Maths?!?!?!?!??

Question

if straight line y=5x-1 does not intersect the curve y=2x^2 + x + p. Find the range of values of p.

solution? thanks

Answer 1

5x -- 1 = 2x^2 + x + p

=> 2x^2 -- 4x + (p + 1) = 0

this has no real roots,

(-- 4)^2 -- 4(2)(p + 1) < 0

8p > 8

p > 1

Answer 2

y=5x+1 and y=2x^2 + x + p don't intersect

so their solution has no real roots

ie 5x+1 =2x^2 +x +p has no real roots

ie 2x^2 -4x +p-1 has no real roots

Ie delta, b^2 -4ac must be negative

16 - 8(p-1) < 0

16 < 8(p-1)

2 < p-1

p > 3

Answer 3

y=5x-1

y=2x^2+x+p



2x^2+x+p=5x-1



2x^2-4x+(p-1)=0

ax^2+bx+c

Doesnt intersect

b^2-4ac<0

(-4)^2-4(2)(p-1)<0

16-8p+8<0

-8p<-24p

when you divide by a negativ value change the signs

so

p>3