Question

solve the qyestion and get 50 pts​

Answer 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 = a² + 2ab + b²

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

Answer 2

Answer:

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) }

(α+1)

2

+(c−1)

(α+1)

2

+

(β+1)

2

+(c−1)

(β+1)

2

Identity : (a + b)² = a² + 2ab + b²

Here, a = α or β , b = 1