Question

Find the zeroes of the quadratic polynomial p(x) = ax²-(a²+ 1)x+a and verify relationship between the zeroes and the coefficients.​

Answer 1

Step-by-step explanation:

Factorization: A(X2 ...

We have been given. a(x2 + 1) - x(a2 + 1) = 0. ax2 - (a2 + 1)x + a = 0. ax2 - a2x - x + a = 0. ax( x - a) - 1(x - a) = 0. (ax - 1)(x - a) = 0. Therefore

Answer 2

Quadratic Polynomial p(x)=ax²-(a²+1)x+a

${\mapsto\sf a {x}^{2} -( {a}^{2} +1)x+a=0}$

${\mapsto\sf a {x}^{2} -{a}^{2}x -x+a=0}$

${\mapsto\sf ax( x -a )-1(x-a)=0}$

${\mapsto\sf ( x -a )(ax + 1)=0}$

${\mapsto\sf ( x -a ) = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (ax - 1)=0}$

${\mapsto\sf x = a \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ax = 1}$

${\mapsto\sf x = a \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = \dfrac{ 1}{a} }$

Let alpha be (a) and beta be (1/a).

In Quadratic Equation ax²-(a²+1)x+a =0

On comparing with ax²+bx+c=0

$\sf{Sum \: \: of \: \: zeres (Alpha+beta) = \dfrac{-b}{a} = \dfrac{-Coefficient \: of \: x}{-Coefficient \: of \: {x}^{2} } }$

$\sf{a + \dfrac{ 1}{a} = \dfrac{ - \{ - ( {a}^{2} + 1) \} }{a} }$

$\sf{ \dfrac{ {a}^{2} + 1}{a} = \dfrac{ {a}^{2} + {1}}{a} }$

$\sf{Product \: of \: zeros \: (Alpha×Beta)=\dfrac{c}{a}=\dfrac{ Constant \: term}{Coefficient \: of \: {x}^{2} }}$

$\sf{a \times \dfrac{ 1}{a} = \dfrac{a}{a} }$

$\sf{a = a }$

Verified