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June 3rd, 2014, 03:42 AM  #1 
Newbie Joined: Sep 2013 Posts: 6 Thanks: 0  quadratic equations
Good day everyone. How do we go about solving this problem: what must be added to $\displaystyle 9a^2 30ab$ so that the resulting expression will be a perfect square I checked the back of the book and found that the answer is $\displaystyle 25b^2$, but now I believe there's a proper way of reaching the answer. Is there some sort of formula for solving that? Last edited by skipjack; June 3rd, 2014 at 11:22 AM. 
June 3rd, 2014, 03:51 AM  #2 
Member Joined: May 2014 From: India Posts: 87 Thanks: 5 Math Focus: Abstract maths!  Answer
You might know that (a+b)^2 = a^2 + 2*a*b + b^2. Here a^2 = 9a^2, so a=3a. Here 2*a*b = 30ab, so b = 5b, since 30ab/2*3a. Therefore last term b^2 = (5b)^2 = 25b^2. This makes it a perfect square > (3a+5b)^2 
June 3rd, 2014, 07:19 AM  #3 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
You may want to use different letters for the identity then are being used in the excersize. (x+y)^2 = x^2 + 2xy + y^2, with x^2 + 2xy + y^2 a perfect square If a and b are integers, So, we have $\displaystyle x^2 + 2xy + y^2 = 9a^2  30ab + \text{something}$. someting is a constant c multiplied by b^2 i.e. c*b^2. In the constant factors of the terms of x^2 + 2xy + y^2, we see a ratio 1:2:1. Divide the constants in the terms by this ratio and you'll get a geometric sequence. 9/1, 30/2 = 15, c/1 = c. The third element in this sequence is such that c/15 = 15/9. i.e. c = (15)^2/9 = 25. So c*b^2 = 25 * b^2 must be added to 9a^2  30ab to get a perfect square for all integer a and b. 
June 3rd, 2014, 07:56 AM  #4  
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40  Quote:
This can lead to all sorts of problems, especially in examinations. It simply isn't correct. Your method marks would be thrown out of the window. (pa+qb)^2=(pa+qb)(pa+qb)=(p^2)*(a^2)+paqb+paqb+(q^ 2)*(b^2) =(p^2)(a^2)+2paqb+(q^2)(b^2)  In this case (p^2)(a^2)=9a^2, therefore p^2= 9, therefore p would be 3 or 3.  2pqab=30ab, therefore 2pq=30. 10p=30, if p is 3. 10p=30 if p=3. 2pq=10p, pq=5p, so q would be 5. 2pq=10p pq=5p, so q would be 5.  If p=3, q=5. If p=3, q=5.  Let's say p=3, so that q=5. (pa+qb)^2=(3a5b)^2=(3a5b)(3a5b)=9a^215ab15ab+25b^2 =9a^230ab+25b^2.  Alternatively: Let's say p=3 so that q=5. (5b3a)^2=(5b3a)(5b3a)=25b^215ab15ab+9a^2=9a^230ab+25b^2  Last edited by perfect_world; June 3rd, 2014 at 08:03 AM.  
June 3rd, 2014, 08:06 AM  #5  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, Kagiso! Quote:
Complete the square . . . $\;\;9a^2  30ab$ $\qquad \Rightarrow\;9\left(a^2  \tfrac{10}{3} ab\right)$ $\qquad \Rightarrow\;9\left(a^2  \tfrac{10}{3}ab \color{red}{+ \tfrac{25}{9}b^2}\right) $ $\qquad \Rightarrow\;9\left(a  \tfrac{5}{3}b\right)^2$ $\qquad \Rightarrow\;(3a  5)^2$ Therefore: $\:9(\tfrac{25}{9}b^2) \:=\:25b^2$ must be added.  
June 3rd, 2014, 08:34 AM  #6 
Member Joined: Feb 2012 From: Hastings, England Posts: 83 Thanks: 14 Math Focus: Problem Solving 
$\displaystyle 9a^230ab$ If it is a perfect square then you know you will have both a and b terms the same in both brackets, and as it is a minus it must be a term involving an a term minus a b term, all squared, hence $\displaystyle (3a cb)(3acb)$ Where c is whatever value is before the b, Multiply this out gives you $\displaystyle 9a^23acb3acb+c^2 b^2$ Simplify gives $\displaystyle 9a^26acb+c^2 b^2$ Now we know that 6c = 30 as we want 30 ab , hence c =5 Substitute that into our $\displaystyle c^2 b^2$ means we have $\displaystyle 5^2 b^2=25b^2$ 
June 3rd, 2014, 11:28 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2202 
If you add 30ab to the expression, the result is 9a², which is the square of 3a. That's just one of infinitely many valid answers. 
June 3rd, 2014, 12:07 PM  #8 
Newbie Joined: Sep 2013 Posts: 6 Thanks: 0  

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