C1 2015 5 Edexcel

5

(a)

We have no real roots if the discriminant of the quadratic is negative. That is, when:

4^2 - 4(p - 1)(p - 5) < 0

Dividing by 4 and moving (p - 1)(p - 5) to the RHS we have:

4 < (p - 1)(p - 5)

So:

p^2 - p - 5p + 5 > 4

We therefore have:

p^2 - 6p + 1 > 0

(b)

The easiest way is completing the square, we have:

p^2 - 6p + 1 = (p - 3)^2 - 9 + 1 = (p - 3)^2 - 8

So:

p^2 - 6p + 1 > 0

is equivalent to:

(p - 3)^2 > 8

This is the case if either p - 3 > \sqrt 8 = 2 \sqrt 2 or p - 3 < -2 \sqrt 2. So either p > 3 + 2 \sqrt 2 or p < 3 - 2 \sqrt 2.

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