Question

If alpha and bita are the zeroes of the polynomial 3x^2+ 2x +3, then find 1/alpha square + 1/bita square
plz ans fast!!​

Answer 1

Step-by-step explanation:

Given:-

$\sf \alpha\:and\:\beta\:are\;zeros\:of\:the\;polynomial\:3x^2+2x+3=0$

To find:-

$\sf \dfrac{1}{\alpha^2}+\dfrac{1}{\beta^2}$

Solution:-

Here

• a=3
• b=2
• c=3

$\boxed{\sf \alpha+\beta=\dfrac{-b}{a}}$

$\\ \tt{:}\longrightarrow \dfrac{-2}{3}$

And

$\boxed{\sf \alpha\beta=\dfrac{c}{a}}$

$\\ \tt{:}\longrightarrow \dfrac{3}{3}$

$\\ \tt{:}\longrightarrow 1$

Now

$\\ \tt{:}\longrightarrow \dfrac{1}{\alpha^2}+\dfrac{1}{\beta^2}$

$\\ \tt{:}\longrightarrow \dfrac{\alpha^2+\beta^2}{\alpha^2\beta^2}$

$\\ \tt{:}\longrightarrow \dfrac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}$

$\\ \tt{:}\longrightarrow \dfrac{\left(\dfrac{-2}{3}\right)^2-2(1)}{(1)^2}$

$\\ \tt{:}\longrightarrow \dfrac{\dfrac{4}{9}-2}{1}$

$\\ \tt{:}\longrightarrow \dfrac{\dfrac{4-18}{9}}{1}$

$\\ \tt{:}\longrightarrow \dfrac{-14}{9}$

$\\ \\ \therefore\sf \dfrac{1}{\boldsymbol{\alpha}^2}+\dfrac{1}{\boldsymbol{\beta}^2}=\dfrac{-14}{9}$

Answer 2

$\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}$

• $\alpha$ and $\beta$ are the zeroes of the polynomial 3x² + 2x + 3.

$\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}$

• $\sf \dfrac{1}{{\alpha}^{2}} \: + \: \dfrac{1}{{\beta}^{2}}$

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}$

First,

• Polynomial = 3x² + 2x + 3

We know that,

• $\boxed{\pink{\sf \alpha \: + \: \beta \: = \: \dfrac{- \: b}{a}}}$

• $\boxed{\pink{\sf \alpha \beta \: = \: \dfrac{c}{a}}}$

Now,

• In 3x² + 2x + 3,

• a = + 3

• b = + 2

• c = + 3

Then,

• $\sf \alpha \: + \: \beta \: = \: \dfrac{- \: b}{a} \: = \: \dfrac{- \: 2}{3}$

• $\sf \alpha \beta \: = \: \dfrac{c}{a} \: = \: \dfrac{3}{3} \: = \: 1$

Now,

• $\sf \dfrac{1}{{\alpha}^{2}} \: + \: \dfrac{1}{{\beta}^{2}}$

Taking LCM,

• $\sf \dfrac{{\alpha}^{2} \: + \: {\beta}^{2}}{{\alpha}^{2} {\beta}^{2}}$

We know that,

• $\boxed{\pink{\sf {\alpha}^{2} \: + \: {\beta}^{2} \: = \: (\alpha \: + \: \beta)^2 \: - \: 2 \alpha \beta}}$

• $\boxed{\pink{\sf {\alpha}^{2} {\beta}^{2} \: = \: (\alpha \beta)^2}}$

Re-writing,

• $\sf \dfrac{(\alpha \: + \: \beta)^2 \: - \: 2 \alpha \beta}{(\alpha \beta)^2}$

Here,

• $\sf \alpha \: + \: \beta \: = \: \dfrac{- \: 2}{3}$

• $\sf \alpha \beta \: = \: \dfrac{3}{3} \: = \: 1$

Substituting the values,

• $\sf \dfrac{\bigg(\dfrac{- \: 2}{3}\bigg)^2 \: - \: 2 \: * \: 1}{(1)^2}$

• $\sf \dfrac{\bigg(\dfrac{- \: 2}{3}\bigg)^2 \: - \: 2}{1}$

• $\sf \bigg(\dfrac{- \: 2}{3}\bigg)^2 \: - \: 2$

• $\sf \dfrac{4}{9} \: - \: 2$

• $\sf \dfrac{4}{9} \: - \: \dfrac{18}{9}$

• $\sf \dfrac{4 \: - \: 18}{9}$

• $\sf \dfrac{- \: 14}{9}$

Therefore,

• $\boxed{\purple{\sf \dfrac{1}{{\alpha}^{2}} \: + \: \dfrac{1}{{\beta}^{2}} \: = \: \dfrac{- \: 14}{9}}}$

$\Large{\bf{\blue{\mathfrak{\dag{\underline{\underline{Extra \: Information:-}}}}}}}$

• Any Quadratic polynomial is of the form ax² + bx + c where a and b simultaneously cannot be equal to 0.

• If $\alpha$ and $\beta$ are the zeroes of a quadratic polynomial then,

• $\sf \alpha \: + \: \beta \: = \: \dfrac{- \: b}{a}$

• $\sf \alpha \beta \: = \: \dfrac{c}{a}$