Question

help please

Answer 1

Is this an medical emergency?

Answer 2

factor the following expression
\[x^2-15\]

Answer 3

make (root over 15)^2 and apply a^2-b^2

Answer 4

Well, whenever you have the subtraction of a variable and a number, you can always factor by a method of difference of squares. Now, it may not always look pretty when you do so, but it can be done. So of course we do have that condition, variable expression minus a constant. When you have that, the difference of squares formula is:
\[a^{2} - b^{2} = (a-b)(a+b)\]
So basically, you're taking the square root of each term, the x^2 and the 15, having one factor have a plus sign in the middle and the other have a minus side in the middle. If that makes sense.

Answer 5

i thought of something like \[(x-\sqrt{15 })(x+\sqrt{15)}\]

Answer 6

That's correct : )

Answer 7

oooh yay!! thanks

Answer 8

Good job : )

Answer 9

what about this one
\[2x^2+3x-1\]

Answer 10

this one got me stressed up

Answer 11

u hv to apply quadritic eq. formula..

Answer 12

oooh ok

Answer 13

So, you have a quadratic of the form ax^2 + bx + c with a not equal to 1. So the first thing you would want to check is if it's factorable under normal conditions or if we have to apply quadratic formula. Now, when a is not equal to 1, this would turn into a factoring by grouping problem (assuming it can be factored in the first place). So in order to check if it can be factored, we want to multiply a and c, in this case 2 and -1, to give us -2. If this equation is factorable by normal means, we should be able to find factors of -2 that add up to b, in case 3. Unfortunately, we cannot do this (but you should know how to check and at least start this factoring process).

So because we cannot do the steps above, we need quadratic formula, if you recall what that is.