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You are not logged in. #1 20051218 05:52:07
Factorising quadratics... againI have problems when trying to factorise a quadratic where the coefficient of x^2 is < 1 i.e. Aloha Nui means Goodbye. #2 20051218 06:02:55
Re: Factorising quadratics... againYou can (what is the word for taking divisor out of brackets),so Last edited by krassi_holmz (20051218 06:05:43) IPBLE: Increasing Performance By Lowering Expectations. #3 20051218 06:06:57
Re: Factorising quadratics... againso you're saying I should divide the whole equation through by the coefficient of x^2 and then factorise normally, multiplying the coefficient back in at the end? Aloha Nui means Goodbye. #4 20051218 06:25:30
Re: Factorising quadratics... againYes. It must work. IPBLE: Increasing Performance By Lowering Expectations. #5 20051218 06:26:55
Re: Factorising quadratics... again(what is the word for taking divisor out of brackets) > factorise Why did the vector cross the road? It wanted to be normal. #6 20051218 06:35:58
Re: Factorising quadratics... againYes, your method is truely better in a ways. IPBLE: Increasing Performance By Lowering Expectations. #7 20051218 06:36:01
Re: Factorising quadratics... againMerry Christmas Krassi and Mathsy Aloha Nui means Goodbye. #8 20051218 06:44:28
Re: Factorising quadratics... againMerry Christmas Ricky and Mathsy! Last edited by krassi_holmz (20051218 06:46:30) IPBLE: Increasing Performance By Lowering Expectations. #9 20051218 06:55:42
Re: Factorising quadratics... againThe picture of yours that I see has the hat on! Are you trying to change it back to your normal one? Aloha Nui means Goodbye. #10 20051218 07:06:37
Re: Factorising quadratics... againI'm still having a little trouble with this. Last edited by rickyoswaldiow (20051218 07:08:01) Aloha Nui means Goodbye. #11 20051218 07:16:56
Re: Factorising quadratics... again
Merry Christmas Ricky and krazzi! Why did the vector cross the road? It wanted to be normal. #12 20051218 07:20:53
Re: Factorising quadratics... againI'll try something... Last edited by krassi_holmz (20051218 07:22:30) IPBLE: Increasing Performance By Lowering Expectations. #13 20051218 07:41:32
Re: Factorising quadratics... againHere is it: IPBLE: Increasing Performance By Lowering Expectations. #14 20051218 07:48:36
Re: Factorising quadratics... againHere's another questionwhen the IPBLE: Increasing Performance By Lowering Expectations. #15 20060108 20:43:47
Re: Factorising quadratics... againIn my opinion the best way is to write the coefficient of x^2 in front of the brackets and then factorise expression in the brackets using completing the square or other method). I prefer completting the square as it always work very well and you do not have to wast your time for finding e * f = c etc. Work well with any question of factorizing quadratics, especially when you have to deal with questions such as 6a^2  23ab + 20b^2 #16 20060108 21:10:05
Re: Factorising quadratics... againAnother easy way is to use determinant(det) b^2  4ac #17 20060108 21:37:52
Re: Factorising quadratics... againThat's the easiest way in general, because it's guaranteed to work, you're just plugging into a formula every time. But this question asked you to factorise, so that method wouldn't be allowed. Why did the vector cross the road? It wanted to be normal. 