \\ -3\left(x^4-8 x^3+24 x^2-32 x+16\right) \\ -\left(x^3-6 x^2+12 x-8\right)+11\left(x^2-4 x+4\right) \\ -12(x-2)+4 \\ =x^5-13 x^4+63 x^3-135 x^2+108 x \end{gathered} $

Hence, the required equation

$ x^5-13 x^4+63 x^3-135 x^2+108 x=0 $

Let the side length as ' $a$ '

So, $a^2=575$

$ \Rightarrow a=\sqrt{575} $

and $529<575<576$

So, $\sqrt{575}=23+\frac{575-529}{576-529}=23+\frac{46}{47}$

$ \text { = } 23.9792 $

$ \begin{array}{ll} \Rightarrow & \alpha+\beta=5 \gamma \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow & \alpha \beta=-6 \delta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

$ \begin{aligned} &\begin{aligned} \Rightarrow \quad \gamma+\delta & =5 \alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ \gamma \delta & =-6 \beta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned}\\ &\text { From Eqs. (ii) and (iv), we get } \end{aligned} $

$ \begin{gathered} \alpha \beta \gamma=6 \gamma \delta \\ \Rightarrow \quad \alpha \beta \gamma=6 \times 6 \beta \\ \Rightarrow \quad \alpha \gamma=36 \\ \alpha^2-5 \alpha \gamma-6 \delta=0 \\ \alpha^2-6 \delta=180 \\ \gamma^2-6 \beta=180 \\ -\quad+\quad- \\ \hline \alpha^2-\gamma^2=6(\delta-\beta) \end{gathered} $

Now, by Eqs. (i) and (iii), we get

$ \begin{aligned} \alpha+\beta-\gamma-\delta & =5 \gamma-5 \alpha \\ 6 \alpha-6 \gamma & =\delta-\beta \\ 6(\alpha-\gamma) & =(\delta-\beta) \\Now, (\alpha+\gamma) & =36 \end{aligned} $

∴ Now adding Eqs. (i) and (ii), we get

$ \alpha+\beta+\gamma+\delta=5(\alpha+\gamma)=5 \times 36=180 $

$ \begin{aligned} & \alpha+\beta+\gamma=-p \\ & \alpha \beta+\beta \gamma+\gamma \alpha=q \\ & \alpha \beta \gamma=-r \\ \text { Now, } & \alpha+\beta=-p-\gamma \\ & \beta+\gamma=-p-\alpha \\ & \gamma+\alpha=-p-\beta \\ \therefore \quad & (\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) \\ & =-(p+\alpha)(p+\beta)(p+\gamma) \\ = & -\left(p^3+p^2(\alpha+\beta+\gamma)\right. +p(\alpha \beta+\beta \gamma+\gamma \alpha)+\alpha \beta \gamma\} \\ = & -\left(p^3+p^2(-p)+p(q)-r\right) \\ = & r-p q \end{aligned} $

$x^2-7 x+10=0$

$ \begin{aligned} & \because \quad a=1 \\ & \therefore \text { Difference of roots }=\frac{\sqrt{D}}{|a|} \\ & \quad=\frac{\sqrt{49-40}}{1}=\sqrt{9}=3 \end{aligned} $

For quadratic equation

$ x^2-17 x+k=0 $

Difference of roots $=\sqrt{289-4 k}$

According to the questions,

$ 3=\sqrt{289-4 k} $

Squaring both sides, we get

$ \begin{array}{ll} \Rightarrow & 9=289-4 k \Rightarrow 4 k=280 \\ \Rightarrow & k=70 \end{array} $

∴ Divisor of $k=14$

$ \begin{aligned} & &\begin{aligned} & \therefore|x|^2-5|x|+6=0 \\ & \Rightarrow \quad|x|^2-2|x|-3|x|+6=0 \\ & \Rightarrow \quad|x|(|x|-2)-3(|x|-2)=0 \\ & \Rightarrow \quad(|x|-2)(|x|-3)=0 \\ & \Rightarrow \quad|x|=2,|x|=3 \\ & \Rightarrow \quad x= \pm 2, x= \pm 3 \\ & \therefore \text { Product of all roots }=2 \times(-2) \times 3(-3) \\ & \quad=4 \times 9=36 \end{aligned} \end{aligned} $

Let the given equation be $P(x) = 5x^3 - 4x^2 + 3x - 2 = 0$.

The roots of the equation are $\alpha, \beta, \gamma$.

According to Vieta's formulas, we have:

1. Sum of the roots: $\alpha + \beta + \gamma = -(\frac{-4}{5}) = \frac{4}{5}$.

2. Sum of the products of the roots taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{3}{5}$.

3. Product of the roots: $\alpha\beta\gamma = -(\frac{-2}{5}) = \frac{2}{5}$.

We want to find the value of $\alpha^3 + \beta^3 + \gamma^3$.

Method 1: Using the identity for sum of cubes.

We know the identity:

$\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2 - (\alpha\beta+\beta\gamma+\gamma\alpha))$.

First, let's find $\alpha^2+\beta^2+\gamma^2$:

$\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)$.

Substitute the values from Vieta's formulas:

$\alpha^2+\beta^2+\gamma^2 = \left(\frac{4}{5}\right)^2 - 2\left(\frac{3}{5}\right)$

$= \frac{16}{25} - \frac{6}{5}$

$= \frac{16}{25} - \frac{30}{25}$

$= -\frac{14}{25}$.

Now, substitute all the required values into the sum of cubes identity:

$\alpha^3 + \beta^3 + \gamma^3 = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2 - (\alpha\beta+\beta\gamma+\gamma\alpha)) + 3\alpha\beta\gamma$.

$\alpha^3 + \beta^3 + \gamma^3 = \left(\frac{4}{5}\right)\left(-\frac{14}{25} - \frac{3}{5}\right) + 3\left(\frac{2}{5}\right)$.

$\alpha^3 + \beta^3 + \gamma^3 = \left(\frac{4}{5}\right)\left(-\frac{14}{25} - \frac{15}{25}\right) + \frac{6}{5}$.

$\alpha^3 + \beta^3 + \gamma^3 = \left(\frac{4}{5}\right)\left(-\frac{29}{25}\right) + \frac{6}{5}$.

$\alpha^3 + \beta^3 + \gamma^3 = -\frac{116}{125} + \frac{6}{5}$.

To sum these fractions, find a common denominator, which is 125:

$\alpha^3 + \beta^3 + \gamma^3 = -\frac{116}{125} + \frac{6 \times 25}{5 \times 25}$.

$\alpha^3 + \beta^3 + \gamma^3 = -\frac{116}{125} + \frac{150}{125}$.

$\alpha^3 + \beta^3 + \gamma^3 = \frac{150 - 116}{125}$.

$\alpha^3 + \beta^3 + \gamma^3 = \frac{34}{125}$.

Method 2: Using Newton's Sums.

Since $\alpha, \beta, \gamma$ are roots of $5x^3 - 4x^2 + 3x - 2 = 0$, they satisfy the equation:

$5\alpha^3 - 4\alpha^2 + 3\alpha - 2 = 0$

$5\beta^3 - 4\beta^2 + 3\beta - 2 = 0$

$5\gamma^3 - 4\gamma^2 + 3\gamma - 2 = 0$

Summing these three equations:

$5(\alpha^3 + \beta^3 + \gamma^3) - 4(\alpha^2 + \beta^2 + \gamma^2) + 3(\alpha + \beta + \gamma) - 2(1+1+1) = 0$.

$5(\alpha^3 + \beta^3 + \gamma^3) - 4(\alpha^2 + \beta^2 + \gamma^2) + 3(\alpha + \beta + \gamma) - 6 = 0$.

Let $p_1 = \alpha + \beta + \gamma = \frac{4}{5}$.

Let $p_2 = \alpha^2 + \beta^2 + \gamma^2$. We calculated this above: $p_2 = -\frac{14}{25}$.

Let $p_3 = \alpha^3 + \beta^3 + \gamma^3$.

Substitute $p_1$ and $p_2$ into the sum equation:

$5p_3 - 4\left(-\frac{14}{25}\right) + 3\left(\frac{4}{5}\right) - 6 = 0$.

$5p_3 + \frac{56}{25} + \frac{12}{5} - 6 = 0$.

To combine the constant terms, find a common denominator (25):

$5p_3 + \frac{56}{25} + \frac{12 \times 5}{5 \times 5} - \frac{6 \times 25}{25} = 0$.

$5p_3 + \frac{56}{25} + \frac{60}{25} - \frac{150}{25} = 0$.

$5p_3 + \frac{56 + 60 - 150}{25} = 0$.

$5p_3 + \frac{116 - 150}{25} = 0$.

$5p_3 - \frac{34}{25} = 0$.

$5p_3 = \frac{34}{25}$.

$p_3 = \frac{34}{25 \times 5}$.

$p_3 = \frac{34}{125}$.

Both methods yield the same result.

The final answer is $\boxed{\frac{34}{125}}$.

Given, $6 x^3+7 x^2-4 x-2=0$

Let $x$ be the root and new root $x^{\prime}=x-n$

$ \begin{aligned} & x=x^{\prime}+n \\ & 6\left(x^{\prime}+n\right)^3+7\left(x^{\prime}+n\right)^2-4\left(x^{\prime}+n\right)-2=0 \end{aligned} $

Coefficient of $x^{\prime}$ :

$ \begin{aligned} & 18 n^2+14 n-4=0 \\ \Rightarrow \quad & 9 n^2+7 n-2=0 \\ \Rightarrow \quad & 9 n^2+9 n-2 n-2=0 \end{aligned} $

$ \Rightarrow \quad 9 n(n+1)-2(n+1)=0 \Rightarrow n=-1, \frac{2}{9} $

Product of all possible values of $n =-2 / 9$

$ \begin{aligned} & \text { } a x^2+b x+c=0, a \neq b \neq c \\ & a, b, c \in\{0,1,2,4\}, a \neq 0 \end{aligned} $

For unequal real roots

$ D>0 \Rightarrow b^2>4 a c $

Let $b=4,4 a c<16$

$ \begin{aligned} & \quad \begin{array}{l} \Rightarrow a c<4 \\ c=0, a=1,2 \rightarrow 2 \\ b=2,4 a c<4 \Rightarrow a c<1 \\ c=0, a=1,4 \rightarrow 2 \end{array} \\ & b=1,4 a c<1 \Rightarrow a c<\frac{1}{4} \end{aligned} $

$ \begin{gathered} c=0, a=2,4 \rightarrow 2 \\ \therefore \text { Total cases }=2+2+2=6 \end{gathered} $

Given, equation is

$ \sqrt{3 x^2+x+5}=x-3 $

Here, $3 x^2+x+5 \geq 0$ (since square root is non-negative)

$ \Rightarrow \quad x-3 \geq 0 $

$ \Rightarrow \quad x \geq 3 $

Now, squaring to both sides of given equation, we get

$ \begin{aligned} & \left(\sqrt{3 x^2+x+5}\right)^2=(x-3)^2 \\ \Rightarrow \quad & 3 x^2+x+5=x^2-6 x+9 \\ \Rightarrow \quad & 2 x^2+7 x-4=0 \end{aligned} $

Using quadratic formula,

$ \begin{aligned} x & =\frac{-7 \pm \sqrt{7^2-4 \cdot 2 \cdot(-4)}}{2 \cdot 2} \\ & =\frac{-7 \pm \sqrt{49+32}}{4}=\frac{-7 \pm 9}{4} \\ \Rightarrow \quad x & =\frac{2}{4} \text { and } x=\frac{-16}{4} \\ \Rightarrow \quad x & =\frac{1}{2} \text { and } x=-4 \end{aligned} $

Since domain, $x \geq 3$, so $x=-4$

And $x=\frac{1}{2}$ are not possible.

$\therefore$ There is zero solution.

Given, $\frac{x^2-1}{(x-4)(x-3)} \geq 1$

$ \begin{aligned} & \Rightarrow \quad \frac{x^2-1}{(x-4)(x-3)}-1 \geq 0 \\ & \Rightarrow \quad \frac{x^2-1-(x-4)(x-3)}{(x-4)(x-3)} \geq 0 \end{aligned} $

Since, the inequality is $\frac{7 x-13}{x^2-7 x+12} \geq 0$

So, the critical points at $7 x-13=0$

$ \Rightarrow \quad x=\frac{13}{7} $

and at $x=4$ and $x=3$, the expression is undefined.

$ \therefore x \in\left[\frac{13}{7}, 3\right) \cup(4, \infty) $

Given, equation is

$ 2 x^3+3 x^2-5 x-7=0 $

Since, $\alpha, \beta$ and $\gamma$ are the roots of the given equation.

$\therefore$ Sum of roots $=\alpha+\beta+\gamma=\frac{-3}{2}$

Sum of products of roots

$ =\alpha \beta+\beta \gamma+\gamma \alpha=-\frac{5}{2} $

Products of roots $=\alpha \beta \gamma=-\left(\frac{-7}{2}\right)=\frac{7}{2}$

$ \begin{gathered} \frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=\frac{\beta^2 \gamma^2+\gamma^2 \alpha^2+\alpha^2 \beta^2}{\alpha^2 \beta^2 \gamma^2} \\ \quad=\frac{\left(\beta \gamma^2\right)+(\gamma \alpha)^2+(\alpha \beta)^2}{(\alpha \beta \gamma)^2} \end{gathered} $

$ \begin{aligned} & =\frac{(\alpha \beta+\beta \gamma+\gamma \alpha)^2-2 \alpha \beta \gamma(\alpha+\beta+\gamma)}{(\alpha \beta \gamma)^2} \\ & =\frac{\left(\frac{-5}{2}\right)^2-2\left(\frac{7}{2}\right)\left(\frac{-3}{2}\right)}{\left(\frac{7}{2}\right)^2} \\ & =\frac{\frac{25}{4}+\frac{21}{2}}{\frac{49}{4}}=\frac{\frac{67}{4}}{\frac{49}{4}} \\ & =\frac{67}{4} \times \frac{4}{49}=\frac{67}{49} \\ & \therefore \quad \frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=\frac{67}{49} \end{aligned} $

Given equation

$ a x^4+b x^3+c x^2+d x+e=0 $

Let $\alpha, \alpha, \beta, \frac{1}{\beta}$ be root of equation

$ \begin{aligned} & \therefore \quad \alpha+\alpha+\beta+\frac{1}{\beta}=-b / a \\ & \Rightarrow \quad\left(\beta+\frac{1}{\beta}\right)=-\frac{b}{a}-2 \alpha \\ & \Rightarrow \quad \alpha \cdot \alpha \cdot \beta \cdot \frac{1}{\beta}=\frac{e}{a} \Rightarrow \alpha^2=e / a \\ & \Rightarrow \quad \alpha^2+2(\alpha \beta)+\frac{2 \alpha}{\beta}+1=c / a \\ & \alpha^2+2 \alpha\left(\beta+\frac{1}{\beta}\right)+1=c / a \\ & \Rightarrow \quad \alpha^2+2 \alpha\left(-\frac{b}{a}-2 \alpha\right)+1=c / a \\ & \Rightarrow \quad-3 \alpha^2-\frac{2 \alpha b}{a}+1=c / a \\ & \Rightarrow \quad \frac{c}{a}+3 \alpha^2+\frac{2 \alpha b}{a}=1 \\ & \Rightarrow \quad 3\left(\frac{e}{a}\right)+2 \sqrt{\frac{e}{a}} \times \frac{b}{c}+\frac{c}{a}=1 \\ & \Rightarrow \quad 3 e+2 b \sqrt{\frac{e}{a}}+c=a \end{aligned} $

Let $f(x)=(a-3) x^2+12 x+(a+6)$ to quadratic always be positive

$ \begin{array}{ll} & a-3>0 \\ \Rightarrow & a>3 \\ \text { and } & D=(12)^2-4(a-3)(a+6)<0 \\ & =144-4\left(a^2+3 a-18\right)<0 \\ \Rightarrow & a^2+3 a-54>0 \\ \Rightarrow & a^2+9 a-6 a-54>0 \\ \Rightarrow & (a+9)(a-6)>0 \end{array} $

$ a \in(-\infty,-9) \cup(6, \infty) $

Combine with $a>3$

$\Rightarrow \quad$ We take $a \in(6, \infty)$

Least positive integeral value

$ \alpha=7, l=6 $

put these values

$ \begin{aligned} & (\alpha-3) x^2+12 x+(l+2) \\ & =4 x^2+12 x+8=0 \\ \Rightarrow \quad & x^2+3 x+2=0 \\ \Rightarrow \quad & (x+1)(x+2)=0 \end{aligned} $

Hence, $x=-1,-2$

$\because x^2+2 a x+b=0$

$ \begin{aligned} \Rightarrow \quad x & =\frac{-2 a \pm \sqrt{(2 a)^2-4 b}}{2} \\ & =-a \pm \sqrt{a^2-b} \end{aligned} $

let,

$ \begin{aligned} & \alpha=-a+\sqrt{a^2-b} \\ & \beta=-a-\sqrt{a^2-b} \end{aligned} $

so, $|\alpha-\beta|=2 \sqrt{a^2-b}$

given that $2 \sqrt{a^2-b} \leq 2 m$

$ \begin{array}{ll} \Rightarrow & a^2-b \leq m^2 \\ \Rightarrow & b \geq a^2-m^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Also, for real and distinct roots

$ \begin{array}{cc} & D=4 a^2-4 b > 0 \\ \Rightarrow & b < a^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{array} $

Hence, $a^2-m^2 \leq b

$ \Rightarrow \quad b \in\left[a^2-m^2, a^2\right) $

$ \begin{aligned} &\text { } x^3-2 x^2+3 x-4=0\\ &\begin{aligned} & \text { So, } \alpha+\beta+\gamma=2 \alpha \beta+\beta \gamma+\gamma \alpha=+3 \text {, } \\ & \alpha \beta \gamma=4 \\ & \text { thus, } \alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2 \\ & \qquad \begin{aligned} & -2(\alpha \beta+\beta \gamma+\gamma \alpha) \\ & =4-2(3)=-2 \\ \text { and } & \alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2 \\ & =(\alpha \beta)^2+(\beta \gamma)^2+(\gamma \alpha)^2 \end{aligned} \\ & \because \alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2 \end{aligned} \end{aligned} $

$ \begin{aligned} & =(\alpha \beta+\beta \gamma+\gamma \alpha)^2-2 \alpha \beta \gamma(\alpha+\beta+\gamma) \\ & =(3)^2-2(4)(2)=9-16=-7 \end{aligned} $

and $\alpha^2 \beta^2 \gamma^2=(\alpha \beta \gamma)^2=4^2=16$

Hence, the cubic equation with roots $\alpha^2, \beta^2, \gamma^2$ is

$ \begin{aligned} & x^3-\left(\alpha^2+\beta^2+\gamma^2\right) x^2+\left(\alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2\right)x-\alpha^2 \beta^2 \gamma^2 \\ & =x^3-(-2) x^2+(-7) x-16 \quad x-\alpha^2 \beta^2 \gamma^2 \\ & =x^3+2 x^2-7 x-16=0 \end{aligned} $

$ \begin{aligned} & \because x^3+p x^2+q x+r=0 \\ & \alpha+\beta+\gamma=-p, \alpha \beta+\beta \gamma+\gamma \alpha=q, \\ & \quad \alpha \beta \gamma=-r \end{aligned} $

$ \begin{aligned} So,\,\,\alpha^3 & +\beta^3+\gamma^3=(\alpha+\beta+\gamma)^3 \\ & -3(\alpha+\beta+\gamma)(\alpha \beta+\beta \gamma+\gamma \alpha)+3 \alpha \beta \gamma \\ = & (-p)^3-3(-p)(q)+3(-r) \\ = & -p^3+3 p q-3 r \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \because \alpha+\beta=5 \Rightarrow \frac{-b}{1}=5 \Rightarrow b=-5 \ldots \text { (i) } \\ & \text { and } \alpha^3+\beta^3=60 \\ & \Rightarrow \quad(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)=60 \\ & \Rightarrow \quad(5)^3-3 \times c(5)=60 \\ & \Rightarrow \quad 125-15 c=60 \\ & \Rightarrow \quad 15 c=125-60=65 \\ & \Rightarrow \quad c=\frac{65}{15}=\frac{13}{3} \\ & \therefore 3 c+2=3 \times \frac{13}{3}+2=13+2=15 \\ & \quad=(-3)(-5)=-3 b \text { (from Eq. }(\mathrm{i})) \end{aligned} \end{aligned} $

Given, $\alpha, \beta$ and $\gamma$ are roots of equation

$ \begin{aligned} & x^3+a x^2+b x+c=0 \\ & \alpha+\beta+\gamma=-a \\ & \alpha \beta+\beta \gamma+\gamma \alpha=b \\ & \alpha \beta \gamma=-c \end{aligned} $

Now, $(\alpha+\beta-2 y)(\beta+\gamma-2 \alpha)(\gamma+\alpha-2 \beta)$

$ \begin{aligned} & =(\alpha+\beta+\gamma-3 \gamma)(\alpha+\beta+\gamma-3 \alpha)(\alpha+\beta+\gamma-3 \beta) \\ & =-(\alpha+3 \alpha)(a+3 \beta)(a+3 \gamma) \end{aligned} $

$ \begin{aligned} & =-\left(a^3+3(\alpha+\beta+\gamma) a^2\right.+9(\alpha \beta+\beta \gamma+\gamma \alpha) a+27(\alpha \beta \gamma) \\ & =-\left(a^3-3 a^3+9 a b-27 c\right) \\ & =2 a^3-9 a b+27 c \end{aligned} $

Given,

$ \begin{aligned} & x^4+2 x^3-7 x^2-8 x+12=0 \\ \Rightarrow & (x-1)(x-3)(x-2)(x+2)=0 \\ \Rightarrow & x=1,3,2-2 \end{aligned} $

Since sum of two roots of Eq. (i) is zero.

So, $-2+2=0$

Hence, sum of the squares of the other two roots

$ (1)^2+(3)^2=10 $

$f(x)=a x^2+b x+c$

substitute $f(x)$ into the given equation

$ \begin{aligned} & a x^2+b x+c+a\left(\frac{1}{x}\right)^2+b\left(\frac{1}{x}\right)+c \\ & =\left(a x^2+b x+c\right)\left(a\left(\frac{1}{x}\right)^2+b\left(\frac{1}{x}\right)+c\right) \\ & \begin{array}{r} \Rightarrow a x^2+b x+c+\frac{a}{x^2}+\frac{b}{x}+c \\ =\left(a x^2+b x+c\right)\left(\frac{a}{x^2}+\frac{b}{x}+c\right) \end{array} \\ & \begin{array}{r} \Rightarrow a x^4+b x^3+c x^2+a+b x+c x^2 \\ =\left(a x^2+b x+c\right)\left(a+b x+c x^2\right) \end{array} \\ & \Rightarrow a x^4+b x^3+2 c x^2+b x+a \\ & =a^2 x^2+a b x^3+a c x^4+a b x+b^2 x^2+b c x^3 +a c+b c x+c^2 x^2 \end{aligned} $

Compare coefficients : $a=a c$,

$ \begin{aligned} b & =a b+b c \\ 2 c & =a^2+b^2+c^2, b=a b+b c, a=a c \end{aligned} $

From $a=a c$, we have $c=1$ or $a=0$, if $a=0$ then $f(x)$ is not quadratic, so $c=1$ from $b=a b+b c$, we have $b=a b+b$, so $a b=0$, since $a \neq 0, b=0$

From $2 c=a^2+b^2+c^2$, we have

$ 2=a^2+0+1 $

So, $\quad a^2=1$ and $a= \pm 1$

thus, $f(x)=x^2+1$ or $f(x)=-x^2+1$

Since, $f(-1)=0$, we have $f(x)=-x^2+1$

$ \begin{aligned} & \Rightarrow \quad f(x)=-x^2+1 \\ & \Rightarrow \text { Range }=(-\infty, 1] \end{aligned} $

If $ \alpha \neq 0 $ and $ 0 $ are the roots of

$ x^2 - 5kx + (6k^2 - 2k) = 0 $, then find $ \alpha $.

Step 1: Use the fact that 0 is a root.

If $ x = 0 $ is a root, it should satisfy the equation:

$ 0^2 - 5k(0) + (6k^2 - 2k) = 0 $

$ \Rightarrow 6k^2 - 2k = 0 $

$ \Rightarrow 2k(3k - 1) = 0 $

So $ k = 0 $ or $ k = \frac{1}{3} $.

But if $ k = 0 $, the equation becomes $ x^2 = 0 $, giving a double root 0 — contradicting that one root is nonzero ($\alpha \neq 0$).

Hence,

$ k = \frac{1}{3}. $

Step 2: Substitute $ k = \frac{1}{3} $ into the equation.

$ x^2 - 5\left(\frac{1}{3}\right)x + \left(6\left(\frac{1}{3}\right)^2 - 2\left(\frac{1}{3}\right)\right) = 0 $

$ \Rightarrow x^2 - \frac{5}{3}x + \left(\frac{6}{9} - \frac{2}{3}\right) = 0 $

$ \Rightarrow x^2 - \frac{5}{3}x + \left(\frac{2}{3} - \frac{2}{3}\right) = 0 $

Wait — compute the constant term carefully:

$ \frac{6}{9} - \frac{2}{3} = \frac{2}{3} - \frac{2}{3} = 0. $

So the equation becomes:

$ x^2 - \frac{5}{3}x = 0 $

$ x (x - \frac{5}{3}) = 0 $

Step 3: Identify the roots.

Roots are $ x = 0 $ and $ x = \frac{5}{3} $.

Given that one root is 0 and another is $ \alpha \neq 0 $,

$ \boxed{\alpha = \frac{5}{3}}. $

✅ Final Answer:

$ \boxed{\alpha = \frac{5}{3}} $

Option C.

$ \begin{aligned} & \text { } \frac{8 x^2-14 x-9}{3 x^2-7 x-6}-2>0 \\ & \Rightarrow \frac{8 x^2-14 x-9-6 x^2+14 x+12}{3 x^2-7 x-6}>0 \\ & \Rightarrow \frac{2 x^2+3}{3 x^2-9 x+2 x-6}>0 \\ & \Rightarrow \frac{2 x^2+3}{3 x(x-3)+2(x-3)}>0 \\ & \Rightarrow \frac{2 x^2+3}{(x-3)(3 x+2)}>0 \end{aligned} $

$ x \in\left(-\infty, \frac{-2}{3}\right) \cup(3, \infty) $

$x\left(x^3-6 x^2+11 x-6\right)=0$

$ \begin{aligned} & \Rightarrow \quad x(x-1)(x-2)(x-3)=0 \\ & \Rightarrow \quad x=0,1,2,3 \end{aligned} $

The least roots is 0 .

Let the roots of $x^3+\alpha x^2+\beta x+6=0$ be $r_1, r_2, r_3$

The roots of $x^3+\alpha x^2+\beta x+6=0$

increased by 1 are $r_1+1, r_2+1, r_3+1$

One of these roots is 0 , so $r_i+1=0$. for some $i^0$.

$ \Rightarrow r_i=-1 $

Since $r_i=-1$ is root of

$ \begin{aligned} & x^3+\alpha x^2+\beta x+6=0 \\ & \Rightarrow(-1)^3+\alpha(-1)^2+\beta(-1)+6=0 \\ & \Rightarrow-1+\alpha-\beta+6=0 \\ & \Rightarrow \alpha-\beta+5=0 \end{aligned} $

$ \begin{aligned} & \text { } \alpha+\beta+\gamma=a \\ & \alpha \beta+\beta \gamma+\gamma \alpha=a \\ & \begin{array}{l} \alpha \beta \gamma=1 \\ \Rightarrow \alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha) \end{array} \\ & \quad \begin{aligned} & \Rightarrow \alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2=(\alpha \beta+\beta \gamma+\gamma \alpha)^2 -2 \alpha \beta \gamma(\alpha+\beta+\gamma) \\ & \quad=a^2-2 a \\ & \therefore \alpha^2 \beta^2 \gamma^2=(\alpha \beta \gamma)^2=1^2=1 \end{aligned} \end{aligned} $

Now, equate the coefficients of the original and new equations.

$ \begin{gathered} a^2-2 a=a \\ a^2-2 a=a \\ 1=1 \end{gathered} $

Solve for $\boldsymbol{a}$

$ a^2-2 a=a $

$ \begin{aligned} & \Rightarrow \quad a^2-3 a=0 \\ & \Rightarrow \quad a(a-3)=0 \\ & \because \quad a \neq 0 \therefore a=3 \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & (2 k-1) x^2-2(3 k-2) x+4 k>0, \forall x \in \mathbb{R} \\ & \Rightarrow \quad 4(3 k-2)^2-4(2 k-1) 4 k<0 \\ & \Rightarrow \quad 9 k^2-12 k+4-8 k^2+4 k<0 \\ & \Rightarrow \quad k^2-8 k+4<0, \\ & \quad k=\frac{8 \pm \sqrt{64-16}}{2}=\frac{8 \pm 4 \sqrt{3}}{2} \\ & \Rightarrow \quad k \in(4-2 \sqrt{3}, 4+2 \sqrt{3}) \\ & \quad k=1,2,3, \ldots, 7 \\ & \therefore \quad \text { Sum of integral value }=28 \end{aligned} \end{aligned} $

$f(x)=18 x^3-33 x^2+20 x-4=0

$ f^{\prime}(x)=54 x^2-66 x+20 $$x=\alpha$ is repeated roots, $f^{\prime}(\alpha)=0$ $ \begin{array}{ll} \text { and } & 54 \alpha^2-66 \alpha+20=0 \\ \Rightarrow & 27 \alpha^2-33 \alpha+10=0 \\ \Rightarrow & 27 \alpha^2-18 \alpha-15 \alpha+10=0 \\ \Rightarrow & (9 \alpha-5)(3 \alpha-2)=0 \Rightarrow \alpha=\frac{2}{3} \text { or } \frac{5}{9} \end{array} $