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 April 17th, 2011, 09:12 AM #1 Member   Joined: Apr 2011 Posts: 46 Thanks: 0 quadratic equation hi the equation is t^2 - 5t + 6 =0 I got (t-2)(t-3) factorizing t-2=0 t-3=0 t=2 t= 3 then the second part of the question is by examining the sign of the discriminant b^2 - 4ac in the quadratic formula show that x^2 +4x +5 =o has no real solutions a=1 b=4 c=5 16-4(1)(5)= 4 so it has real solutions. Can anyone tell if I went wrong anywhere?
April 17th, 2011, 09:41 AM   #2
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Quote:
 Originally Posted by daveyf 16-4(1)(5)=4 so it has real solutions. Can anyone tell if I went wrong anywhere?
16 - 4(1)(5) = -4. not 4. So it has no real solutions.

 April 17th, 2011, 12:51 PM #3 Member   Joined: Apr 2011 Posts: 46 Thanks: 0 Re: quadratic equation ah ya negative cheers
 April 17th, 2011, 03:34 PM #4 Senior Member   Joined: Nov 2010 Posts: 502 Thanks: 0 Re: quadratic equation No problem! Good luck.

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