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June 3rd, 2014, 03:42 AM   #1
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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.
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June 3rd, 2014, 03:51 AM   #2
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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
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June 3rd, 2014, 07:19 AM   #3
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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.
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June 3rd, 2014, 07:56 AM   #4
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Quote:
Originally Posted by Rishabh View Post
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.
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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.
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June 3rd, 2014, 08:34 AM   #6
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$\displaystyle 9a^2-30ab$

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 $\displaystyle 5^2 b^2=25b^2$
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June 3rd, 2014, 11:28 AM   #7
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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.
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June 3rd, 2014, 12:07 PM   #8
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Quote:
Originally Posted by soroban View Post

Complete the square . . .
I thought of completing the square, but I wasn't sure about it...
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