Quadratic Equations

We have,

$\sqrt{\frac{5 x}{x-2}}+\sqrt{\frac{x-2}{5 x}}=\frac{29}{10}$

Let $\quad \sqrt{\frac{5 x}{x-2}}=y$

Then, $y+\frac{1}{y}=\frac{29}{10}$

$ \begin{array}{ll} \Rightarrow & \frac{y^2+1}{y}=\frac{29}{10} \\ \Rightarrow & 10 y^2+10=29 y \\ \Rightarrow & 10 y^2-29 y+10=0 \\ \Rightarrow & 10 y^2-25 y-4 y+10=0 \\ \Rightarrow & 5 y(2 y-5)-2(2 y-5)=0 \\ \Rightarrow & (5 y-2)(2 y-5)=0 \end{array} $

$ \begin{aligned} & \Rightarrow \quad y=\frac{2}{5}, \frac{5}{2} \\ & \text { Now, if } y=\frac{2}{5} \Rightarrow \sqrt{\frac{5 x}{x-2}}=\frac{2}{5} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \frac{5 x}{x-2}=\frac{4}{25} \\ \Rightarrow & 125 x=4 x-8 \\ \Rightarrow & 121 x=-8 \\ \Rightarrow & x=-8 / 121 \end{array} $

$ \text { And if } y=\frac{5}{2} \Rightarrow \sqrt{\frac{5 x}{x-2}}=\frac{5}{2} $

$ \begin{array}{ll} \Rightarrow & \frac{5 x}{x-2}=\frac{25}{4} \\ \Rightarrow & \frac{x}{x-2}=\frac{5}{4} \\ \Rightarrow & 5 x-10=4 x \\ \Rightarrow & x=10 \end{array} $

$ \begin{aligned} & \therefore \text { Let } \alpha=10 \text { and } \beta=\frac{-8}{121} \quad(\because \alpha>\beta) \\ & \\ & \quad \begin{aligned} \therefore &\,\,\,\,\,\,\,\,\, \sqrt{\alpha^2-11^4 \beta^2} \\ = & \sqrt{(10)^2-11^4\left(\frac{-8}{121}\right)^2} \\ = & \sqrt{100-64} \\ = & \sqrt{36}=6 \end{aligned} \end{aligned} $

Let $y=\frac{9 \cdot 3^{2 x}+6 \cdot 3^x+4}{9 \cdot 3^{2 x}-6 \cdot 3^x+4}$

Put, $3^x=t$

$ \begin{aligned} & \therefore \quad y=\frac{9 \cdot t^2+6 t+4}{9 t^2-6 t+4} \\ & \Rightarrow y\left(9 t^2-6 t+4\right)=9 t^2+6 t+4 \\ & \Rightarrow 9 t^2(y-1)-6 t(y+1)+(4 y-4)=0 \\ & \Rightarrow \quad(y-1) t^2 \frac{-6}{9}(y+1) t+\frac{4}{9}(y-1)=0 \\ & \Rightarrow(y-1) t^2-\frac{2}{3}(y+1) t+\frac{4}{9}(y-1)=0 \end{aligned} $

$\because t$ is real

$ \therefore D \geq 0 $

$ \begin{array}{lr} \Rightarrow \frac{4}{9}(y+1)^2-4(y-1) \frac{4}{9}(y-1) \geq 0 \\ \Rightarrow \frac{4}{9}(y+1)^2-\frac{16}{9}(y-1)^2 \geq 0 \\ \Rightarrow (y+1)^2-4(y-1)^2 \geq 0 \\ \Rightarrow y^2+1+2 y-4 y^2-4+8 y \geq 0 \\ \Rightarrow -3 y^2+10 y-3 \geq 0 \\ \Rightarrow 3 y^2-10 y+3 \leq 0 \\ \Rightarrow (y-3)(3 y-1) \leq 0 \\ \Rightarrow \frac{1}{3} \leq y \leq 3 \\ \text { Hence, minimum value }=\frac{1}{3} . \end{array} $

Let the roots of the given cubic equation $ x^3 - p x^2 + p x - 1 = 0 $ be $ \alpha, \beta, \gamma $.

From the standard relationships between coefficients and roots, we get:

$ \begin{aligned} \alpha + \beta + \gamma &= p, \\ \alpha\beta + \beta\gamma + \gamma\alpha &= p, \\ \alpha\beta\gamma &= 1. \end{aligned} $

The question says that the new equation, whose roots are the squares of the original roots (i.e., $ \alpha^2, \beta^2, \gamma^2 $), is the same as the given equation itself. Therefore, the symmetric sums of these squared roots must also be:

$ \begin{aligned} \alpha^2 + \beta^2 + \gamma^2 &= p, \\ \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 &= p, \\ \alpha^2\beta^2\gamma^2 &= 1. \end{aligned} $

Now, using the identity:

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

Substitute the known values:

$ p^2 = p + 2p $

Simplify this to get:

$ p^2 = 3p $

Since $ p \neq 0 $, divide both sides by $ p $:

$ p = 3 $

We have,

$ \begin{aligned} & \lambda x^2+13 x+7=0 \\ \therefore \quad & D=(13)^2-4(\lambda)(7)=169-28 \lambda \end{aligned} $

For rational roots, $D=$ perfect square

So, $\lambda \in(-3,7)$ has values $-2,0,6$ so that $D$ become perfect square.

∴ Required sum $=-2+0+6=4$

We have, $f(x)=1-2 x-5 x^2$

$ \begin{aligned} & =-5\left[x^2+\frac{2}{5} x-\frac{1}{5}\right] \\ & =-5\left[\left(x+\frac{1}{5}\right)^2-\frac{1}{5}-\frac{1}{25}\right] \\ & =-5\left(x+\frac{1}{5}\right)^2+1+\frac{1}{5} \\ & =-5\left(x+\frac{1}{5}\right)^2+\frac{6}{5} \\ & \therefore \quad f(x) \leq \frac{6}{5} ; \quad \therefore \quad \alpha=\frac{6}{5} \\ & \text { Again, } g(x)=x^2-2 x+r \\ & \quad=\quad(x-1)^2+r-1 \\ & \therefore \quad g(x) \geq(r-1) \\ & \therefore \quad \beta=r-1 \end{aligned} $

So, $5 \alpha x^2+\beta x+6=6 x^2+(r-1) x+6$

Since, $5 \alpha x^2+\beta x+6>0$ and coefficient of $x^2>0$

$ \begin{array}{lr} \therefore & D<0 \\ \Rightarrow & (r-1)^2-4 \times 6 \times 6<0 \\ \Rightarrow & (r+11)(r-13)<0 \\ \Rightarrow & r \in(-11,13) \end{array} $

$ \begin{aligned} & \text { Let } p(x)=x^4+x^3-4 x^2+x-1 \\ & \quad \begin{array}{l} =(x-1)\left(x^3+2 x^2-2 x-1\right) \\ =(x-1)(x-1)\left(x^2+3 x+1\right) \\ =(x-1)(x-1)\left[x-\left(\frac{-3+\sqrt{5}}{2}\right)\right] \\ \quad\left[x-\left(\frac{-3-\sqrt{5}}{2}\right)\right] \end{array} \end{aligned} $

∴ Roots of the equation, $p(x)=0$ are

$ x=1,1, \frac{-3+\sqrt{5}}{2}, \frac{-3-\sqrt{5}}{2} $

∴ Required ratio

$ \begin{aligned} & =\frac{(1)^2+(1)^2+\left(\frac{-3+\sqrt{5}}{2}\right)^2+\left(\frac{-3-\sqrt{5}}{2}\right)^2}{1 \times\left(\frac{-3+\sqrt{5}}{2}\right)\left(\frac{-3-\sqrt{5}}{2}\right)} \\ & =\frac{1+1+\frac{5+9-6 \sqrt{5}}{4}+\frac{5+9+6 \sqrt{5}}{4}}{\frac{9-5}{4}} \\ & =\frac{8+5+9-6 \sqrt{5}+5+9+6 \sqrt{5}}{4}=\frac{36}{4}=9 \\ & \therefore \text { Ratio }=9: 1 \end{aligned} $

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} $