Question

quadratic equation class 10
pls solve with detailed explanation i have a lot doubt about it​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: \alpha , \: \beta \: are \: zeroes \: of \: the \: polynomial \: {2x}^{2} - 4x + 5$

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm\implies \: \alpha + \beta = - \dfrac{( - 4)}{2} = 2$

And

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm\implies \: \alpha \beta = \dfrac{5}{2}$

Now, Consider

$\rm \: { \alpha }^{2} + { \beta }^{2}$

$\rm \: = { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta - 2 \alpha \beta$

$\rm \: = \: {( \alpha + \beta )}^{2} - 2 \alpha \beta$

$\rm \: = \: {2}^{2} - 2 \times \dfrac{5}{2}$

$\rm \: = \: 4 - 5$

$\rm \: = \: - \: 1$

Hence,

$\rm\implies \: \:\boxed{\tt{ \: \rm \: { \alpha }^{2} + { \beta }^{2} = \: - \: 1 \: }}$

So, Option (c) is correct.

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ADDITIONAL INFORMATION

$\rm \: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta )}^{3} - 3 \alpha \beta ( \alpha + \beta ) \:$

$\rm \: {( \alpha - \beta )}^{2} = {( \alpha + \beta )}^{2} - 4 \alpha \beta$

$\rm \: { \alpha }^{4} + { \beta }^{4} = {\bigg( {( \alpha + \beta )}^{2} - 2 \alpha \beta \bigg) }^{2} - 2 {( \alpha \beta )}^{2}$

Answer 2

$\underline{ \pink{ \rm GIVEN \: \: DATA\: \: : }}$

$\rm p(x) = 2 {x}^{2} - 4x + 5$

$\rm \alpha \: \: and \: \: \beta \: \: \: are \: \: \: zeros \: \: \: of \: \: p(x)$

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$\underline{ \purple{ \rm TO \: \: FIND \: \: : }}$

$\rm the \: \: value \: \: of \: \: { \alpha }^{2} + { \beta }^{2}$

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$\underline{ \red{ \rm SOLUTION \: \: : }}$

the values of a,b,c in the given polynomial are :

a = 2 , b = -4 , c = 5

Method 1 :

by using,

$\boxed{ \orange\ast \: \: \boxed{ \boxed{ \green{ \rm { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2( \alpha \beta )}}}}$

we obtain,

$\rm⇒ \: \: ( 2 ) ^{2} - \cancel2( \frac{5}{ \cancel2} )$

$\rm⇒ \: \: 4 - 5$

$\bf ⇒ \: \: - 1$

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Method 2 :

by using,

$\boxed { \color{darkgreen}{\ast} \: \: \boxed{ \boxed{ \rm { \orange{{ \alpha }^{2} + { \beta }^{2} = \frac{ {b}^{2} - 2ac}{ {a}^{2} } }}}}}$

we obtain,

$\rm⇒ \: \: \frac{16 - 2(2)(5)}{4}$

$\rm⇒ \: \: \frac{16 - 20}{4}$

$\bf ⇒ \: \: - 1$

$\color{gold} ▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄$