Quadratic Equation and Inequalities

To find the sum of the fourth powers of the roots $\alpha_\theta$ and $\beta_\theta$ of the quadratic equation $2x^2 + (\cos \theta)x - 1 = 0$, we start analyzing the expression $\alpha_\theta^4 + \beta_\theta^4$.

The equation can be rewritten with its roots using: $\alpha + \beta = -\dfrac{\cos \theta}{2}, \quad \alpha \beta = -\dfrac{1}{2}$ We need to calculate $\alpha^2 + \beta^2$ and $\alpha^2 \beta^2$: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\dfrac{\cos \theta}{2}\right)^2 - 2\left(-\dfrac{1}{2}\right) = \dfrac{\cos^2 \theta}{4} + 1$ $\alpha^2 \beta^2 = (\alpha \beta)^2 = \left(-\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$ Substitute these into: $\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2 \beta^2 = \left(\dfrac{\cos^2 \theta}{4} + 1\right)^2 - \dfrac{1}{2}$ Maximize and minimize $\left(\dfrac{\cos^2 \theta}{4} + 1\right)^2$: Zero of $\cos\theta$ leads to: $\left(\dfrac{0^2}{4} + 1\right)^2 = 1$ Max value $\cos^2 \theta = 1$: $\left(\dfrac{1}{4} + 1\right)^2 = \left(\dfrac{5}{4}\right)^2 = \dfrac{25}{16}$ Substitute back: $\text{Max: } \dfrac{25}{16} - \dfrac{1}{2} = \dfrac{25}{16} - \dfrac{8}{16} = \dfrac{17}{16}$ $\text{Min: } 1 - \dfrac{1}{2} = \dfrac{1}{2}$ Finally, compute $16(M + m)$: $16\left(\dfrac{17}{16} + \dfrac{1}{2}\right) = 16\left(\dfrac{17}{16} + \dfrac{8}{16}\right) = 16 \times \dfrac{25}{16} = 25$

as $f(x)$ is a polynomial of degree two let it be

on satisfying given conditions we get

hence $f(x)=1 \pm x^2$ also range $\in(-\infty, 1]$ hence

let roots of this equation be $\alpha \& \beta$ then $\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta$

To solve the given equation, start by rewriting the expression: $(x^2 - 9x + 11)^2 - (x - 4)(x - 5) = 3$ First, simplify the second part of the expression: $(x - 4)(x - 5) = x^2 - 9x + 20$ Now the equation becomes: $(x^2 - 9x + 11)^2 - (x^2 - 9x + 20) = 3$ Introduce a substitution for simplification: $\text{Let } t = x^2 - 9x$ Thus, the equation transforms to: $t^2 + 22t + 121 - t - 20 - 3 = 0$ Simplify further: $t^2 + 21t + 98 = 0$ Factor the quadratic: $(t + 14)(t + 7) = 0$ This gives: $t = -7 \quad \text{or} \quad t = -14$ Address each case where $t = x^2 - 9x$: $x^2 - 9x = -7$ $x^2 - 9x + 7 = 0$ Solving this quadratic equation, we find the roots: $x = \dfrac{9 \pm \sqrt{81 - 4 \times 7}}{2} = \dfrac{9 \pm \sqrt{53}}{2}$ $x^2 - 9x = -14$ $x^2 - 9x + 14 = 0$ Solving this quadratic equation: $x = \dfrac{9 \pm \sqrt{81 - 4 \times 14}}{2} = \dfrac{9 \pm \sqrt{25}}{2}$ $x = \dfrac{9 \pm 5}{2} = 7 \quad \text{or} \quad x = 2$ The rational roots from the second equation are 7 and 2.

Thus, the product of all the rational roots is: $7 \times 2 = 14$

Only 2 solutions.

To find the sum of the squares of all the roots of the equation $x^2 + |2x - 3| - 4 = 0$: Case I: $x \geq \dfrac{3}{2}$ For $x \geq \dfrac{3}{2}$, the expression $|2x - 3|$ becomes $2x - 3$.

Thus, the equation becomes: $x^2 + 2x - 3 - 4 = 0$ Simplifying gives: $x^2 + 2x - 7 = 0$ Solving this quadratic equation, we find: $x = 2\sqrt{2} - 1$ Case II: $x For$ x $x^2 + 3 - 2x - 4 = 0$ Simplifying gives: $x^2 - 2x - 1 = 0$ Solving this quadratic equation, we obtain: $x = 1 - \sqrt{2}$ Sum of the Squares of the Roots The sum of the squares of the roots is: $(2\sqrt{2} - 1)^2 + (1 - \sqrt{2})^2$ Calculating each term: $(2\sqrt{2} - 1)^2 = (2\sqrt{2})^2 - 2 \cdot 2\sqrt{2} \cdot 1 + 1^2 = 8 - 4\sqrt{2} + 1 = 9 - 4\sqrt{2}$ $(1 - \sqrt{2})^2 = 1^2 - 2 \cdot 1 \cdot \sqrt{2} + (\sqrt{2})^2 = 1 - 2\sqrt{2} + 2 = 3 - 2\sqrt{2}$ Adding these results: $(9 - 4\sqrt{2}) + (3 - 2\sqrt{2}) = 12 - 6\sqrt{2}$ This simplifies to $6(2 - \sqrt{2})$.

Thus, the sum of the squares of the roots is $6(2 - \sqrt{2})$.

$$\begin{aligned} & |\mathrm{x}-2|^2+2|\mathrm{x}-2|-|\mathrm{x}-2|-2=0 \\ & \Rightarrow(|\mathrm{x}-2|+2)(|\mathrm{x}-2|-1)=0 \\ & \Rightarrow|\mathrm{x}-2|=1 \\ & \Rightarrow \mathrm{x}=2 \pm 1=3,1 \\ & \Rightarrow \text { sum of square of roots }=9+1=10 \\ & \mathrm{x}^2-2|\mathrm{x}-3|-5=0 \\ & \text { Case-I } \mathrm{x}-3 \geq 0 \\ & \Rightarrow \mathrm{x}^2-2 \mathrm{x}+1=0 \\ & \Rightarrow(\mathrm{x}-1)^2=0 \\ & \Rightarrow \mathrm{x}=1 \\ & \text { But } \mathrm{x} \geq 3 \\ & \Rightarrow \mathrm{x} \in \phi \\ & \text { Case-II } \mathrm{x}-30 \Rightarrow \text { Real & distinct roots } \\ & \mathrm{f}(\mathrm{x})=\mathrm{x}^2+2 \mathrm{x}-11 \\ & \mathrm{f}(3)>0, \frac{-\mathrm{p}}{2 \mathrm{a}}=-1

Given: $P_{10} = 123$ $P_9 = 76$ $P_8 = 47$ $P_1 = 1$ We know that: $P_n = \alpha^n + \beta^n$ According to Newton’s identities, we have the relation: $P_{10} = P_9 + P_8$ This implies: $P_{10} - P_9 - P_8 = 0$ From $P_1 = 1$, it follows that: $\alpha + \beta = 1$ $\alpha \beta = 1$ Now, we are tasked with finding the quadratic equation whose roots are $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$.

For such a quadratic equation: Using the relationship between roots and coefficients, the equation with roots $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$ is: $x^2 - \left(\dfrac{1}{\alpha} + \dfrac{1}{\beta}\right)x + \dfrac{1}{\alpha \beta} = 0$ Substitute the known sum and product of $\alpha$ and $\beta$: $x^2 - \left(\dfrac{\alpha + \beta}{\alpha \beta}\right)x + \dfrac{1}{\alpha \beta} = 0$ Given that $\alpha + \beta = 1$ and $\alpha \beta = 1$, we can simplify: $x^2 - \left(\dfrac{1}{1}\right)x + \dfrac{1}{1} = 0$ This simplifies to: $x^2 + x - 1 = 0$ Thus, the quadratic equation having roots $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$ is: $x^2 + x - 1 = 0$

To find the set of all values of $p \in \mathbb{R}$ for which both roots of the equation $x^2-(p+2)x+(2p+9)=0$ are negative real numbers, follow these steps: Discriminant Condition: The equation's discriminant $D$ must be non-negative for real roots: $(p+2)^2 - 4(2p+9) \geq 0$ Simplifying this: $p^2 + 4p + 4 - 8p - 36 \geq 0 \quad \Rightarrow \quad p^2 - 4p - 32 \geq 0$ This can be factored as: $(p-8)(p+4) \geq 0$ Meaning $p \in (-\infty, -4] \cup [8, \infty)$.

…… (1) Sum of Roots Condition: The sum of the roots (which is $p+2$) must be negative: $p + 2 Product of Roots Condition: The product of the roots$ (2p + 9) $must be positive:$ 2p + 9 > 0 \quad \Rightarrow \quad p > -\frac{9}{2} $…… (3) Determine the Valid Interval: Combine the results from conditions (1), (2), and (3). From conditions (1) and (2), we find$ p Intersection of $(-\infty, -4]$ and $(-\dfrac{9}{2}, -2)$ gives: $p \in \left(-\dfrac{9}{2}, -4\right]$ Calculate $\beta - 2\alpha$: With $\alpha = -\dfrac{9}{2}$ and $\beta = -4$, compute: $\beta - 2\alpha = -4 - 2\left(-\dfrac{9}{2}\right) = -4 + 9 = 5$ Therefore, the difference $\beta - 2\alpha$ is 5.

For $x$ to be integer $4+n$ must be perfect squares

$\left\{25,36, \ldots 10^2\right\} \in S \Rightarrow 5^2, 6^2, \ldots 10^2 \Rightarrow 6$ values of $n$