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Ooooh. Difference of two squares
I will let @Zedditup explain because everyone should get a turn to help.
Oh. Oh ok lol
Okay, so this is similar to the previous problem. The only thing is that this is MUCH easier just in the process. I don't know if there's a long process, but the way I was taught was just a few seconds of thinking and you have your answer. Basically, as there is no number, or x, between them, it's counted as 0. So what two numbers can multiply to get -100 and add together to make 0?
-50 and 2 and 0+0=0
well its easier than that difference of squares \[\huge\rm a^2 -b^2 =(a+b)(a-b)\] take square root of first and 2nd term write ur answer in two parenthses (sqrt of 1st term + sqrt of 2nd term) (sqt of first term - sqrt of 2nd term)
No. Go about it a different way. Think about the last problem. -8 and positive 3 multiplied to -24 and added to -5, which were the numbers in the original equation. What can get you multiplied to -100, but added to 0?
using the property as Nnesha suggested.
>_> umm ok thank you all
OS is lagging for me
Wait Darkmoon, did you understand it? If not, just stick with me lol
@ambujverma Please do not hand out direct answers and refer to @Zedditup's response. That is rude as @Zedditup was trying to explain step by step it is also against the COC. Please do not do it again.
Eh yeah I want the explanation
Alright, so where we left off: Similar to the last problem, -8 and 3 multiplied together got you -24. They added together and got you -5. This problem is just the same. Think of it as x^2+0-100, though. What two numbers, when multiplied together, form -100 and, when added, give you 0?
Ok wait so 10 times negative 10 AHHH I see what they did now.....
Ok I understand....
Alright. So what did you get for your answer?
Well I thought back to the old problem and how the answer for it was just -8 and 3 because if you multiply those things happen so I thought and at first I thought it was 2 times -50 but then I thought about 10 times 10 is also 100 so then I got (x-10) (x+10)
That is correct ;)