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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
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Quote:
 Originally Posted by Rishabh 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
I would not advise anyone to attach a^2 to 9a^2 or a to 3a.

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)

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In this case (p^2)(a^2)=9a^2, therefore p^2= 9, therefore p would be -3 or 3.

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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.

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If p=-3, q=5. If p=3, q=-5.

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Let's say p=3, so that q=-5.

(pa+qb)^2=(3a-5b)^2=(3a-5b)(3a-5b)=9a^2-15ab-15ab+25b^2

=9a^2-30ab+25b^2.

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Alternatively: Let's say p=-3 so that q=5.

(5b-3a)^2=(5b-3a)(5b-3a)=25b^2-15ab-15ab+9a^2=9a^2-30ab+25b^2

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Last edited by perfect_world; June 3rd, 2014 at 08:03 AM.

June 3rd, 2014, 08:06 AM   #5
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Hello, Kagiso!

Quote:
 What must be added to $9a^2- 30ab$ so that the resulting expression will be a perfect square?

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 $\displaystyle 9a^2-30ab$ 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 $\displaystyle (3a- cb)(3a-cb)$ Where c is whatever value is before the b, Multiply this out gives you $\displaystyle 9a^2-3acb-3acb+c^2 b^2$ Simplify gives $\displaystyle 9a^2-6acb+c^2 b^2$ 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 $\displaystyle 5^2 b^2=25b^2$ Thanks from perfect_world
 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. Thanks from Hoempa
June 3rd, 2014, 12:07 PM   #8
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Quote:
 Originally Posted by soroban Complete the square . . .
I thought of completing the square, but I wasn't sure about it...

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