Help plz!The equation x2 + 5kx + 2k = 0, where k is a constant..

The equation x2 + 5kx + 2k = 0, where k is a constant, has real roots.

(a) Prove that k(25k – 8) > 0.

(b) Hence find the set of possible values of k.

(c) Write down the values of k for which the equation x2 + 5kx + 2k = 0 has equal roots

Help would be much appreciated!

Reply 1

Kunster1337

a=1 b = 5k c= 2K?? :s

Reply 2

Aired

(a) As it has real roots, b² -4ac > 0

25k² - 4.1.2k > 0

25k² - 8k > 0
k(25 - 8) > 0

Reply 3

Fenchurch

how didnt i get the first bit :| lol n00b me got that, i need help on next bits thoguh (b n c) :p:

Reply 4

bobbob

b) k > 25/8 (as if k < 8/25, it's a negative number)
c) Where b² - 4ac = 0 it has equal roots

k(25k – 8) = 0

So where k = 0, or where k = 8/25

Reply 5

silent ninja

bobbob

b) k > 25/8 (as if k < 8/25, it's a negative number)
c) Where b² - 4ac = 0 it has equal roots

k(25k – 8) = 0

So where k = 0, or where k = 8/25

for (b) k is less than or equal to 0, or greater than or equal to 8/25. draw the graph of k(25k - 8), indeed for any questions like this, if in doubt.