i cant even explain what it is... heres one...
b(squared)-48
helppp
If you are solving for b:
b^2 - 48 = 0 (set equal to 0)
b^2 = 48 (drag 48 over)
b = 7 (take square root)
This is if you're solving for b. You didn't really tell what to solve for. But this is just what I'm guessing the question is.
Hope that helps!
Ethan
There's nothing other than b so I'd guess that's what you have to solve for. Usually when you have a question like this it would say
b+2 = 4
or b(squared)=9
There's no equal sign in your question, so EthanDavis just assumed that it was equal to zero. Then he just did the algebra steps, moved over 48 to the other side and changed sign, and then solved.
However....the square root of 48 is not 7, the square root of 49 is seven. The square root of 48 is 6.93.....
my teacher says this is called factoring trinomials... & ther isnt supposed to be a sign... ;[ im sosoo lost can you exlpain please? ok, so the teacher also said something like you use FOIL and the distrubutive property & factor . i dont know.. IM SO CONFUSED ;[[
Oh, you're supposed to factor it...well, that would have been helpful..
There's a step by step on using that method on this website..
Intermediate Algebra Tutorial on Factoring Trinomials
But are you sure it's not b^2-49?
That would be (n+4)(n-4)
Just think of this form:
$\displaystyle ax^2+bx+c$
So in that form, your question $\displaystyle n^2-16$, it's like this
$\displaystyle (1)x^2+(0)x+(-16)$
A, b and c are just constants..x is the variable (n in your question)
You want something in this form...
( ..... )( ..... )
The first terms in each bracket need to multiply to form the first term in what you started with...so if you have n(squared)-16, you know that n times n equals n(squared), so now you have this
(n ..... )(n ..... )
And then you need to look for two numbers that multiply to form "c", but add together to form b. c is -16, and b is 0 in your example. So 4 and -4 fit the circumstances...So then you have
(n+4)(n-4)
With these consider your standard form $\displaystyle ax^2+bx+c$
In this case where there is no "b" term you are going to end up with the form (x+some number)*(x-some number)
So in the case $\displaystyle n^2-16$ consider the number 16, can you take the square root and get a whole number? In this case it would be four, so you can factor it thus,
$\displaystyle (n+4)(n-4)$
You can test this result by foiling it back out and it should look the same.
Have you learned the quadratic formula? If you can't do it simply, then you can use the quadratic formula.
$\displaystyle \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$
But you can't do that here because b^2-4ac gives you a negative value, which you can't take the square root of..
..so there's no roots for this...