The straight line with equation y=3x-7 does not cross or touch the curve with equation y=2px^2-6px+4p, where p is a constant.(a) Show that 4p^2-20p+9<0 (b) Hence find the set of possible values for p.

(a) If we consider the intersection of these two lines then, 3x-7=2px^2-6px+4p.This can be rearranged into the form ax^2+bx+c=0 such that, 2px^2+(-6p-3)x+4p+7, where a=2p, b=-6p-3 and c=4p+7.However, since the straight line and the curve do not infact cross or touch (two or one distinct root respectively), there will be no real roots/solutions to this quadratic equation. Using the determinant, it follows that b^2-4ac<0 and we can now substitute our values for a, b and c into this inequality.Hence, b^2-4ac=(-6p-3)^2-4(2p)(4p+7)=36p^2+36p+9-32p^2-56p=4p^2-20p+9So, 4p^2-20p+9<0 QED(b) If we factorise the quadratic then, (2p-9)(2p-1)<0So the critical values for p are when 2p-9=0, 2p-1=0 so p=9/2, p=1/2If we draw out the graph of y=(2p-9)(2p-1) then we can see that y<0 for 1/2<p<9/2