Question

If alpha beta are the roots of x+1/x=10/3,then relation between alpha and beeta.​

Answer 1

$\large{\underline{\underline{\red{\sf{GIVEN:}}}}}$

• $\tt{\alpha \:and\: \beta \:are\: zeroes \:of \:equ^{n}}$
• $\tt{The\:equ^{n}\:is\:x+\dfrac{1}{x}=\dfrac{10}{3}}$

$\large{\underline{\underline{\red{\sf{TO\:FIND:}}}}}$

• $\tt{Relation\: between\:\alpha\:and\:\beta}$

$\large{\underline{\underline{\red{\sf{ANSWER:}}}}}$

$\sf{Given \:Quadratic \:equation\: is \:x +\dfrac{ 1}{x }=\dfrac{ 10}{3}}$.

So , firstly let's simplify the equation ,

$\tt{\implies x+\dfrac{1}{x}=\dfrac{10}{3}}$

$\tt{\implies \dfrac{x^2+1}{x}=\dfrac{10}{3}}$

$\tt{\implies 3(x^2+1)=10x}$

$\tt{\implies 3x^2+3=10x }$

$\tt{\implies 3x^2-10x+3=0}$

Now we converted this equⁿ in standard form of a quadratic equation that is ax²+bx+c .

Now we know a relationship between zeroes and coefficients as ,

$\large{\underline{\boxed{\purple{\rm{\leadsto Product\:of\: Zeroes=\dfrac{Coefficient\:of\: constant\:term}{Coefficient\:of\:x^2}=\dfrac{c}{a}}}}}}$

Now here zeroes are alpha and beta.

Here ,

• $\orange{\sf{Coefficient\:of\:c\:=\:3}}$
• $\orange{\sf{Coefficient\:of\:x^2\:=\:3}}$

$\tt{\implies \alpha\beta=\dfrac{3}{3}}$

$\tt{\implies \alpha\beta=1}$

$\underline{\green{\tt{\underset{\purple{Required\: relationship}}{\underbrace{\dag\alpha=\dfrac{1}{\beta}}}}}}$

Answer 2

Answer:

Required relations between $\alpha \: and \: \beta$ are $\alpha = \frac{10}{3} - \beta \: and \: \alpha = \frac{1}{ \beta }$

Step-by-step explanation:

Given equation

$x + \frac{1}{x} = \frac{10}{3} \\ \frac{ {x}^{2} + 1 }{x} = \frac{10}{3} \\ 3 {x}^{2} + 3 = 10x \\ 3 {x}^{2} - 10x + 3 = 0....(1)$

We know if $a {x}^{2} + bx + c = 0$ is a equation then sum of roots of the equation is $- \frac{b}{a}$

and product of roots of the equation is $\frac{c}{a}$

We are comparing equation (1) with $a {x}^{2} + bx + c = 0$

and get,

$a = 3 \\ b = - 10 \\ c = 3$

It is also given that $\alpha \: and \: \beta$

are roots of equation (1).

Then by rule,

$\alpha + \beta = -( \frac{ - 10}{3}) \\ = \frac{10}{3} \\ \alpha = \frac{10}{3} - \beta$

and

$\alpha \beta = \frac{3}{3} = 1 \\ \alpha = \frac{1}{ \beta }$

Equation related two more questions:

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