Quadratic Equations

Given,

$\alpha_1, \beta_1, \gamma_1, \delta_1$ are the roots of equation

$ a x^4+b x^3+c x^2+d x+e=0 \text { and } $

$\alpha_2, \beta_2, \gamma_2, \delta_2$ are the roots of equation

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

∴ Clearly,

$\alpha_1, \beta_1, \gamma_1, \delta_1$ are the reciprocal of roots of

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

Also, $\delta_1>\gamma_1>\beta_1>\alpha_1>0$

and $\delta_2>\gamma_2>\beta_2>\alpha_2>0$

$\therefore \delta_1$ and $\alpha_2$ are reciprocal

Similarly $\gamma_1$ and $\beta_2$,

$\gamma_2$ and $\beta_1$ and $\alpha_1$ and $\delta_2$ are reciprocal

$ \begin{aligned} & \therefore \quad \alpha_1 \delta_2=1, \beta_1 \gamma_2=1, \beta_2 \gamma_1=1, \alpha_2 \delta_1=1 \end{aligned} $

$ \begin{aligned} & \alpha_1-\delta_2=2 \\ & \therefore \quad \alpha_1-\frac{1}{\alpha_1}=2 \Rightarrow \alpha_1^2-2 \alpha_1-1=0 \\ & \text { and } \delta_1-\alpha_2=4 \\ & \\ & \delta_1-\frac{1}{\delta_1}=4=\delta_1^2-4 \delta_1-1=0 \end{aligned} $

$\therefore \alpha_1$ and $\delta_1$ are roots of

$ \begin{gathered} a x^4+b x^3+c x^2+d x+e=0 \\ \therefore\left(x^2-2 x-1\right)\left(x^2-4 x-1\right) \\ =a x^4+b x^3+c x^2+d x+e \end{gathered} $

Put $x=1$,

$ \begin{gathered} (1-2-1)(1-4-1)=a+b+c+d+c \\ 8=a+b+c+d+e \end{gathered} $

We have,

${x^2} - (k - 2)x + ({k^2} + 3k + 5) = 0$

x₁, x₂ are roots of equations

$\therefore$ ${x_1} + {x_2} = k - 2,{x_1}{x_2} = {k^2} + 3k + 5$

$x_1^2 + x_2^2 = {({x_1} + {x_2})^2} - 2{x_1}{x_2}$

$ = {(k - 2)^2} - 2({k^2} + 3k + 5)$

$ = {k^2} - 4k + 4 - 2{k^2} - 6k - 10$

$ = - ({k^2} + 10k + 6)$

$ = - [{(k + 5)^2} - 19]$

$ = 19 - {(k + 5)^2} \le 19$

$\therefore$ Maximum value of $x_1^2 + x_2^2 = 19$

Let f(x) = x¹⁰⁰

g(x) = x² $-$ 3x + 2

q(x) = quotient

r(x) = remainder = (ax + b)

$\therefore$ f(x) = g(x) q(x) + r(x)

x¹⁰⁰ = (x² $-$ 3x + 2) q(x) + ax + b

Put x = 1 $\to$ 1 = a + b ..... (i)

Put x = 2 $\Rightarrow$ 2¹⁰⁰ = 2a + b ..... (ii)

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

a = 2¹⁰⁰ $-$ 1, b = 2 $-$ 2¹⁰⁰

Remainder = (2¹⁰⁰ $-$ 1)x + (2 $-$ 2¹⁰⁰)

= (2^{99 + 1} $-$ 1)x $-$ (2⁹⁹ $-$ 1)

$\therefore$ k = 99