Quadratic Equations

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

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} $

We start with the equation: $6 x^4-5 x^3+13 x^2-5 x+6=0$.

Divide both sides by $x^2$ (where $x \neq 0$) to get:

$ 6 x^2 - 5x + 13 - \frac{5}{x} + \frac{6}{x^2} = 0 $

Rewrite the expression by grouping terms: $ 6\left(x^2 + \frac{1}{x^2}\right) - 5\left(x + \frac{1}{x}\right) + 13 = 0 $

Let $t = x + \frac{1}{x}$. Then $x^2 + \frac{1}{x^2} = t^2 - 2$.

Substitute these into the equation: $ 6(t^2 - 2) - 5t + 13 = 0 $

Simplify: $ 6t^2 - 12 - 5t + 13 = 0 \\ 6t^2 - 5t + 1 = 0 $

This is a quadratic equation in $t$. The discriminant ($D$) is:

$D = (-5)^2 - 4\cdot6\cdot1 = 25 - 24 = 1$ (which is greater than 0).

This means there are two real solutions for $t$: $ t = \frac{5 \pm 1}{12} = \frac{6}{12} \text{ or } \frac{4}{12} \\ t = \frac{1}{2} \quad \text{or} \quad \frac{1}{3} $

So, $x + \frac{1}{x} = \frac{1}{2}$ or $x + \frac{1}{x} = \frac{1}{3}$.

Now solve for $x$ in each case:

If $x + \frac{1}{x} = \frac{1}{2}$, multiply both sides by $x$: $x^2 + 1 = \frac{1}{2} x$. Rearranged: $ 2x^2 - x + 2 = 0 $

If $x + \frac{1}{x} = \frac{1}{3}$, multiply both sides by $x$: $x^2 + 1 = \frac{1}{3} x$. Rearranged: $ 3x^2 - x + 3 = 0 $

For both equations, the discriminant is less than 0. This means there are no real solutions for $x$.

Therefore, all the solutions for $x$ are complex numbers.

Given, $x^2-6(k-1) x+4(k-2)=0$

$ \begin{aligned} & \alpha+\beta=6(k-1)=0 \Rightarrow k=1 \\ & \alpha \beta=4(k-2)<0 \Rightarrow k<2 \end{aligned} $

for $k=1, x^2-4=0 \Rightarrow x^2=4$

$ \therefore \quad x= \pm 2 \Rightarrow \alpha=+2 \text { and } \beta=-2 $

Then, $2 x^2-\alpha x+6 \beta(\alpha+1)=0$ become

$ \begin{array}{r} 2 x^2-2 x+6(-2)(3)=0 \\ x^2-x-18=0 \end{array} $

Now, product of roots $=-18$

We have, $a x^2+b x+c<0$

If $a<0$, then $\Delta<0$

$ a<0 \text { and } b^2-4 a c<0 $

$ \therefore b^2<4 a c $

And $c x^2+a x+b$ and $a x^2+b x+c$ have same extreme value

$ \because \quad-\frac{a}{2 c}=-\frac{b}{2 a} \Rightarrow a^2=b c $

Now, the minimum value of $c x^2+a x+b$ is $-\frac{\Delta}{4 a}=-\frac{a^2-4 b c}{4 c}=-\frac{b c-4 b c}{4 c}=-\left(-\frac{3 b c}{4 c}\right)=\frac{3 b}{4}$

Because $b^2=$ positive and $a<0 \Rightarrow c<0$

$\therefore \frac{3 b}{4}$ is the maximum value of $c x^2+a x+b$

$f(x)=x^4-17 x^3+k x^2-247 x+210$

Its factor: $x^2-5 x+6$

$\because(x-2)(x-3)$ is also a factor.

$ \begin{aligned} & \mathrm{f}(2)=16-17(8)+k(4)-(247) 2+210=0 \\ & \Rightarrow \quad 16-136+4 k-494+210=0 \\ & \Rightarrow \quad 4 k-404=0 \Rightarrow k=101 \\ & \because f(x)=x^4-17 x^3+101 x^2-247 x+210 \\ & \quad=\left(x^2-5 x+6\right)\left(x^2-12 x+35\right) \end{aligned} $

$ \begin{aligned} &\text { Given, } f(x)=x^2-5 x+4\\ &\begin{aligned} & f(x)>10 \\ & x^2-5 x+4>10 \\ & x^2-5 x-6>0 \end{aligned} \end{aligned} $

$ \begin{aligned} & (x-6)(x+1)>0 \\ & x<-1 \text { or } x>6 \\ & n(S)={ }^{20} C_1=20 \\ & n(E)={ }^{14} C_1=14 \\ & P(E)=\frac{14}{20}=\frac{7}{10} \end{aligned} $

$\sqrt{2} x^2-b x+(8-2 \sqrt{5})=0$

Let it's root's be $\alpha$ and $\beta$,

$ \alpha+\beta=\frac{b}{\sqrt{2}}, \quad \alpha \beta=\frac{8-2 \sqrt{5}}{\sqrt{2}} $

The harmonic mean of the roots,

$ \begin{array}{r} H=\frac{2 \alpha \beta}{\alpha+\beta}=4 \\ \alpha \beta=2(\alpha+\beta) \\ \frac{8-2 \sqrt{5}}{\sqrt{2}}=2 \cdot\left(\frac{b}{\sqrt{2}}\right) \\ 2 b=2(4-\sqrt{5}) \\ b=4-\sqrt{5} \end{array} $

We have,

$ \begin{aligned} 2 k x^2-(4 k+1) x+2 & < 0 \\ \because 2 k x^2-4 k x-x+2 & < 0 \\ 2 k x(x-2)-1(x-2) & < 0 \\ (x-2)(2 k x-1) & < 0 \\ 2 < x & < \frac{1}{2 k} \end{aligned} $

Expression is negative for exactly 3 integral values of $x$ which is $x=3,4,5$

$ \begin{array}{ll} & 5 < x \leq 6 \\ \because & \frac{1}{2 k} \leq 6 \Rightarrow k \geq \frac{1}{12} \\ \therefore & \frac{1}{2 k} > 5 \Rightarrow k < \frac{1}{10} \\ \therefore & k \in\left[\frac{1}{12}, \frac{1}{10}\right) \end{array} $

Given, $x^3-13 x^2+k x+189=0$

$ \begin{aligned} & \alpha+\beta+\gamma=13, \alpha \beta+\beta \gamma+\gamma \alpha=k \\ & \text { and } \quad \alpha \beta \gamma=-189 \text { and } \beta-\gamma=2 \\ & \quad[\because \gamma=\beta-2] \\ & \,\,\,\,\,\,\,\, \quad \alpha+\beta+\beta-2=13 \\ & \Rightarrow \quad(15-2 \beta) \beta(\beta-2)=-189 \\ & \Rightarrow \quad \beta\left(15 \beta-2 \beta^2-30+4 \beta\right)=-189 \\ & \Rightarrow \quad-2 \beta^3+19 \beta^2-30 \beta+189=0 \\ & \Rightarrow \quad 2 \beta^3-19 \beta^2+30 \beta-189=0 \\ & \Rightarrow \quad(\beta-9)\left(2 \beta^2-\beta+21\right)=0 \\ & \Rightarrow \quad \beta=9,2 \beta^2-\beta+21 \neq 0 \forall \beta \in R \\ & \therefore \quad \beta=9, \gamma=7 \end{aligned} $

and $\quad \alpha=-3$

$ k=\alpha \beta+\beta \gamma+\gamma \alpha=-27+63-21=15 $

Now, $\frac{\beta+\gamma}{k+\alpha}=\frac{9+7}{15-3}=\frac{16}{12}=\frac{4}{3}$

$ \therefore \quad \beta+\gamma: k+\alpha=4: 3 $

We have, $\frac{7 x^2-5 x-18}{2 x^2+x-6}<2$

$ \begin{aligned} & \frac{7 x^2-5 x-18}{2 x^2+x-6}-2<0 \\ & \frac{7 x^2-5 x-18-4 x^2-2 x+12}{2 x^2+x-6}<0 \\ & \frac{3 x^2-7 x-6}{2 x^2+x-6}<0 \\ & \frac{3 x^2-9 x+2 x-6}{2 x^2+4 x-3 x-6}<0 \\ & \frac{(3 x+2)(x-3)}{(2 x-3)(x+2)}<0 \end{aligned} $

Case I

$ \begin{aligned} & (3 x+2)(x-3)<0 \text { and } \\ & (2 x-3)(x+2)>0 \\ & \Rightarrow x \in\left(\frac{-2}{3}, 3\right) \text { and } x \in(-\infty,-2) \cup\left(\frac{3}{2}, 0\right) \\ & \Rightarrow x \in\left(\frac{3}{2}, 3\right) \end{aligned} $

$ \begin{aligned} &\text { Case II }\\ &\begin{array}{ll} & (3 x+2)(x-3)>0 \\ \text { and } \quad & (2 x-3)(x+2)<0 \end{array} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad x \in\left(-\infty,-\frac{2}{3}\right) \cup(3, \infty) \text { and } \\ & \Rightarrow \quad x \in\left(-2, \frac{3}{2}\right) \\ & \Rightarrow \quad x \in\left(-2,-\frac{2}{3}\right) \\ & \text { Hence, } x \in\left(-2,-\frac{2}{3}\right) \cup\left(\frac{3}{2}, 3\right) \end{aligned} $

To determine the set of all values of $ k $ for which the inequality $ x^2 - (3k+1)x + 4k^2 + 3k - 3 > 0 $ holds for all real values of $ x $, we proceed as follows.

First, recall that the inequality $ f(x) > 0 $ for all $ x $ requires that the quadratic equation has no real roots, which corresponds to the discriminant $ D $ being less than zero. For a quadratic equation $ ax^2 + bx + c $, the discriminant is given by $ b^2 - 4ac $.

In this scenario, the quadratic in $ x $ is given by:

$ a = 1, \quad b = -(3k+1), \quad c = 4k^2 + 3k - 3. $

The discriminant $ D $ is:

$ D = (3k+1)^2 - 4(1)(4k^2 + 3k - 3). $

Simplifying, we get:

$ D = (3k+1)^2 - 16k^2 - 12k + 12. $

Expanding:

$ D = 9k^2 + 6k + 1 - 16k^2 - 12k + 12. $

Simplifying further:

$ D = -7k^2 - 6k + 13. $

To ensure $ D < 0 $, we need:

$ 7k^2 + 6k - 13 > 0. $

To solve this inequality, factor the quadratic expression:

$ 7k^2 + 13k - 7k - 13 = (7k+13)(k-1). $

The solution to $ (7k + 13)(k - 1) > 0 $ can be found using sign analysis. The critical points are $ k = -\frac{13}{7} $ and $ k = 1 $.

The intervals to consider are:

$ k < -\frac{13}{7} $

$ -\frac{13}{7} < k < 1 $

$ k > 1 $

Test each interval:

For $ k < -\frac{13}{7} $ and $ k > 1 $, both factors $ (7k + 13) $ and $ (k - 1) $ have the same sign (either both positive or both negative), making the product positive.

For $ -\frac{13}{7} < k < 1 $, the factors have opposite signs, making the product negative.

Thus, the solution set for $ k $ is:

$ k \in \left(-\infty, -\frac{13}{7}\right) \cup (1, \infty). $

To find the cubic equation whose roots are the squares of the roots of the given equation, we start by analyzing the equation:

$ 12x^3 - 20x^2 + x + 3 = 0 $

Let's consider $\alpha$, $\beta$, and $\gamma$ as the roots of this equation. From the properties of polynomial roots, we have:

$\alpha + \beta + \gamma = \frac{20}{12} = \frac{5}{3}$

$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{1}{12}$

$\alpha\beta\gamma = -\frac{3}{12} = -\frac{1}{4}$

Next, we need to find relationships involving the squares of these roots $\alpha^2$, $\beta^2$, and $\gamma^2$.

Sum of squares of the roots:

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

$ = \left(\frac{5}{3}\right)^2 - 2 \times \frac{1}{12} = \frac{25}{9} - \frac{1}{6} = \frac{50 - 3}{18} = \frac{47}{18} $

Product of squares of the roots:

$ \alpha^2 \beta^2 \gamma^2 = (\alpha\beta\gamma)^2 = \left(-\frac{1}{4}\right)^2 = \frac{1}{16} $

Sum of products of squares of roots taken two at a time:

$ \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) $

$ = \left(\frac{1}{12}\right)^2 - 2 \times \left(-\frac{1}{4}\right) \times \frac{5}{3} $

$ = \frac{1}{144} + \frac{5}{6} = \frac{1 + 120}{144} = \frac{121}{144} $

Using these calculated values, the cubic equation whose roots are the squares of the original roots is given by:

$ x^3 - \frac{47}{18}x^2 + \frac{121}{144}x - \frac{1}{16} = 0 $

Multiplying through by 144 to clear fractions, we obtain:

$ 144x^3 - 376x^2 + 121x - 9 = 0 $

Let $\alpha, \beta$ and $\gamma$ be the roots of equation $x^3+3 x^2-10 x-24=0$

$ \alpha+\beta+\gamma=-3, \alpha \beta+\beta \gamma+\gamma \alpha=-10 $

and $\alpha \beta \gamma=24$

$ \Rightarrow \quad \alpha=-2, \beta=-4, \gamma=3 $

As, $\alpha(\beta+\gamma), \beta(\gamma+\infty)$ and $\gamma(\alpha+\beta)$ are the roots of equation

$ \begin{aligned} x^3 & +p x^2+q x+r=0 \\ -p & =\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta) \\ & =-2(-4+3)-4(-2+3)+3(-2-4) \\ & =2-4-18 \Rightarrow-20 \\ \therefore \quad p & =20 \\ q & =2(-4)+(-4)(-18)+(-18) 2 \\ & =-8+72-36=28 \end{aligned} $

Given,

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

where, $a$ is a rational number

Then,

discriminant, $D=B^2-4 A C$

$ \begin{aligned} & =(-3 a)^2-4 \times 1 \times\left(a^2-2 a-4\right) \\ & =9 a^2-4 a^2+8 a+16 \\ & =5 a^2+8 a+16 \\ D & =\underbrace{5 a^2+16+8 a}_{+\mathrm{ve}} \end{aligned} $

So, $D>0$

So, roots of equation will be different real number.

$ \begin{aligned} &\text { Given, }-1<\frac{2 x^2+a x+2}{x^2+x+1}<3\\ &\begin{aligned} & \text { Now, }-1<\frac{2 x^2+a x+2}{x^2+x+1} \\ & \Rightarrow \quad-x^2-x-1<2 x^2+a x+2 \\ & \Rightarrow \quad 0<3 x^2+(a+1) x+3 \\ & \Rightarrow \quad D<0 \\ & \Rightarrow \quad(a+1)^2-4 \times 3 \times 3<0 \\ & \Rightarrow \quad(a+1)^2-36<0 \Rightarrow(a+1)^2-6^2<0 \\ & \Rightarrow \quad(a+1-6)(a+1+6)<0 \\ & \Rightarrow \quad(a-5)(a+7)<0 \\ & \text { and } \quad \frac{2 x^2+a x+2}{x^2+x+1} < 3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & 2 x^2+a x+2 < 3 x^2+3 x+3 \\ \Rightarrow & 0 < x^2+(3-a) x+1 \Rightarrow D<0 \\ \Rightarrow & (3-a)^2-4 \times 1 \times 1<0 \\ \Rightarrow & (3-a)^2-2^2<0 \\ \Rightarrow & (3-a-2)(3-a+2)<0 \\ \Rightarrow & (-a+1)(5-a)<0 \end{array} $

$ \begin{aligned} & \text { } \left.x^2+x-3\right) \\ & \begin{array}{l} 3 x^5-4 x^4+5 x^3-3 x^2+6 x-8 \\ \left(3 x^3-7 x^2+21 x-45\right. \end{array} \end{aligned} $

Given, $x^2-6 a x+2-2 a+9 a^2=0$

$ \Rightarrow \quad x^2-6 a x+\left(2-2 a+9 a^2\right)=0 $

For both roots to exceed 3, $x=3$ should be a root or both roots should be greater than 3.

Sum of roots $=-\frac{b}{a}=-\left(\frac{-6 a}{1}\right)=6 a$

$ 6 a>18 \Rightarrow a>3 $

Product of roots $=\frac{c}{a}=\frac{9 a^2-2 a+2}{1}$

Since, $a>3$ so product, also should be greater than 3 .

That leads us to the correct condition

$ a>\frac{11}{9} $

We begin with the polynomial:

$ x^5 - 5x^3 + 5x^2 - 1 = (x-1)^3(x^2 + 3x + 1) $

We are given that $\alpha$ and $\beta$ are distinct negative roots of the equation. Therefore, $\alpha$ and $\beta$ must be the roots of the quadratic equation:

$ x^2 + 3x + 1 = 0 $

Given that $\alpha$ and $\beta$ are negative, we establish that:

$ \alpha \beta = 1 \quad \text{(as derived from the product of the roots of } x^2 + 3x + 1 = 0\text{)} $

We need to find the polynomial equation for which $\sqrt{-\alpha}$ and $\sqrt{-\beta}$ are roots. Note:

$ \sqrt{-\alpha} \cdot \sqrt{-\beta} = \sqrt{\alpha \cdot \beta} = \sqrt{1} = 1 $

Thus, the product of $\sqrt{-\alpha}$ and $\sqrt{-\beta}$ is 1. The equation with these roots and integer coefficients is:

$ x^4 - 3x^2 + 1 = 0 $

The product of the roots for this equation equals 1, satisfying the condition $\sqrt{-\alpha} \cdot \sqrt{-\beta} = 1$, confirming that $\sqrt{-\alpha}$ and $\sqrt{-\beta}$ are roots of the equation.

Given,

$ \begin{aligned} & x^2-5 x+\lambda=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & x^2-8 x-2 \lambda=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

$\because \alpha$ is a common factor of Eqs (i) and (ii), we get

And $\beta, \gamma$ are the other roots of them,

So

$ \begin{aligned} & \alpha^2-5 \alpha+\lambda=0 \\ & \alpha^2-8 \alpha-2 \lambda=0 \\ &\,\,\,\,\,\, -(+)\,\,\,\,\,\,\,(+) \\ & \hline \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3 \alpha+3 \lambda=0 \Rightarrow \lambda=-\alpha \end{aligned} $

$ \text { Then, } \quad \alpha^2-5 \alpha-\alpha=0 \Rightarrow \alpha^2-6 \alpha=0 $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad \alpha(\alpha-6)=0 \Rightarrow \alpha=0,6 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \alpha \neq 0 \\ & \Rightarrow \alpha=6 \text { and } \lambda=-6 \end{aligned}\\ &\text { Now, from Eq. (i), we get } \end{aligned} $

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

$ \begin{aligned} & \therefore \quad \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\beta=-1 \\ & \text { From Eq. (ii), we get } \end{aligned} $

$ \begin{array} & x^2-8 x-2 \times(-6) & =0 \\ \Rightarrow \quad x^2-8 x+12 & =0 \\ \Rightarrow \quad(x-6)(x-2) & =0 \Rightarrow x=6,2 \end{array} $

$ \therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad \gamma=2 $

$ \begin{aligned} &\text { }\\ &\begin{aligned}Now,\,\,\,\,\, \alpha+\beta+\gamma+\lambda & =6+(-1)+2+(-\theta) \\ & =1 \end{aligned} \end{aligned} $

To solve the given polynomial equation $x^4 - x^3 - 6x^2 + 4x + 8 = 0$ with two equal roots, we proceed as follows:

The equation can be expressed as:

$ x^3(x+1) - 2x^2(x+1) - 4x(x+1) + 8(x+1) = 0 $

Factoring out $(x+1)$ gives:

$ (x+1)(x^3 - 2x^2 - 4x + 8) = 0 $

Further factoring results in:

$ (x+1)(x-2)(x^2-4) = 0 $

This simplifies to:

$ x+1=0, \quad x-2=0, \quad x^2-4=0 $

Hence, the roots are:

$ x = -1, \quad x = 2, \quad x = \pm 2 $

The distinct roots of the equation are $-2$, $-1$, and $2$. Since one root, $2$, repeats, we determine the remaining roots as $\alpha = -2$ and $\beta = -1$.

To find $\alpha^2 + \beta^2$, compute:

$ \alpha^2 + \beta^2 = (-2)^2 + (-1)^2 = 4 + 1 = 5 $

Thus, $\alpha^2 + \beta^2 = 5$.

We have,

$ a(b-c) x^2+b(c-a) x+c(a-b)=0 $

Using hit and trial method,

Put $x=1$ in Eq. (i)

$ \begin{array}{r} a(b-c)+b(c-a)+c(a-b)=0 \\ a b-a c+b c-a b+a c-b c=0 \end{array} $

$\therefore$ one root is 1 .

Now, let another root be $\alpha$

Then, product of root, $\alpha \cdot 1=\frac{c(a-b)}{a(b-c)}$

$ \Rightarrow \quad \alpha=\frac{c(a-b)}{a(b-c)} $

Hence, $\frac{c(a+b)}{a(b-c)}, 1$