Question

show that the constant p cannot lie between 1 and 2 if the equation px^2-p+10=2(p+2)x has real roots

Answer 1

Use the discriminant property
if we have real roots, then b^2 - 4ac > 0

px^2-p+10=2(p+2)x
px^2 - p + 10 = 2px + 4x
px^2 - 2px - 4x + 10 = 0
px^2 + x(-2p-4) + 10 = 0

b^2 - 4ac > 0
(-2p-4)^2 - 4(p)(10) > 0
4p^2 + 16p + 16 - 40p > 0
4p^2 - 24p + 16 > 0
p^2 - 6p + 4 > 0
p^2 - 6p + 9 > -4+9
(p-3)^2 > 5
p-3 > √5 or -p+3 > √5
p > 3+√5 or p < 3-√5
p > 5.236.. or p < .7639..

so a number between 1 and 2 does not satisfy this condition