Minuet Posted January 6, 2011 Report Share Posted January 6, 2011 This might sound like a silly/basic question to many, but I'm anything but bright in maths.As much as I try, and despite the fact that I've tried reading from various books, I cannot for the love of God understand perfect square factorisation.Apparently, there's an easy way of solving it with the GDC. Can someone please explain both methods, with and without GDC? Reply Link to post Share on other sites More sharing options...
The Economist Posted January 6, 2011 Report Share Posted January 6, 2011 you mean that, for instance, you do not understand the calculation of (x+5)2 or you mean the completion of square process? Reply Link to post Share on other sites More sharing options...
Minuet Posted January 6, 2011 Author Report Share Posted January 6, 2011 Nope, it's completion of square. The rule stated in my book is (a+b)^2=a^2+2ab+b^2 Reply Link to post Share on other sites More sharing options...
The Economist Posted January 6, 2011 Report Share Posted January 6, 2011 (edited) the rule that you state above is not the completion of square. Anyway, can you explain a bit more the problem you are facing? When completing the square by the way, it means that any quadratic expression y = ax^2 + bx + c can be expressed into the form y = a(x-k)^2 + Ibut from what I understand this is not your problem.If you have (x+5)^2 this can be written as x^2+2(5)x+5^2 = x^2 + 10x + 25. Is it the reverse that you don't get?*sorry if I'm not very helpful Edited January 6, 2011 by Jirashimosu 1 Reply Link to post Share on other sites More sharing options...
ILM Posted January 6, 2011 Report Share Posted January 6, 2011 The perfect square is in the form of:(ax(square)+2abx+b(square))By normal factorisation you will get(ax + b) squareexample:x (square) +2x + 1first of all look to the third term is it a square (or you can find the two dividends of it that can when added giv you the middle number)you know that 1= 1*1and you know that 1+1=2so you can make two brakets(x+1) (x+1)This can be written in a form of (x+1)squareLook for the third term and its dividendsNote if the third term is a square, it should not be a perfect square:Example :x(square)+ 10 x +1616= 4*4but in this case 4+4 =/10but 8+2 = 10Other perfect square root is when (ax-b) squarethen the exansion is ax(square) -2axb+b(square)I Do not know how to solve it by GDP, but i will search for Is that what you wantAsk me further questionIf you have an example in specific it will be better 1 Reply Link to post Share on other sites More sharing options...
The Economist Posted January 6, 2011 Report Share Posted January 6, 2011 obviously I have failed to understand that such extensive answer was needed.anyway, if you are using the casio calc (fx - 9860GII). you go to equation mode (from the main menu, you press A or else it is the square saying "EQUA axn+ ...= 0), then you press F2 (it is for polynomial eqs) and if it is a quadratic you press F1 (the respective button for a second degree eq). You want to have the equation written in its expanded form, for example: (x+2)^2 you first need to expand the square and write it in the form:x2 + 4x + 4 (it is the expansion of (x+2)2) and then at the first bracket (above which there is an "a") you write the coefficient of x2 which in this case is 1. Then by pressing the EXE button (at the bottom right of the calc), it moves you to a block above which there is the letter "b" and here you must write the coefficient of x, which in the above example is 4. Following the procedure explained above, you press again EXE and in the next block write the 4 (which is the final part of the equation). After this press again EXE and the calc will show you the double root the equation has (since the discriminant is 0) which is -2. 1 Reply Link to post Share on other sites More sharing options...
Minuet Posted January 6, 2011 Author Report Share Posted January 6, 2011 inm, nope, that's not my question.Jirashimosu, my apologies! The formula I gave is the wrong one; I got confused. The one you gave is the one I don't understand. For further elaboration, here's the worked solution they gave in the book:x^2 + 10x + 25= x^2 + (2)(x)(5) + 5^2= (x+5)^2wut? Reply Link to post Share on other sites More sharing options...
ILM Posted January 6, 2011 Report Share Posted January 6, 2011 (edited) Work it to understand it;(a+b)^2= (a+b) (a+b)then using simple foill methodeTo use foil you mulitiplyFirst symbol of each bracke= a.a= a^2Outer symbol= a. b= abthen inner=a.b=abthen last= b.b= b^2add them to get:a^2+ab+ab+b^2so it equals to:a^2+2ab+b^2 which is the same as what your book says:(a+b)^2=a^2+2ab+b^2 In the example:x^2 + 10x + 25= x^2 + (2)(x)(5) + 5^2= (x+5)^2You found that 25 is the squae of 5and 5+ 5= 10then you can write it as= x^2 + (2)(x)(5) + 5^2 (here a=x , b= 5)the you will change it from a^2+2ab+b^2 shape to (a+b)^2 (you know that a=x and b=5)so (x+5)^2is this what are searching for.. Edited January 6, 2011 by inm Reply Link to post Share on other sites More sharing options...
The Economist Posted January 6, 2011 Report Share Posted January 6, 2011 No worries but the formula you gave was correct. Assuming that you have the example: x2 + 10x + 25. Consider the formula (a + b)2 = a2 + 2ab + b2. You should be aware that it works in both ways which means that if you have a2 + 2ab + b2 it can be converted to (a+b)2.In this example, x^2 + 10x + 25, you should find which represents the factor "a" and which one the factor "b". The second step your book gives you is so that it is clearer for you which is "a" and which is b" as you can see from the initial formula, these two factors are squared and their product multiplied by two gives you 10. Once you understand which one is "a" and which one is "b", you can write it in the form (a + b)2 but instead of "a" writing "x" and instead of "b" writing 5.Better? Reply Link to post Share on other sites More sharing options...
Minuet Posted January 6, 2011 Author Report Share Posted January 6, 2011 I think I have a slight idea. Is it like...x^2 + 6x + 9= x^2 + 2x3 + 3^2...Um, now I'm lost. I don't really understand the next step. Did I even do that first step correctly? Reply Link to post Share on other sites More sharing options...
The Economist Posted January 6, 2011 Report Share Posted January 6, 2011 yes!! it is correct! so, in your example, which one is represented by "a" and which one by "b" (shown in the formula: (a + b)2 = a2 + 2ab + b2)? Reply Link to post Share on other sites More sharing options...
ILM Posted January 6, 2011 Report Share Posted January 6, 2011 You reached a good step, now you should ask you self what is:a and ba = is the first term which is under the squarein your case a=xexamples:4x^2= 2^2x^2 then a== 2xx^4= (x^2)^2 then a = x^2b= the square root of the third termif it is 9, then b=3if it is 16, then b= 4if it is y^2, then b=yin your question b=3so your result will(a+b)^2a=x, b= 3then your result(x+3)^2 Reply Link to post Share on other sites More sharing options...
Minuet Posted January 6, 2011 Author Report Share Posted January 6, 2011 Um, so...x^2 + 3x2 + 3^2 = (2 + 3)^2? That's it? I'm sorry but I don't really get it Can you explain more? Like...in detail; I don't really understand from just equations. Reply Link to post Share on other sites More sharing options...
ILM Posted January 6, 2011 Report Share Posted January 6, 2011 First of all)) how can you factorise any quadratic formula:example: x^2+ 7X+ 6You should look at third term and its dividends, in the example it is 6:6= 2.3=6.1you will choose the one in which when added give you the coffiecient of the second term (7)2+3=56+1= 7you will choose the second oneso open two brakets( ) ( )since you have x^2(x + ) (x+ )since 6 and 1, satisfy you formulait will be(x+1) (x+6)check it Using foil((x+1) (x+6)= x.x + 6.x + 1.x+ 1.6 first outer inner last(x+1) (x+6)= x^2+ 6x +1x +6 since you have 6x and 1 x with the same degreeyour final result =x^2+ 7x+6in perfect square you will have the same thin but the two brakets are equal to each otherto have (a+b) (a+b) = (a+b)^2Do you understand that Reply Link to post Share on other sites More sharing options...
Minuet Posted January 6, 2011 Author Report Share Posted January 6, 2011 Yes! Thank you very much just to make sure, this also applies for when x^2 has a coefficient, or no? Reply Link to post Share on other sites More sharing options...
ILM Posted January 6, 2011 Report Share Posted January 6, 2011 (edited) Yes, diiffently, it seems that you understand , to make sure solve this:1) x^2 + 20 x + 64and2) x^2 + 16 x + 64even if you do not know, i will find in which step you have a problem Edited January 6, 2011 by inm Reply Link to post Share on other sites More sharing options...
heyit'salison Posted April 24, 2011 Report Share Posted April 24, 2011 x^2 + 10x + 25.which is likea^2 + bx + cthe way i'd put it is take the square root of "c"which means, the square root of 25 = 5and now take the square root of "a"which in this case, "a" is 1x2so the root of 1x2 is 1x which is the same as xtherefore you have x and 5since the 5 is positive you put a + in front of it = + 5now combine (x + 5)and then all you have to do is square it.therefore you get (x + 5)2when you expand you will see (x + 5)(x + 5)which through normal expansion comes to x2 + 10x + 25thus i have proved my steps Reply Link to post Share on other sites More sharing options...
Guest JimmyK Posted April 24, 2011 Report Share Posted April 24, 2011 I think these two guidelines will help a lot (you may consider them tricks but they are not tricks)when you have a quadratic polynomial x^2 + bx + c, you can factorise it into (x + k) (x + n) by:1) looking at the bx. The bx is always the sum of the k and n2) c is always the multiplication of k and n.Thus, what you can do, is to look for all the possible values that can give k+n=b and all the values that can give kn=c. Then, choose the correct one (i.e, the one that matches in both operations). That's a quick way to solve factorisations and I always do like this 1 Reply Link to post Share on other sites More sharing options...
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