Maths Problem anyone?

Question

If the equation px^2 - p + 10 = 2(p+2)x has real solution(s), show that p cannot lie between 1 and 2.

Answer 1

px^2-p+10=2(p+2)x
rewriting
px^2-2(p+2)x-(p-10)=0
here a=p b=-2(p+2) c=-(p-10)
b^2-4ac=4(p+2)^2+4p(p-10)
=4(p^2+4p+4)+4p^2-40p
=4p^2+16p+16+4p^2-40p
=8p^2-24p+16
for real roots this must be positive
8(p^2-3p+2)=0
(p-2)(p-1)=0
so p must lie outside of 2 and 1 if (p-2)(p-1) is>0

Answer 2

transfer the rhs to the left
px^2 -2(p+2)x -p +10 = 0
comparing with ax^2+bx+c =0
a = p
b = -2(p+2)
c= 10-p
this has real root if b^2 >=4ac
or 4(p+2)^2 > = 4p(10-p)
or (p+2)^2 >= p(10-p)
or p^2+4p+4 >= 10p-p^2
or 2p^2 - 6p +4 >=0
or p^2-3p+2 >=0
or (p-1)(p-2) >=0
so both are positive or both -ve
p-1 >0 and p-2 >0 =>p>2
p-1 <0 and p-2 <0 =>p <1
so p <1 or >2 so p caanot lie between 1 and 2

Answer 3

Do a bit of algebraic manipulation and get a zero RHS quadratic.
then find the discriminant
(for general quadratic eq. ax^2 +bx +c this is b^2 -4ac)
if the roots are to be real this is greater than or equal to 0
so get a quadratic with factors of p and map out all possibilities, consider them and if any yield less than zero take em out.
You get the answer.

Answer 4

Agree with rfamilyme answer.

Aloha

Answer 5

whats is the topic?