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Answer 1

Answer:

Explanation:

We have polynomial  f ( x ) =  x2  - p x  + q

And

Roots are α  and β

And

we know from relationship between zeros and coefficient . 

Sum of zeros  = −Coefficient of xCoefficient of x2

So,

α  + β  =  - p                                         ------ ( 1 )

Taking whole square on both hand side , we get

( α  + β  )2 = p 2                                  ------ ( 2 )

⇒α2 + β2 +  2 α β = p2⇒α2 + β2 +  2 α β − 2αβ + 2αβ=  p2⇒α2 + β2 −  2 α β  +4 αβ=  p2⇒(α  − β) 2  +4 αβ=  p2                     −−−− ( 3 )

And

Products of zeros  = Constant termCoefficient of x2

So,

α  β  =  q        , Substitute that value in equation 3 , we get

⇒(α− β)2 +  4 (q ) = p2⇒(α− β)2 +4 q=  p2⇒(α− β)2  = p2 − 4 q              −−−− ( 4 )   

Now we add equation 2 and 4 and get

(α + β)2 + (α − β)2 = p2 +  p2 − 4 q= 2 p2 − 4 q

And we multiply equation 2 and 4 and get

(α + β)2 × (α − β)2 = p2( p2 − 4 q)= p4 − 4 p2q

And we know formula for polynomial when sum of zeros and product of zeros we know :

Polynomial  =  k [ x2  - ( Sum of zeros ) x  + ( Product of zeros ) ]   , Here k is any non zero real number.

Substitute values , we get

Quadratic polynomial  =  k [ x2  - ( 2 p2 - 4 q) x  + ( 2 p4 - 4 p2q) ] 

                            

= x2  - ( 2 p2 - 4 q) x  + ( 2 p4 - 4 p2q) [ taking k = 1 ]                                         ( Ans )

Hope this information will clear your doubts about topic.

please brainliest..................................................................

Answer 2

SOLUTION:-

Given:

$\alpha \: and \: \beta \: are \: the \: zeroes \: of \: the \: quadratic \: polynomial \: f(x) = {x}^{2} - px + q$

To prove:

$= > \frac{ { \alpha }^{2} }{ { \beta }^{2} } + \frac{ { \beta }^{2} }{ { \alpha }^{2} } = \frac{ {p}^{4} }{ {q}^{2} } - \frac{4 {p}^{2} }{q} + 2$

Proof:

We know that sum of zeroes:

$\alpha + \beta = \frac{ - ( - p)}{1} \\ \\ = > \alpha + \beta = p$

Sum of products:

$= > \alpha \beta = \frac{q}{1} = > q$

According to the question:

$\frac{ { \alpha }^{2} }{ { \beta }^{2} } + \frac{ { \beta }^{2} }{ { \alpha }^{2} } \\ \\ = > \frac{ { \alpha }^{4} + { \beta }^{4} }{ { \alpha }^{2} { \beta }^{2} } \\ \\ = > \frac{( { \alpha }^{2} + { \beta }^{2} ) {}^{2} - 2 { \alpha }^{2} { \beta }^{2} }{ { \alpha }^{2} { \beta }^{2} } \\ \\ = > \frac{( \alpha + \beta ) {}^{2} - 2 \alpha \beta {}^{2} - 2( { \alpha \beta )}^{2} }{( { \alpha \beta )}^{2} } \\ \\ = > \frac{( {p}^{2} - 2q) {}^{2} - 2(q) {}^{2} }{ {q}^{2} } \\ \\ = > \frac{ {p}^{4} + 4 {q}^{2} - 4 {p}^{2}q - 2 {q}^{2} }{ {q}^{2} } \\ \\ = > \frac{ {p}^{4} - 4 {p}^{2}q + 2 {q}^{2} }{ {q}^{2} } \\ \\ = > \frac{ {p}^{4} }{ {q}^{2} } - \frac{4 {p}^{2} q}{q {}^{2} } + \frac{2 {q}^{2} }{ {q}^{2} } \\ \\ = > \frac{ {p}^{4} }{ {q}^{2} } - \frac{4 {p}^{2} }{q} + 2$

Hence,

Proved.

Hope it helps ☺️