help

  • 0 votes

find the values of k for which kx^2 +8x + 5 = 0 has real roots

thnx =)

Posted Mon 1st October, 2012 @ 12:50 by Akeel 786

6 Answers

  • 2 votes

as it has real roots this means that you should be using the quadratic formula, the discriminant, which its b^2-4ac. and then make it =0 and then solve it as in finding K... you'll get a fraction as an answer, or you could just put it in decimals, whichever way you like. 

Just in case you havent done in quadratic equation... in the example you gave a=K b=8 c=5. soo there you.  

Answered Mon 1st October, 2012 @ 21:27 by Liliana Martin
  • 0 votes

thnx u helped a lot

Answered Tue 2nd October, 2012 @ 09:49 by Akeel 786
  • 0 votes

 Liliana Martin

ermm kinda stuk again =(

for the discriminant (b^2-4ac) i gt 64 - 20k ----->> iz dat ryt????

what do i do next?!?!?!?!?!?!?!?!?! (BTW i aint allowed to use a calc)

thnx XD

Answered Tue 2nd October, 2012 @ 17:25 by Akeel 786
  • 0 votes

8^2 + 4xkx5 > 0 as has real roots not one root i thought? As one real root is = 0 isnt it?
Anyway...
64 + 20k > 0
20k > -64
K > -64/20
K > -16/5
Thats what i got? Hope it helps even if its wrong you might be able to see where to go next?

Answered Mon 29th October, 2012 @ 18:04 by Megan Huntington
  • 0 votes

kx^2 + 8x +5 = 0

b^2 - 4ac

(8)^2 - (4)(k)(5)

64 -20k = 0

64 = 20k

k = 64/20

k = 32/10

k = 3.2

That's how I would do it, but I don't really understand the discriminant much :/

Answered Mon 29th October, 2012 @ 18:08 by Joanne
Edited by Joanne on Mon 29th October, 2012 @ 18:08
  • 0 votes

thnx people ^_^ helped a lot

Answered Mon 12th November, 2012 @ 19:58 by Akeel 786