# Math Help - Perfect Square Trinomial

Find two values of b that will make $$9x^2+bx+25[tex] a perfect square trinomial. The formulas for Perfect Square Trinomial are: [tex](a+b)^2=a^2+2ab+b^2[tex] [tex](a-b)^2=a^2-2ab+b^2[tex] As I solved: [tex]9x^2+bx+25[tex] [tex](3x)^2+2(3x)(5)+5^2[tex] [tex]9x^2+6x(5)+25[tex] [tex]9x^2+30x+25[tex] Then...according to this one of those two values would be 30 but I'm not sure about it for I didn't find anything. I just replaced the given values [tex]9x^2[tex] and [tex]25[tex] into the middle term of the formula. And I couldn't find the other number. 2. Originally Posted by Alienis Back Find two values of b that will make [tex]9x^2+bx+25[tex] a perfect square trinomial. The formulas for Perfect Square Trinomial are: [tex](a+b)^2=a^2+2ab+b^2[tex] [tex](a-b)^2=a^2-2ab+b^2[tex] As I solved: [tex]9x^2+bx+25[tex] [tex](3x)^2+2(3x)(5)+5^2[tex] [tex]9x^2+6x(5)+25[tex] [tex]9x^2+30x+25[tex] Then...according to this one of those two values would be 30 but I'm not sure about it for I didn't find anything. I just replaced the given values [tex]9x^2[tex] and [tex]25[tex] into the middle term of the formula. And I couldn't find the other number. (3x + 5)^2 = ...... So b = 30 is correct. (3x - 5)^2 = ..... So b = ....... 3. ## Just one more question Why didn't it show the problem without the tags...mmm...I mean this "[tex]" things. I'd like to learn how to write problems using them. Is there any thread on the subject? 4. Originally Posted by Alienis Back Why didn't it show the problem without the tags...mmm...I mean this "[tex]" things. I'd like to learn how to write problems using them. Is there any thread on the subject? $(a+b)^2=a^2+2ab+b^2$ $(a-b)^2=a^2-2ab+b^2$ come from [tex](a+b)^2=a^2+2ab+b^2$$ and $$(a-b)^2=a^2-2ab+b^2$$.

See http://www.mathhelpforum.com/math-help/latex-help/

5. ## Let's see if it works...

Find two values of b that will make $9x^2+bx+25$ a perfect square trinomial.

The formulas for Perfect Square Trinomial are:
$(a+b)^2=a^2+2ab+b^2$
$(a-b)^2=a^2-2ab+b^2$

As I solved:

$9x^2+bx+25$
$(3x)^2+2(3x)(5)+5^2$
$9x^2+6x(5)+25$
$9x^2+30x+25$

Then...according to this one of those two values would be 30 but I'm not sure about it for I didn't find anything. I just replaced the given values $9x^2$ and $25$ into the middle term of the formula. And I couldn't find the other number.

6. WOW...That's Great!!