what answers do I get if I have m^2 > 16 ?

miker2808

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I'm not that great with math.. and I'm on the very beginning to study this kind of math.. and I never had to work with detla >/=/< 0
I reached the answer where m^2-16>0
and I do not get it why in the answer I get m < -4 and m > 4 if anybody could explain why this is what I get? (I know about the plus minus when using square root)
 
I'm not that great with math.. and I'm on the very beginning to study this kind of math.. and I never had to work with detla >/=/< 0
I reached the answer where m^2-16>0
and I do not get it why in the answer I get m < -4 and m > 4 if anybody could explain why this is what I get? (I know about the plus minus when using square root)

So: \(\displaystyle m^2-16>0 \\ m^2 >16 \) . Since you have m to the power of 2, there are always two possible answers.

For example: \(\displaystyle b^2=4 \).
You should solve the task using the absolute value when you have variable to the second power or to any power that is an even number. So, you would solve it as follows: \(\displaystyle b^2=4 \\ |b|=2 \). I guess that you know when there is an absolute value the variable can assume the positive and the negative value, so: \(\displaystyle b_1=-2 \) and \(\displaystyle b_2=2 \) .

You could solve your task in the same way: \(\displaystyle m^2-16>0 \\ m^2 >16 \). Now, take out the square root, but you need to put absolute value since the exponent is an even number(2): \(\displaystyle |m|>4 \). Everything should be clear now. If \(\displaystyle m>4 \), it fits. And if \(\displaystyle m<-4 \), it fits because m is to the second power. If you take for example \(\displaystyle -5 \), then you would have \(\displaystyle |m|>4 \\ |-5|>4 \\ 5 >4 \) which is true.
Do you see where I'm going?

Or you can even solve it by sketching a graph. Assume that you have a function and that is \(\displaystyle f(m)=m^2-16 \).
Now you find m for which f(m)=0. \(\displaystyle f(m)=0 \\ m^2-16=0 \\ m^2 = 16 \\ m_1=-4 \, m_2=4 \). Now you draw only x-axis and note down the points for which f(m)=0. Since a=1 (a>0; a is a coefficient next to the member that has exponent 2), then parabola (it's opening) is turned upwards. Now you need to sketch that parabola. It is not important how narrow/wide it is, just that it crosses those points for which f(m)=0. Now y-axis represents values of a function. If function graph is over the x-axis then values are larger than zero( f(m)>0) and if the graph is under the x-axis, then values are less than zero ( f(m)<0 ). So, you look for m, when values of a function \(\displaystyle f(m)=m^2-16 \) are larger than zero: \(\displaystyle f(m)=m^2 -16 >0 \). And that happens when your graph is over the x-axis. And that is when x<-4 and x>4.



All should be clear. I am not sure what is confusing you further. If you place any number m< -4 in m^2-16, then m^2-16>0. As simple as that.
 
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