Math, asked by cuteangel12, 11 months ago

solve the sum and get 50 points ......​

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Answered by Anonymous
1

Answer:

1

Step-by-step explanation:

Given equation: x² - p(x + 1) - c = 0

Solving the given equation further,

→ x² - px - p - c = 0

→ x² - px - (p + c) = 0

On comparing with ax² + bx + c = 0, we get

a = 1, b = - p, c = - (p + c)

As we know that,

α + β = - b/a = - (- p)/1 = p ...(i)

αβ = c/a = - (p + c)/1 = - p - c ...(ii)

Now, from (ii), we can write :

αβ = - p - c

→ c = - p - αβ

→ c = - (α + β) - αβ [ from (i) ]

→ c = - α - β - αβ ...(iii)

Question :  \dfrac{( \alpha  + 1)^{2} }{ {( \alpha  + 1)}^{2} + (c - 1) }  +  \dfrac{ { (\beta + 1) }^{2} }{ {( \beta  + 1)}^{2} + (c - 1) }

Identity : (a + b)² = + 2ab +

Here, a = α or β , b = 1

=  \dfrac{ {( \alpha  + 1)}^{2} }{ { (\alpha )}^{2}  +2( \alpha )(1) +  {(1)}^{2}   + (c - 1)}  +  \dfrac{ {(  \beta   + 1)}^{2} }{ { ( \beta  )}^{2}  +2(  \beta  )(1) +  {(1)}^{2}   + (c - 1)}

=  \dfrac{ {( \alpha  + 1)}^{2} }{ { \alpha }^{2} + 2 \alpha  + 1 + c - 1 }  +  \dfrac{  {( \beta  + 1)}^{2}  }{ { \beta }^{2} + 2 \beta  + 1 + c - 1 }

=  \dfrac{ {( \alpha  + 1)}^{2} }{ { \alpha }^{2}  + 2 \alpha  + c}  +  \dfrac{ {( \beta  + 1)}^{2} }{ { \beta }^{2} + 2 \beta  + c }

Putting the value of c, we get

=  \dfrac{ {( \alpha  + 1)}^{2} }{ { \alpha }^{2} + 2 \alpha  + ( -  \alpha  -  \beta  -  \alpha  \beta ) }  +  \dfrac{ {(  \beta   + 1)}^{2} }{ {  \beta  }^{2} + 2  \beta   + ( -  \alpha  -  \beta  -  \alpha  \beta ) }

=  \dfrac{ {( \alpha  + 1)}^{2} }{ { \alpha }^{2} + 2 \alpha  -  \alpha  -  \beta  -  \alpha \beta }  +   \dfrac{ {(  \beta   + 1)}^{2} }{ {  \beta  }^{2} + 2  \beta   -  \alpha  -  \beta  -  \alpha \beta }

=  \dfrac{ {( \alpha  + 1)}^{2} }{ { \alpha }^{2} +  \alpha  -  \beta  -  \alpha  \beta  }  +  \dfrac{ {( \beta  + 1)}^{2} }{ { \beta }^{2} +  \beta  -  \alpha  -  \alpha  \beta  }

=  \dfrac{ {( \alpha  + 1)}^{2} }{ \alpha ( \alpha  + 1) -  \beta ( \alpha  + 1)}  +  \dfrac{ { (\beta  + 1)}^{2} }{ \beta ( \beta  + 1) -  \alpha ( \beta  + 1)}

=  \dfrac{ { (\alpha + 1) }^{2} }{( \alpha  -  \beta )( \alpha  + 1)}  +  \dfrac{ {( \beta  + 1)}^{2} }{( \beta  -  \alpha )( \beta  - 1)}

Multiply the numerator and denominator of  \dfrac{ {( \beta  + 1)}^{2} }{( \beta  -  \alpha )( \beta  + 1)} by - 1, we get

=  \dfrac{ {( \alpha  + 1)}^{2} }{( \alpha  -  \beta )( \alpha  + 1)} -  \dfrac{ {( \beta  + 1)}^{2} }{( \alpha  -  \beta )( \beta  + 1)}

=  \dfrac{ \alpha  + 1}{ \alpha  -  \beta }  -  \dfrac{ \beta  + 1}{ \alpha   -  \beta }

=  \dfrac{ \alpha  + 1 - ( \beta  + 1)}{ \alpha  -  \beta }

=  \dfrac{ \alpha  + 1 -  \beta  - 1}{ \alpha  -  \beta }

=  \dfrac{ \alpha  -  \beta }{  \alpha  -  \beta  }

= 1

Answered by lalito1729
0

Answer:

.

Step-by-step explanation:

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