Question

find all the values of P so that 6 lies between the roots of x^2 +2( p -3)x+9 ?

Answer 1

SOLUTION

Hence, values of p satisfying (1), (2) &(3) is p E (-infinite , -3/4) U (9,infinite)

Refer to the attachment.

hope it helps ☺️

Answer 2

Answer:

p =  ( 3 ,  ∞  ) U  ( 9 ,  ∞ ).

Step-by-step explanation:

Given :

$\displaystyle p(x)=x^2 +2( p -3)x+9$

Roots lies between 6.

We have to find value of p.

Using discriminant formula here  

$\displaystyle \text{ b^2-4ac\geq 0 for real values}$

When roots lies between 6 means

$\displaystyle b^2-4ac\geq6$

Now put the values here we get

$\displaystyle [2(p-3)]^2-4\times1\times9\geq6\\\\\displaystyle [2(p-3)]^2-(2\times3)^2\geq6\\\\\displaystyle [2(p-3)]^2-(6)^2\geq6$

Using identity

$\displaystyle (a^2-b^2)=(a+b)(a-b)$

$\displaystyle (2(p-3)+6)(2(p-3)-6)\geq6\\\\\displaystyle (2(p-3)+6\geq6\\\\\displaystyle (2(p-3)\geq0\\\\\displaystyle (p-3)\geq0\\\\\displaystyle \text{ p\geq3 \ OR}$

p € ( 3 ,  ∞  )

$\displaystyle 2(p-3)-6\geq6\\\\\displaystyle 2(p-3)\geq12\\\\\displaystyle (p-3)\geq6\\\\\displaystyle p\geq9$

p € ( 9 ,  ∞ )

In question it is said that root lie between 6.

So we get  p =  ( 3 ,  ∞  ) U  ( 9 ,  ∞ ).