My Math Forum (http://mymathforum.com/math-forums.php)
-   Algebra (http://mymathforum.com/algebra/)

 Kagiso June 3rd, 2014 03:42 AM

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?

 Rishabh June 3rd, 2014 03:51 AM

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

 Hoempa June 3rd, 2014 07:19 AM

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.

 perfect_world June 3rd, 2014 07:56 AM

Quote:
 Originally Posted by Rishabh (Post 196142) 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)

----------------------------------------------------------------------------------------

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

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

----------------------------------------------------------------------------------------

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

----------------------------------------------------------------------------------------

 soroban June 3rd, 2014 08:06 AM

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.

 nmenumber1 June 3rd, 2014 08:34 AM

$\displaystyle 9a^2-30ab$

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 $\displaystyle 5^2 b^2=25b^2$

 skipjack June 3rd, 2014 11:28 AM

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.

 Kagiso June 3rd, 2014 12:07 PM

Quote:
 Originally Posted by soroban (Post 196177) Complete the square . . .
I thought of completing the square, but I wasn't sure about it...

 All times are GMT -8. The time now is 05:47 AM.