Quadratic Equations

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 $

$f(x)=1+x^2$ is the polynomial satisfying the given condition

$ \begin{aligned} & f\left(\frac{2}{3}\right)=1+\left(\frac{2}{3}\right)^2=1+\frac{4}{9}=\frac{13}{9} \\\\ & f\left(\frac{3}{2}\right)=1+\left(\frac{3}{2}\right)^2=1+\frac{9}{4}=\frac{13}{4} \end{aligned} $

$ \begin{aligned} & f\left(\frac{2}{3}\right)+f\left(\frac{3}{2}\right)=\frac{13}{9}+\frac{13}{4}=13\left(\frac{1}{9}+\frac{1}{4}\right) \\\\ & \quad=\frac{13 \cdot 13}{36}=\left(\frac{13}{6}\right)^2 \\\\ & \therefore \quad \sqrt{f\left(\frac{2}{3}\right)+f\left(\frac{3}{2}\right)}=\sqrt{\left(\frac{13}{6}\right)^2}=\frac{13}{6} \end{aligned} $

We have $\alpha^2-\alpha+1=0$

$ \begin{array}{ll} \Rightarrow & \alpha+\frac{1}{\alpha}=1 \Rightarrow \alpha^2+\frac{1}{\alpha^2}+2=1 \\\\ \Rightarrow & \alpha^2+\frac{1}{\alpha^2}=-1 \Rightarrow \alpha^4+\frac{1}{\alpha^4}+2=1 \\\\ \Rightarrow & \alpha^4+\frac{1}{\alpha^4}=-1 \Rightarrow\left(\alpha+\frac{1}{\alpha}\right)^3=1 \\\\ \Rightarrow & \alpha^3+\frac{1}{\alpha^3}+3\left(\alpha+\frac{1}{\alpha}\right)=1 \\\\ \Rightarrow & \alpha^3+\frac{1}{\alpha^3}=-2 \\\\ \Rightarrow & 1^3+(-1)^3+(-2)^3+(-1)^3 \\\\ \Rightarrow & 1-1-8-1=-9 \end{array} $

Given the quadratic equation $x^2 + ax + b = 0$, we know from the problem statement:

The sum of the roots, $\alpha + \beta$, is $\frac{1}{2}$.

The sum of the cubes of the roots, $\alpha^3 + \beta^3$, is $\frac{37}{8}$.

From Vieta's formulas:

$\alpha + \beta = -a = \frac{1}{2}$. Thus, $a = -\frac{1}{2}$.

The product of the roots $\alpha \beta = b$.

We are given that $\alpha^3 + \beta^3 = \frac{37}{8}$. Using the identity for the sum of cubes:

$ \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) $

Substituting the known values:

$ \alpha^3 + \beta^3 = \left(\frac{1}{2}\right)^3 - 3b \left(\frac{1}{2}\right) $

Substitute the given value for $\alpha^3 + \beta^3$:

$ \frac{37}{8} = \frac{1}{8} - \frac{3b}{2} $

Rearranging gives us:

$ \frac{3b}{2} = \frac{1}{8} - \frac{37}{8} = \frac{-36}{8} = \frac{-9}{2} $

Thus,

$ 3b = -9 \quad \Rightarrow \quad b = -3 $

To find $a - \frac{1}{b}$:

$ a - \frac{1}{b} = -\frac{1}{2} - \frac{1}{-3} = -\frac{1}{2} + \frac{1}{3} $

Calculate the above expression:

$ a - \frac{1}{b} = \frac{-3}{6} + \frac{2}{6} = \frac{-1}{6} $

$4 x^3-3 x^2+2 x-1=0 \quad \text { [given] } $

$ \begin{array}{r} \alpha+\beta+\gamma=\frac{3}{4} \\\\ \alpha \beta+\beta \gamma+\gamma \alpha=\frac{1}{2} \end{array} $

and $\quad \alpha \beta \gamma=\frac{1}{4}$

We know that

$\alpha^3+\beta^3+\gamma^3-3 \alpha \beta \gamma$

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

$\Rightarrow \alpha^3+\beta^3+\gamma^3-\frac{3}{4}=\frac{3}{4}\left(-\frac{7}{16}-\frac{1}{2}\right)$.

$\left[\because(\alpha+\beta+\gamma)^2=\left(\alpha^2+\beta^2+\gamma^2\right)\right.+2(\alpha \beta+\beta \gamma+\gamma \alpha)]$

$ \left(\frac{3}{4}\right)^2=\left(\alpha^2+\beta^2+\gamma^2\right)+2 \cdot \frac{1}{2} $

$ \left.\alpha^2+\beta^2+\gamma^2=\frac{9}{16}-1=-\frac{7}{16}\right] $

$ \begin{aligned} \Rightarrow \quad & \alpha^3+\beta^3+\gamma^3=-\frac{3}{4}\left(\frac{7+8}{16}\right)+\frac{3}{4} \\\\ & =-\frac{45}{64}+\frac{3}{4}=\frac{-45+48}{64} \\\\ & \alpha^3+\beta^3+\gamma^3=\frac{3}{64} \end{aligned} $

$16 x^4+16 x^3-4 x-1=0$

$ \begin{aligned} & \Rightarrow \quad\left(16 x^4-1\right)+\left(16 x^3-4 x\right)=0 \\\\ & \Rightarrow \quad\left(4 x^2+1\right)\left(4 x^2-1\right)+4 x\left(4 x^2-1\right)=0 \\\\ & \Rightarrow \quad\left(4 x^2-1\right)\left(4 x^2+1+4 x\right)=0 \\\\ & \Rightarrow \quad\left(4 x^2-1\right)(2 x+1)^2=0 \Rightarrow x= \pm \frac{1}{2},-\frac{1}{2} \\\\ & \begin{aligned} \therefore \alpha= & \frac{1}{2}, \beta=-\frac{1}{2}, \gamma=-\frac{1}{2} \text { and } \delta=-\frac{1}{2} \end{aligned} \\\\ & \begin{aligned} \frac{1}{\alpha}= & 2, \frac{1}{\beta}=-2, \frac{1}{\gamma}=-2 \text { and } \frac{1}{\delta}=-2 \end{aligned} \\\\ & \begin{aligned} \therefore \frac{1}{\alpha^4} & +\frac{1}{\beta^4}+\frac{1}{\gamma^2}+\frac{1}{\delta^4} \\\\ & =2^4+(-2)^4+(-2)^4+(-2)^4 \\\\ & =4 \times 16=64 \end{aligned} \end{aligned} $

To find the solution set of the inequality $3^{x} + 3^{1-x} - 4 < 0$, we begin by rewriting it as:

$ 3^{x} + \frac{3}{3^{x}} - 4 < 0 $

Let $3^{x} = t$. Since $3^x$ is always positive, we have $t > 0$.

Substitute into the equation:

$ t + \frac{3}{t} - 4 < 0 $

Multiply the entire inequality by $t$ (noting that $t > 0$ so the inequality sign does not change):

$ t^2 - 4t + 3 < 0 $

Factoring the quadratic, we get:

$ (t-3)(t-1) < 0 $

The solution to this inequality is:

$ t \in (1, 3) $

Since $t = 3^{x}$, substitute back:

$ 3^{x} \in (1, 3) $

Taking the logarithm with base 3 of both sides gives:

$ x \in \left(\log_3{1}, \log_3{3}\right) $

Since $\log_3{1} = 0$ and $\log_3{3} = 1$, we find:

$ x \in (0, 1) $

To find the common solution set for the inequalities $x^2 - 4x \leq 12$ and $x^2 - 2x \geq 15$, we'll begin by rewriting them in a standard form:

$x^2 - 4x - 12 \leq 0$

$x^2 - 2x - 15 \geq 0$

Next, we factor each equation:

$ (x-6)(x+2) \leq 0 $

$ (x-5)(x+3) \geq 0 $

Now, we determine the solution sets for each inequality:

For $ (x-6)(x+2) \leq 0 $, the interval is $ x \in [-2, 6] $.

For $ (x-5)(x+3) \geq 0 $, the intervals are $ x \in (-\infty, -3] \cup [5, \infty) $.

To find the common solution set, we identify where these intervals overlap:

The intersection of $ [-2, 6] $ and $(- \infty, -3] \cup [5, \infty)$ is $ x \in [5, 6] $.

Thus, the common solution set for the given inequalities is $x \in [5, 6]$.

Let $ f(x)=3 x^{3}+b x^{2}+b x+3 $

(A) All the roots are negative. So, the number of sign change in $ f(-x) $ should be 3 , which is possible only when $ b $ is positive. According to options (A) $ \rightarrow $ IV.

(B) Clearly for any value of $ b, f(x)=0 $ has root $ x=-1 $.

$ \begin{array}{l} \therefore f(x)=0 \Rightarrow(x+1)\left(3 x^{2}+(b-3) x+3\right)=0 \\ \Rightarrow 3 x^{2}+(b-3) x+3 \neq 0 \end{array} $

[ $ \because $ two roots are complex]

$ \Rightarrow \quad(b-3)^{2}-4(3)(3) < 0 $

$ \Rightarrow \quad b^{2}-6 b-27 < 0 $

$ \Rightarrow \quad(b+3)(b-9) < 0 $

$ \Rightarrow \quad-3 < b < 9 $

Hence, $ \mathrm{B} \rightarrow $ II.

(C) We know that one root is $ x=-1 $

(Negative).

Two roots are complex if $ -3 < b < 9 $.

If $ b=-3 $, then $ f(x)=0 $

$ \Rightarrow \quad(x+1)(x-1)^{2}=0 $

In this case two roots are positive and equal.

(D) All roots will be real and distinct if and only if $ 3 x^{2}+(b-3) x+3=0 $ have two distinct real solutions other, than -1 .

$ \begin{array}{rc} & (b-3)^{2}-4(3)(3) > 0 \\ \Rightarrow & b^{2}-6 b-27 > 0 \\ \Rightarrow & (b+3)(b-9) > 0 \\ \Rightarrow & b \in(-\infty,-3) \cup(9, \infty) \end{array} $

Thus, D $ \rightarrow $ III

Therefore, $ \mathrm{A} \rightarrow \mathrm{IV}, \mathrm{B} \rightarrow \mathrm{II}, \mathrm{C} \rightarrow \mathrm{V}, \mathrm{D} \rightarrow \mathrm{III} $

Given, $ x+\frac{4}{x}=2 \sqrt{3} $

$ x^{2}-2 \sqrt{3} x+4=0 $

Roots, $ x=\sqrt{3} \pm i \Rightarrow \alpha, \beta=\sqrt{3} \pm i $

$ \begin{aligned} & \alpha^{2024}, \beta^{2024}=(\sqrt{3} \pm i)^{2024} \\\\ & =2^{2024}\left[\frac{\sqrt{3}}{2} \pm \frac{1}{2} i\right]^{2024} \\\\ & =2^{2024}\left[\cos \frac{\pi}{6} \pm i \sin \frac{\pi}{6}\right]^{2024} \\\\ & =2^{2024}\left[\cos \frac{2024 \pi}{6} \pm i \sin \frac{2024 \pi}{6}\right] \\\\ \therefore \mid \alpha^{2024} & -\beta^{2024}\left|=2^{2024} \times\left|2 i \sin \left(\frac{2024 \pi}{6}\right)\right|\right. \\\\ & =2^{2024} \times 2 \times\left|\sin \left(337 \pi+\frac{\pi}{3}\right)\right| \\\\ & =2^{2025} \times \frac{\sqrt{3}}{2} \end{aligned} $

Hence, $ \frac{2}{\sqrt{3}}\left|\alpha^{2024}-\beta^{2024}\right|=2^{2025} $

Given, $ 12 x^{1 / 3}-25 x^{1 / 6}+12=0 $

Let $ y=x^{1 / 6} $, then $ y^{2}=x^{1 / 3} $

So, $ 12 y^{2}-25 y+12=0 $

$ \begin{aligned} y & =\frac{4}{3} \text { or } \frac{3}{4} \\\\ x^{1 / 6} & =\frac{4}{3} \text { or } \frac{3}{4} \\\\ x & =\left(\frac{4}{3}\right)^{6} \text { or }\left(\frac{3}{4}\right)^{6} \end{aligned} $

Here, $ \alpha=\left(\frac{4}{3}\right)^{6} $ and $ \beta=\left(\frac{3}{4}\right)^{6} $

$ \sqrt[6]{\frac{\alpha}{\beta}}=\left(\frac{4}{3} \times \frac{4}{3}\right)^{6 \times \frac{1}{6}}=\frac{16}{9} $

Given, $ x^{3}+3 x^{2}-10 x-24=0 $

Factorise

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

Here, $ \alpha=3, \beta=-2 \gamma=-4 $

$ \alpha^{3}+3 \beta^{2}-10 \gamma-24=11 K $

$ 27+3 \times 4-10 \times(-4)-24=11 K $

$ 27+12+40-24=11 K \Rightarrow K=\frac{55}{11}=5 $

Given,

$ 8 x^{3}-42 x^{2}+63 x-27=0 $

Factorise $ (4 x-3)(2 x-3)(x-3)=0 $

Roots are $ x=\frac{3}{4}, \frac{3}{2} $ or 3

So, $ \beta=\frac{3}{4}, \gamma=\frac{3}{2} $ and $ \alpha=3 $

Let $ f(x)=\frac{3}{2} x^{2}+3 x+3 $

Extreme value $ =f\left(\frac{-b}{2 a}\right) $

$ \frac{-b}{2 a}=\frac{-3}{2 \times \frac{3}{2}} \Rightarrow-1 \Rightarrow f(-1)=\frac{3}{2}-3+3=\frac{3}{2} $

Given, $ \frac{2 x^{3}+1}{2 x^{2}-x-6} $

$ \begin{array}{r} \left.2 x^{2}-x-6\right) \frac{}{2 x^{3} \quad \quad \quad.+1}\left(x+\frac{1}{2}\right. \\\\ 2 x^{3}-x^{2}-6 x \\\\ \frac{(-) \quad(+) \quad(+)}{x^{2}+6 x+1} \\\\ x^{2}-\frac{1}{2} x-3 \\\\ \frac{(-)(+) \quad(+)}{\frac{13}{2} x+4} \end{array} $

$ \begin{aligned} \frac{2 x^{3}+1}{2 x^{2}-x-6} & =\left(x+\frac{1}{2}\right)+\frac{\frac{13}{2} x+4}{2 x^{2}-x-6} \\ & =\left(x+\frac{1}{2}\right)+\frac{\frac{13 x}{2}+4}{(2 x+3)(x-2)}\end{aligned} $

Now, $ \frac{\frac{13 x}{2}+4}{(2 x-3)(x+2)}=\frac{A}{2 x+3}+\frac{B}{x-2} $

$ \frac{13 x}{2}+4=B(2 x+3)+A(x-2) $

Put $ x=2 $

$ 17=7 B \Rightarrow B=\frac{17}{7} $

Put $ x=\frac{-3}{2} $

$ \begin{aligned} \frac{-39}{4}+4= & A\left(\frac{-3}{2}-2\right) \\\\ -\frac{23}{4}= & A\left(\frac{-7}{2}\right) \Rightarrow A=\frac{23}{14} \\\\ \therefore \frac{2 x^{3}+1}{2 x^{2}-x-6}= & \left(x+\frac{1}{2}\right) \\\\ & +\frac{23}{14(2 x+3)}+\frac{17}{7(x-2)} \end{aligned} $

Here, $ \quad a=1, b=\frac{1}{2}, p=1, q=3 $

$ \begin{array}{l} A=\frac{17}{7}, B=\frac{23}{14} \\\\ \Rightarrow \quad 51 \mathrm{apB}=51 \times 1 \times 1 \times \frac{23}{14}=\frac{1173}{14} \\\\ \Rightarrow 23 b q A=23 \times \frac{1}{2} \times 3 \times \frac{17}{7}=\frac{1173}{14} \\\\ \therefore \quad 51 a p B=23 b q A \end{array} $

Given $\alpha$ is a root of the equation

$ \begin{aligned} & \frac{x-1}{\sqrt{2 x^2-5 x+2}}=\frac{41}{60}, \text { where }-\frac{1}{2}<\alpha<0 \\\\ \therefore \quad & \frac{\alpha-1}{\sqrt{2 \alpha^2-5 \alpha+2}}=\frac{41}{60} \end{aligned} $

$\begin{aligned} & \Rightarrow \quad \frac{(\alpha-1)^2}{2 \alpha^2-5 \alpha+2}=\left(\frac{41}{60}\right)^2 \\\\ & \Rightarrow 3600\left(\alpha^2-2 \alpha+1\right)=1681\left(2 \alpha^2-5 \alpha+2\right) \\\\ & \Rightarrow 238 \alpha^2+1205 \alpha+238=0 \\\\ & \Rightarrow 238 \alpha^2+1156 \alpha+49 \alpha+238=0 \\\\ & \Rightarrow 34 \alpha(7 \alpha+34)+7(7 \alpha+34)=0 \\\\ & \Rightarrow(34 \alpha+7)(7 \alpha+34)=0 \\\\ & \Rightarrow \alpha=-\frac{7}{34},-\frac{34}{7} \\\\ & \because-\frac{1}{2}<\alpha<0 \\\\ & \therefore \alpha=-\frac{7}{34}\end{aligned}$

Given, roots of the equation

$ x^5+4 x^4-13 x^3-52 x^2+36 x+144=0 $

are $\alpha_1 \beta, \gamma, 2$, and $\varepsilon$, where

$ \alpha<\beta<\gamma<2<\varepsilon $

$ \begin{aligned} & \text { Now, } x^5+4 x^4-13 x^3-52 x^2+36 x+144=0 \\\\ & \Rightarrow \quad(x+4)(x+3)(x+2)(x-2)(x-3)=0 \\\\ & \Rightarrow \quad x=-4,-3,-2,2,3 \\\\ & \therefore \alpha=-4, \beta=-3, \gamma=-2 \text { and } \varepsilon=3 \\\\ & \text { Now, } \alpha+2 \beta+3 \gamma+5 \varepsilon \\\\ & =(-4)+2(-3)+3(-2)+5(3) \\\\ & =-4-6-6+15 \\\\ & =-16+15=-1 \end{aligned} $

We have,

$ \begin{aligned} \qquad 3 x^2+(2 k+1) x-5 k=0 \end{aligned} $

$\because \quad$ roots are real and equal.

$ \begin{aligned} \Rightarrow & D=0 \\\\ \Rightarrow & (2 k+1)^2-4(-5 k) \times 3=0 \\\\ \Rightarrow & 4 k^2+4 k+1+60 k=0 \\\\ \Rightarrow & 4 k^2+64 k+1=0 \\\\ & k=\frac{-64 \pm \sqrt{4096-16}}{2 \times 4} \\\\ & =\frac{-64 \pm \sqrt{4080}}{8}=\frac{-16 \pm \sqrt{255}}{2} \end{aligned} $

$Hence, k=\frac{-16+\sqrt{255}}{2}( \because \frac{1}{2} < k < 0)$

Let $\alpha$ be the common root of

$ \begin{aligned} & 2 x^2+a x-2=0 \text { and } x^2+x+0 \rightarrow 2=0 \\\\ & \Rightarrow 2 \alpha^2+a \alpha-2=0 \quad \ldots \text { (i) } \\\\ & \alpha^2+\alpha+2 a=0 \quad \ldots \text { (ii) } \\\\ & \Rightarrow \frac{\alpha^2}{2 a^2+2}=\frac{\alpha}{-2-4 a}=\frac{1}{2-a} \quad\left[a \neq 2,-\frac{1}{2}\right] \\\\ & \Rightarrow \quad \alpha=\frac{4 a+2}{a-2} \end{aligned} $

So, $\frac{(4 a+2)^2}{(a-2)^2\left(2 a^2+2\right)}=\frac{-1}{a-2}$

$ \begin{gathered} \frac{2(2 a+1)^2}{\left(a^2+1\right)}=2-a \\\\ 8 a^2+2+8 a=2 a^2+2-a^3-a \\\\ a^3+6 a^2+9 a=0 \Rightarrow a\left(a^2+6 a+9\right)=0 \\\\ \Rightarrow a=-3 \quad[\because a \neq 0) \\\\ \therefore a x^2-4 x-2 a=-3 x^2-4 x+6=0 \\\\ 3 x^2+4 x-6=0 \\\\ x=\frac{-4 \pm \sqrt{16+72}}{6} \\\\ =\frac{-2 \pm \sqrt{22}}{3} \end{gathered} $

So, one root of $a x^2-4 x-2 a=0$ is

$ =\frac{-2+\sqrt{22}}{3} . $

We have,

$ \begin{gathered} 2 x^3-3 x^2+5 x-7=0 \\\\ \therefore \quad \alpha+\beta+\gamma=-\left(-\frac{3}{2}\right)=\frac{3}{2} \quad \ldots \text { (i) } \\\\ \alpha \beta+\beta \gamma+\gamma \alpha=\frac{5}{2} \quad \ldots \text { (ii) } \\\\ \alpha \beta \gamma=\frac{7}{2} \quad \ldots \text { (iii) } \end{gathered} $

On squaring both side of Eq. (ii), we get

$ \begin{aligned} & (\alpha \beta+\beta \gamma+\gamma \alpha)^2=\frac{25}{4} \\\\ & \alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2+2 \\\\ & \left(\alpha \beta^2 \gamma+\alpha \beta \gamma^2+\alpha^2 \beta \gamma\right)=\frac{25}{4} \\\\ & \Sigma \alpha^2 \beta^2+2 \alpha \beta \gamma(\alpha+\beta+\gamma)=\frac{25}{4} \end{aligned} $

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

$ \begin{aligned} & \Sigma \alpha^2 \beta^2+2\left(\frac{7}{2}\right)\left(\frac{3}{2}\right)=\frac{25}{4} \\\\ & \Sigma \alpha^2 \beta^2=\frac{25}{4}-\frac{21}{2}=-\frac{17}{4} \end{aligned} $

We have,

$ x^4-x^3-16 x^2+4 x+48=0 \quad \ldots \text { (i) } $

By hit and trial, we get, $x^2-4$ is the factor of

$ \begin{gathered} x^4-x^3-16 x^2+4 x+48=0 \\\\ \Rightarrow \quad \frac{x^4-x^3-16 x^2+4 x+48}{x^2-4} \\\\ \Rightarrow x^2-x-12 \\\\ \Rightarrow x^4-x^3-16 x^2+4 x+48=\left(x^2-4\right) \\\\ \left(x^2-x-12\right) \\\\ \Rightarrow(x-2)(x+2)(x-4)(x+3)=0 \end{gathered} $

$\Rightarrow$ Roots of above Eq. (i) are,

$ -3,-2,2,4 $

$ \text { So, } \begin{aligned} \alpha^4+\beta^4+\gamma^4+\delta^4 & =(-3)^4+\left(-2^4\right. \\\\ + & \left(2^4+(4)^4=369\right. \end{aligned} $

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

To find the algebraic equation of degree 4 whose roots are translations of the roots of the given equation $x^4+5x^3+6x^2+7x+9=0$ by $-1$, we perform a substitution where $x$ becomes $x - 1$.

Given the equation:

$x^4+5x^3+6x^2+7x+9=0,$

we translate the roots by $-1$, meaning we consider $x = t - 1$, which implies $t = x + 1$.

We need to find $f(t-1) = 0$:

$ \begin{aligned} f(t-1) &= (t-1)^4 + 5(t-1)^3 + 6(t-1)^2 + 7(t-1) + 9 \\ &= (t^4 - 4t^3 + 6t^2 - 4t + 1) \\ &\quad + \, 5(t^3 - 3t^2 + 3t - 1) \\ &\quad + \, 6(t^2 - 2t + 1) \\ &\quad + \, 7(t - 1) + 9. \end{aligned} $

Simplifying each expression:

$(t^4 - 4t^3 + 6t^2 - 4t + 1)$

$5(t^3 - 3t^2 + 3t - 1) = 5t^3 - 15t^2 + 15t - 5$

$6(t^2 - 2t + 1) = 6t^2 - 12t + 6$

$7(t - 1) = 7t - 7$

Adding these terms together:

$ \begin{aligned} &t^4 - 4t^3 + 6t^2 - 4t + 1 \\ &+ 5t^3 - 15t^2 + 15t - 5 \\ &+ 6t^2 - 12t + 6 \\ &+ 7t - 7 + 9 = 0 \end{aligned} $

Combine like terms:

$ \begin{aligned} &t^4 + (5t^3 - 4t^3) + (6t^2 - 15t^2 + 6t^2) + (-4t + 15t - 12t + 7t) + (1 - 5 + 6 - 7 + 9) \\ &= t^4 + t^3 - 3t^2 + 6t + 4 = 0. \end{aligned} $

Thus, the required equation is:

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