Let x₁,x₂,x₃ ∈ R-{0} , x₁ + x₂ + x₃≠ 0 and frac{1}{x₁}+frac{1}{x₂}+frac{1}{

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4-2.Quadratic Equations and Inequations

normal

Let $x_1,x_2,x_3 \in R-\{0\} $ ,$x_1 + x_2 + x_3\neq 0$ and $\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}=\frac{1}{x_1+x_2+x_3}$, then $\frac{1}{{x^n}_1+{x^n}_2+{x^n}_3} =\frac{1}{{x^n}_1}+\frac{1}{{x^n}_2}+\frac{1}{{x^n}_3}$ holds good for

all $n \in N$

all odd numbers $n$

all even number $n$

no integer $n$

Solution

Consider a cubic equation whose roots are

$\mathrm{x}_{1}, \mathrm{x}_{2}$ and $\mathrm{x}_{3}$

$f(\mathrm{x})=\mathrm{x}^{3}+\alpha \mathrm{x}^{2}+\beta \mathrm{x}+\gamma$

$=\left(\mathrm{x}-\mathrm{x}_{1}\right)\left(\mathrm{x}-\mathrm{x}_{2}\right)\left(\mathrm{x}-\mathrm{x}_{3}\right)=0$

Given $\left(\mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{x}_{2} \mathrm{x}_{3}+\mathrm{x}_{3} \mathrm{x}_{1}\right)\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}\right)=\mathrm{x}_{1} \mathrm{x}_{2} \mathrm{x}_{3}$

$\beta(-\alpha)=-\gamma$

$\gamma=\alpha \beta$

$\therefore f(\mathrm{x})=0=\mathrm{x}^{3}+\alpha \mathrm{x}^{2}+\beta \mathrm{x}+\alpha \beta$

$=\mathrm{x}^{2}(\mathrm{x}+\alpha)+\beta(\mathrm{x}+\alpha)=(\mathrm{x}+\alpha)\left(\mathrm{x}^{2}+\beta\right)$

$\therefore \quad \mathrm{x}=-\alpha$

Let $\mathrm{x}_{1}=-\alpha ; \mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}=-\alpha \Rightarrow \mathrm{x}_{2}+\mathrm{x}_{3}=0$

$\Rightarrow \quad x_{2}=-x_{3}$

$\therefore \frac{1}{\mathrm{x}_{1}^{\mathrm{n}}+\mathrm{x}_{2}^{\mathrm{n}}+\mathrm{x}_{3}^{\mathrm{n}}}=\frac{1}{\mathrm{x}_{1}^{\mathrm{n}}+\left(\mathrm{x}_{2}\right)^{\mathrm{n}}+\left(-\mathrm{x}_{2}\right)^{\mathrm{n}}}$

$=\frac{1}{x_{1}^{n}}+\frac{1}{x_{2}^{n}}+\frac{1}{\left(-x_{2}\right)^{n}}$

Which is true for all odd integer $n$.

Standard 11

Mathematics

If $\alpha ,\,\beta ,\,\gamma $ are the roots of the equation ${x^3} + 4x + 1 = 0,$ then ${(\alpha + \beta )^{ – 1}} + {(\beta + \gamma )^{ – 1}} + {(\gamma + \alpha )^{ – 1}} = $

medium

Suppose the quadratic polynomial $p(x)=a x^2+b x+c$ has positive coefficient $a, b, c$ such that $b-a=c-b$. If $p(x)=0$ has integer roots $\alpha$ and $\beta$ then what could be the possible value of $\alpha+\beta+\alpha \beta$ if $0 \leq \alpha+\beta+\alpha \beta \leq 8$

normal

(KVPY-2016)

Let $\mathrm{S}$ be the set of positive integral values of $a$ for which $\frac{\mathrm{ax}^2+2(\mathrm{a}+1) \mathrm{x}+9 \mathrm{a}+4}{\mathrm{x}^2-8 \mathrm{x}+32}<0, \forall \mathrm{x} \in \mathbb{R}$. Then, the number of elements in $\mathrm{S}$ is :

hard

(JEE MAIN-2024)

Let $f(x)={{x}^{2}}-x+k-2,k\in R$ then the complete set of values of $k$ for which $y=\left| f\left( \left| x \right| \right) \right|$ is non-derivable at $5$ distinict points is

normal

Let $x$ and $y$ be two $2-$digit numbers such that $y$ is obtained by reversing the digits of $x$. Suppose they also satisfy $x^2-y^2=m^2$ for some positive integer $m$. The value of $x+y+m$ is

normal

(KVPY-2014)

Let $x_1,x_2,x_3 \in R-\{0\} $ ,$x_1 + x_2 + x_3\neq 0$ and $\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}=\frac{1}{x_1+x_2+x_3}$, then $\frac{1}{{x^n}_1+{x^n}_2+{x^n}_3} =\frac{1}{{x^n}_1}+\frac{1}{{x^n}_2}+\frac{1}{{x^n}_3}$ holds good for

Solution

Similar Questions

If $\alpha ,\,\beta ,\,\gamma $ are the roots of the equation ${x^3} + 4x + 1 = 0,$ then ${(\alpha + \beta )^{ – 1}} + {(\beta + \gamma )^{ – 1}} + {(\gamma + \alpha )^{ – 1}} = $

Suppose the quadratic polynomial $p(x)=a x^2+b x+c$ has positive coefficient $a, b, c$ such that $b-a=c-b$. If $p(x)=0$ has integer roots $\alpha$ and $\beta$ then what could be the possible value of $\alpha+\beta+\alpha \beta$ if $0 \leq \alpha+\beta+\alpha \beta \leq 8$

Let $\mathrm{S}$ be the set of positive integral values of $a$ for which $\frac{\mathrm{ax}^2+2(\mathrm{a}+1) \mathrm{x}+9 \mathrm{a}+4}{\mathrm{x}^2-8 \mathrm{x}+32}<0, \forall \mathrm{x} \in \mathbb{R}$. Then, the number of elements in $\mathrm{S}$ is :

Let $f(x)={{x}^{2}}-x+k-2,k\in R$ then the complete set of values of $k$ for which $y=\left| f\left( \left| x \right| \right) \right|$ is non-derivable at $5$ distinict points is

Let $x$ and $y$ be two $2-$digit numbers such that $y$ is obtained by reversing the digits of $x$. Suppose they also satisfy $x^2-y^2=m^2$ for some positive integer $m$. The value of $x+y+m$ is

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