Question

find the zeroes of the quadratic polynomial p(x)=p(x^2+1) -x(p^2+1) and verify the relationship between the zeroes and the coefficient​

Answer 1

Answer:

p(x)= px²+p-xp²-x

px²-(p²+1)x+p

sum of zeroes =-b/a=p²+1/p

product of zeroes =c/a= p/p=1

Answer 2

Appropriate Question:

$\sf{find \: the \: zeroes \: of \: the \: quadratic \: polynomial \: p(x) = p(x^2+1)-x(p^2+1) } \\ \sf{and \: verify \: the \: relationship \: between \: the \: zeroes \: and \: the \: coefficient.}$

Solution:

We Have,

$\tt{p(x) = p( {x}^{2} + 1) - x( {p}^{2} + 1) } \\ \tt{ = {px}^{2} + p - {p}^{2}x - x + p } \\ \tt{ = {px}^{2} - x( {p}^{2} + 1) + p } \: \: \: \: \: \: { \bigg[a = p; \: b = ({p}^{2}+1), \: c=p \bigg] }$

For Zeroes, p(x) = 0

$:\implies \tt{ {px}^{2} - x( {p}^{2} + 1) + p = 0} \\ :\implies \tt{ {px}^{2} - {p}^{2}x - x + p = 0 } \\ :\implies \tt{px(x - p) - 1(x - p) = 0} \\ :\implies \tt{(px - 1)(x - p)} = 0 \\ :\implies \tt{px - 1 = 0 \: \:} \mathfrak{or} \: \: \tt{x - p = 0} \\ :\implies \tt{px = 1} \: \: \mathfrak{or} \: \: \tt{x = p} \\ :\implies \tt{x = \frac{1}{p} } \: \: \: \mathfrak{or} \: \: \: \tt{x = p}$

$\tt{Let \: \alpha = \frac{1}{p} \: and \: \beta = p } \\ \\ \tt{Now, \alpha + \beta = \frac{1}{p} + p } \\ \\ \tt{= \frac{1 + {p}^{2} }{p} } \\ \\ \tt{ = \frac{ - \bigg \{ - \bigg (p^2+1\bigg)\bigg\}}{p} } \\ \\ \tt{ = \frac{Coefficient \: of \: x}{Coefficient \: of \: x^2} }$

Also,

$\tt{ \alpha \beta = \frac{1}{p} \times p = 1 = \frac{ p }{p} = \frac{Constant \: Term}{Coefficient \: of \: x^2} }$