Question

Anyone, who can solve this question with complete explanation..​

Answer 1

QUESTION:

(See the attachment for correct question)

ANSWER:

Given:

$\:\:\:\:\bullet\:\:\:\:\text{Zeroes of x ^2 - p(x + 1) - c are \alpha and \beta .}$

To Do:

$\:\:\:\:\bullet\:\:\:\:\text{Find value of ( \alpha + 1)( \beta + 1)}$

$\:\:\:\:\bullet\:\:\:\:\text{Prove that:}\:\:\:\dfrac{\alpha^2+2\alpha+1}{\alpha^2+2\alpha+c}+\dfrac{\beta^2+2\beta+1}{\beta^2+2\beta+c}=1$

Solution:

$:\longrightarrow p(x)=x^2-p(x+1)-c\\\\:\implies p(x)=x^2-px-p-c\\\\\text{We know that sum of zeroes of a polynomial is,}\\\\:\implies\dfrac{\text{-(coefficient of x)}}{\text{(coefficient of x^2 )}}=\dfrac{-B}{A}\\\\\text{So,}\\\\:\implies\alpha+\beta=\dfrac{-(-p)}{1}\\\\:\implies\alpha+\beta=p- - - -(1)\\\\\text{And,}\\\\\text{We know that product of zeroes of a polynomial is,}\\\\:\implies\dfrac{\text{(constant)}}{(\text{coefficient of x ^2 })}=\dfrac{C}{A}\\\\\text{So,}$

$:\implies\alpha\beta=\dfrac{-p-c}{1}\\\\:\implies\alpha\beta=(-p-c)- - - -(2)\\\\\text{\underline{PART 1}}\\\\:\longrightarrow(\alpha+1)(\beta+1)\\\\\text{On simplifying,}\\\\:\implies\alpha\beta+\alpha+\beta+1\\\\:\implies(\alpha\beta)+(\alpha+\beta)+1\\\\\text{From (1) and (2),}\\\\:\implies(-p-c)+(p)+1\\\\:\implies-p\!\!\!/\,-c+p\!\!\!/\,+1\\\\:\implies\underline{1-c}\\\\\text{\bf{Hence, ( \alpha + 1)( \beta + 1) = 1 - c }}$

$\\\text{\underline{PART 2}}\\\\\text{To Prove:}\\\\:\longrightarrow\dfrac{\alpha^2+2\alpha+1}{\alpha^2+2\alpha+c}+\dfrac{\beta^2+2\beta+1}{\beta^2+2\beta+c}=1\\\\\text{Proof:}\\\\\text{Taking LHS,}\\\\:\longrightarrow\dfrac{\alpha^2+2\alpha+1}{\alpha^2+2\alpha+c}+\dfrac{\beta^2+2\beta+1}{\beta^2+2\beta+c}\\\\\text{We know that, a ^2 +2ab+b ^2 =(a+b) ^2 . So,}\\\\:\implies\dfrac{(\alpha+1)^2}{\alpha^2+2\alpha+c}+\dfrac{(\beta+1)^2}{\beta^2+2\beta+c}- - - -(3)\\\\\text{From (2),}\\\\:\implies\alpha\beta=(-p-c)$

$:\implies c=-p-\alpha\beta\\\\\text{From (1),}\\\\:\implies c=-(\alpha+\beta)-\alpha\beta\\\\:\implies c=(-\alpha-\beta-\alpha\beta)- - - -(4)\\\\\text{Putting value of c from (4) in (3),}\\\\:\implies\dfrac{(\alpha+1)^2}{\alpha^2+2\alpha+c}+\dfrac{(\beta+1)^2}{\beta^2+2\beta+c}\\\\:\implies\dfrac{(\alpha+1)^2}{\alpha^2+2\alpha+(-\alpha-\beta-\alpha\beta)}+\dfrac{(\beta+1)^2}{\beta^2+2\beta+(-\alpha-\beta-\alpha\beta)}$

$:\implies\dfrac{(\alpha+1)^2}{\alpha^2+2\alpha-\alpha-\beta-\alpha\beta}+\dfrac{(\beta+1)^2}{\beta^2+2\beta-\alpha-\beta-\alpha\beta}\\\\:\implies\dfrac{(\alpha+1)^2}{\alpha^2+\alpha-\beta-\alpha\beta}+\dfrac{(\beta+1)^2}{\beta^2+\beta-\alpha-\alpha\beta}\\\\:\implies\dfrac{(\alpha+1)^2}{\alpha(\alpha+1)-\beta(1+\alpha)}+\dfrac{(\beta+1)^2}{\beta(\beta+1)-\alpha(1+\beta)}\\\\:\implies\dfrac{(\alpha+1)^2}{\alpha(\alpha+1)-\beta(\alpha+1)}+\dfrac{(\beta+1)^2}{\beta(\beta+1)-\alpha(\beta+1)}$

$:\implies\dfrac{(\alpha+1)^2}{(\alpha+1)(\alpha-\beta)}+\dfrac{(\beta+1)^2}{(\beta+1)(\beta-\alpha)}\\\\:\implies\dfrac{(\alpha+1)}{(\alpha-\beta)}+\dfrac{(\beta+1)}{(\beta-\alpha)}\\\\:\implies\dfrac{(\alpha+1)}{(\alpha-\beta)}-\dfrac{(\beta+1)}{(\alpha-\beta)}\\\\:\implies\dfrac{(\alpha+1)-(\beta+1)}{(\alpha-\beta)}\\\\:\implies\dfrac{\alpha+1\!\!\!/-\beta-1\!\!\!/}{(\alpha-\beta)}\\\\:\implies\dfrac{\alpha-\beta}{\alpha-\beta}\\\\:\implies 1\,\,=RHS\\\\\text{\bf{HENCE PROVED}}$