Recent content by Lytk

Hi , Sorry I didnt realize I made mistakes posting the question. 1) I had written the equation of the line wrong It is 3x - 4y -12 =0 which makes the slope -3/-4 , which is 3/4 3/4 * -4/3 = -1 2) In (a) the question actually asked for the distance between P(10,k) and L1.

The co-ordinates of two points are A(4,-1) and B(7, -5) The line L1: 3x+4y−12 = is perpendicular to AB. (a) Find, in terms of k, the distance between the point P (10, k) and the line L2 (b) P(10, k) is on a bisector of the angles between the lines L1 and L2 and L2 :5x+12y−20 =0...

$b^2>ac$ $c^2>ab$ I multiply the corresponding sides and I get: $b^2c^2>a^2cb$ which cancels down to $bc>a^2$ $0>a^2 -bc$ (discriminant) $\therefore$ Roots are not realTHANK YOU SO MUCH! This was confusing me for a long time :)

Hey :) so the way to prove that the roots are not real, would be if the discriminant is less than 0. I used the quadratic formula first to find the real roots of the first two equations: The root of $ax^2 + 2bx +c$ = $ \frac{-b\pm\sqrt{b^2-ac}}{a}$ ( Real) ${b^2-ac}$ > 0 The root of...

a, b , c are positive constants and the roots of ax^2 + 2bx+ c and bx^2 + 2cx +a are all real and unequal(unique). Show that the roots of cx^2 + 2ax +b = 0 are NOT real. Help!:)