Question

answer the questions
need explanation​

Answer 1

AnswEr:-

1) 4.

2) 8.

ExplanaTion:-

Given:-

• A quadratic polynomial,

$\tt\longrightarrow4 {x}^{2} - 4x + 1$

$\tt\longrightarrow \alpha\:\:And\:\: \beta$ Are the zeroes of given polynomial.

To Find:-

1. The value of $\tt\dfrac{1}{\alpha}+\dfrac{1}{\beta}$.

2. The value of $\tt\dfrac{1}{\alpha^2}+\dfrac{1}{\beta^2}$.

ID used:-

$\tt\longrightarrow a^2+b^2 = (a+b)^2-2ab$

Concepts Used:-

• In a quadratic polynomial:-

Sum of zeroes,

$\tt\longrightarrow \alpha+\beta = \dfrac{-b}{a}$.

Product of zeroes,

$\tt\longrightarrow\alpha\beta=\dfrac{c}{a}$.

Where,

• a = Coefficient of x².
• b = Coefficient of x.
• c = Constant Term.

So Here,

$\tt\longrightarrow \alpha + \beta = \frac{ - ( - 4)}{4} \\ \\ \tt\longrightarrow \alpha + \beta = \frac{4}{4} \\ \\ \tt\longrightarrow \alpha + \beta = 1...(1) \\ \\ \tt \: and \\ \\ \tt\longrightarrow \alpha \beta = \frac{c}{a} \\ \\ \tt\longrightarrow \alpha \beta = \frac{1}{4} ...(2)$

1:-

Given that $\tt\alpha \:\:And\:\: \beta$ are zeroes and we have to find the value of $\tt\dfrac{1}{\alpha}+\dfrac{1}{\beta}$.

• So lets solve this:-

$\tt\mapsto \frac{1}{ \alpha } + \frac{1}{ \beta } \\ \\ \\ \tt = \dfrac{1( \beta ) + 1( \alpha )}{ \alpha \times \beta } \\ \\ \\ \tt[taking \: lcm] \\ \\ \\ \tt = \frac{ \beta + \alpha }{ \alpha \beta } \\ \\ \\ \tt = \dfrac{ \alpha + \beta }{ \alpha \beta } \\ \\ \\ \tt = \dfrac{1}{ \frac{1}{4} } . \\ \\ \\ \tt[from \: (1) \: and \: (2)]\\ \\ \\ \tt = \dfrac{4}{1} \\ \\ \\ \tt = 4.$

$\therefore$ The required value of $\dfrac{1}{\alpha}+\dfrac{1}{\beta}$ is 4.

2:-

Here we have to find out the value of $\dfrac{1}{\alpha^2}+\dfrac{1}{\beta^2}$ .

• So lets solve this:-

$\tt\longrightarrow \dfrac{1}{ \alpha {}^{2} } + \dfrac{1}{ \beta {}^{2} } \\ \\ \\ \tt = \dfrac{1( \alpha {}^{2}) + 1( \beta {}^{2}) }{ \alpha {}^{2} \beta {}^{2} } \\ \\ \\ \tt[taking \: lcm] \\ \\ \tt = \dfrac{ \alpha {}^{2} + \beta {}^{2} }{( \alpha \beta )( \alpha \beta )} \\ \\ \\ \tt = \dfrac{( \alpha + \beta ) {}^{2} - 2 \alpha \beta }{( \alpha \beta )( \alpha \beta )} \\ \\ \\ \tt(using \: id) \\ \\ \\ \tt = \dfrac{(1) {}^{2 } - \cancel{2}( \frac{1}{\cancel{4}}) }{( \frac{1}{4})( \frac{1}{4} ) } \\ \\ \\ \tt[from \: (1) \: and(2)] \\ \\ \\ \tt = \dfrac{1- \frac{1}{2} }{ \frac{1}{16} } \\ \\ \\ \tt = \dfrac{ \frac{2 - 1}{2} }{ \frac{1}{16} } \\ \\ \\ \tt = \dfrac{ \frac{1}{2} }{ \frac{1}{16} } \\ \\ \\ \tt = \dfrac{1}{2} \times \frac{16}{1} \\ \\ \\ \tt = \dfrac{16}{2} \\ \\ \\ \tt = 8.$

$\therefore$ The required value of $\dfrac{1}{\alpha^2}+\dfrac{1}{\beta^2}$ is 8.

Answer 2

Question :–

▪︎ If ${ \bold{ \: \: \alpha \: \: and \: \: \beta \: \: }}$ are Zero's of 4x² - 4x + 1 = 0 , then find –

$\\ { \bold{ (a) \: \: \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } \: \: \: \: \: \: \: \: \: (b) \: \: \dfrac{1}{ { \alpha }^{2} } + \dfrac{1}{ { \beta }^{2} } }}$

ANSWER :–

$\\ { \bold{ \: (a) \: \: \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } = 4 }}$

$\\ { \bold{ \: (b) \: \: \dfrac{1}{ \alpha {}^{2} } + \dfrac{1}{ \beta {}^{2} } = 8 }}$

EXPLANATION :–

• If a quadratic equation ax² + bx + c = 0 which have two roots ${ \bold{ \: \: \alpha \: \: and \: \: \beta \: \: }}$ then –

$\\ \longrightarrow{ \bold{ sum \: \: of \: \: roots = \alpha + \beta = - \dfrac{b}{a} }}$

$\\ \longrightarrow{ \bold{ product \: \: of \: \: roots = \alpha \beta = \dfrac{c}{a} }}$

• In the given equation –

$\\ \: \: \: \: \: \: \: { \huge{.}} \: \: \: { \bold{ a = 4 }}$

$\\ \: \: \: \: \: \: \: { \huge{.}} \: \: \: { \bold{ b = -4 }}$

$\\ \: \: \: \: \: \: \: { \huge{.}} \: \: \: { \bold{ c = 1 }}$

• So that –

$\\ \longrightarrow{ \bold{ sum \: \: of \: \: roots = \alpha + \beta = - \dfrac{( - 4)}{4} = 1 }}$

$\\ \longrightarrow{ \bold{ product \: \: of \: \: roots = \alpha \beta = \dfrac{1}{4} }}$

$\\ \rule{220}{2}$

$\\ { \bold{ (a) \: \: \: \: = \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } }}$

$\\ { \bold{ \: \: \: \: \: \: \: = \dfrac{ \alpha + \beta }{ \alpha \beta } }}$

• Now put the values –

$\\ { \bold{ \: \: \: \: \: \: \: = \dfrac{ 1 }{ \frac{1}{4} } }}$

$\\ { \bold{ \: \: \: \: \: \: \: = 4}}$

• So that –

$\\ { \bold{ \implies \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } = 4 }}$

$\\ \rule{220}{2}$

$\\ { \bold{ (b) \: \: \: \: = \dfrac{1}{ { \alpha }^{2} } + \dfrac{1}{ { \beta }^{2} } }}$

$\\ { \bold{ \: \: \: \: = \frac{ { \alpha }^{2} + { \beta }^{2} }{ {( \alpha \beta )}^{2} } }}$

$\\ { \bold{ \: \: \: \: \: \: = \frac{ {( \alpha + \beta )}^{2} - 2 \alpha \beta }{ {( \alpha \beta )}^{2} } \: \: \: \: \: \: [ \because \: \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab] }}$

• Now put the values –

$\\ { \bold{ \: \: \: \: \: \: = \frac{ {( 1 )}^{2} - 2 ( \dfrac{1}{4} ) }{ {( \dfrac{1}{4} )}^{2} } }}$

$\\ { \bold{ \: \: \: \: \: \: = \frac{ 1 - \dfrac{1}{2} }{ \dfrac{1}{16} } }}$

$\\ { \bold{ \: \: \: \: \: \: = \frac{ \dfrac{1}{ \cancel 2} }{ \dfrac{1}{ \cancel {16}} } }}$

$\\ { \bold{ \: \: \: \: \: \: = 8}}$

• So that –

$\\ { \bold{ \implies \dfrac{1}{ \alpha {}^{2} } + \dfrac{1}{ \beta {}^{2} } = 8 }}$

$\\ \rule{220}{2}$