Question

5.Find the range of values of p,given that the straight line y=px-1 does not intercept th curve y=x^2+2x+3. @ganeshie8

Answer 1

yes set both equations equal to each other and get the quadratic in standard form

Answer 2

\[px - 1 = x^2+2x+3\]

Answer 3

write it in standard form

Answer 4

what do you get ?

Answer 5

\[-x^2-2x+px-3+1=0\]

Answer 6

\[-x^2-x(2-p)-2=0\]

Answer 7

\[x^2+x(2-p)+2=0\]

Answer 8

the constant term doesn't look right, work it again

Answer 9

\[px - 1 = x^2+2x+3\]

subtracting px both sides gives
\[- 1 = x^2+x(2-p)+3\]

adding 1 both sides gives
\[0 = x^2+x(2-p)+4\]

Answer 10

you want this quadratic equation NOT to have real solutions because you don't want the straight line to touch the curve

Answer 11

whats the condition for a quadratic NOT to have real solutions ?

Answer 12

b2-4ac<0

Answer 13

Yes!

Answer 14

\[(2-p)^2-4(1)(4)<0\]

Answer 15

Excellent! see if you can factor the left hand side

Answer 16

\[4-4p+p^2-16<0\]

Answer 17

\[p^2-4p-12<0\]

Answer 18

\[(p-6)(p+2)<0\]

Answer 19

Looks great! draw a parabola for finding the range