Question

If P(x) = x2 - (2-p)x+ p - 2 assumes both positive and negative value, then the complete set ofvalues of 'p' is

Answer 1

Given:

$\bold{ p(x)= x^2-(2-p)x+p-2}$

To find:

Set value of p= ?

Solution:

In the question it is defined that function p(x) use both positive and negative value both so, in this question we assign -2, -1, 1, 2 to get p(x) value.

$\ Equation: \\ \Rightarrow p(x)= x^2-(2-p)x+p-2\\\\\ when \ x= -2 : \\\\\Rightarrow p(-2) = (-2)^2-(2-p)(-2)+p-2\\\Rightarrow p(-2) = 4+4-2p+p-2\\\Rightarrow p(-2) = 6-p$

$\ when \ x= -1 : \\\\\Rightarrow p(-1) = (-1)^2-(2-p)(-1)+p-2\\\Rightarrow p(-1) = 1+2-p+p-2\\\Rightarrow p(-1) = 1$

$\ when \ x= 1 : \\\\\Rightarrow p(1) = (1)^2-(2-p)(1)+p-2\\\Rightarrow p(1) = 1+-2+p+p-2\\\Rightarrow p(1) = 2p-3$

$\ when \ x= 2 : \\\\\Rightarrow p(2) = (2)^2-(2-p)(2)+p-2\\\Rightarrow p(2) = 4-4+2p+p-2\\\Rightarrow p(2) = 3p-2$

Set of p: {(6-p), 1, (2p-3), (3p-2)}

Answer 2

step-by-step explaination

D=0

b^2=4ac

(2-p)^2=4(p-2)

p^2-4p+4= 4p-8

p^2-8p+12=0

p=2,6

but as we need both positive and negative value so,

p=(-∞,2)U(6,∞)