Referring to the figure, evaluate the expression shown when a = 3, b = 7, c = 2 http://www.ppstest2.com/images2/AL9003.jpg please help i have just a few test left and i need to get them done and i dont have my math mook i really need help
it gave you the values of a,b,c just plug them into the equation and solve
i have to make sure im right because im failing math so im asking for help to see if i got the right anwsers
5
see thats not what i got
Referring to the figure, simplify the expression. http://www.ppstest2.com/images2/AL9004.jpg
\[\sqrt{b ^{2}-4ac}\]\[\sqrt{7^{2}-(4\times3\times2)}\]\[\sqrt{49-24}\]\[\sqrt{25}=5\]
\[\sqrt{45}=\sqrt{5\times9}=\sqrt{5}\sqrt{9}=3\sqrt{5}\]
This problem you have to split up the number into to multiples and one should be a perfect square
a perfect square is a number that is equal to a number times itself
this is helpful when simplifying radicals because it can be brought out in front
are you understanding?
im following
Do you see how i was able to split up the radical into two separate radicals multiplied together. This works only for multiplication and division. Do you know what the term radical is when I use it?
no im dyslexic so its hard me to read it and get it
sry. radical is the square root symbol. That is the name of it
ok
example how would you simplify sqrt(32)?
think about the numbers that make up 32 first
1,32 2,16 4,8
now look at these numbers and are there any perfect squares?
remember the definition of a perfect square it a number that can be multiplied by itself to equal the "perfect square"
1,32?
1 is a perfect square because 1*1=1, but if you use these numbers we are left with our original equation. There must be a better choice...
ok
4,8 is an option. 4 is a perfect square 2*2=4, but if we use these numbers we are left with \[2\sqrt{8}\]and that is not simplified all the way because 8 has perfect squares, 2 and 4. If you got to this point you could just simplify it again and you would get the correct answer. But the fastest way is to use 2,16. 4*4=16
I have to go to class. good luck. feel free to bring me into your future questions by doing this, @ChmE and I will be linked in.
kk thank you!
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