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Is 64 a perfect square?
Is x^2 a square?
is it \[x^2 + 64?\]
What is the definition of a perfect square?
Hmm I say we teach her If she gets it, medal is yours I'm not here for the medal i'm here to get her to understand the question :)
takes the square 8 time 8 is 64
How do you get x^2 in the first place? What two variables do you multiply to get it?
Good! And what two numbers do you multiply to get 64? Obviously 8 right thanks to AA What do these two have in common x^2, 8^2
Good! Taking the square out. You get (x+8)^2 Expanding this: (x+8)^2..... Will that give you x^2 + 64?
Actually it would not (x+8)^2 gives you x^2 + 16x + 64 What does that tell you about factoring x^2 + 64?
It tells you it cannot be factored It is simply impossible Every time you have something like x^2 + P where P is perfect square You cannot factor it. Does that make sense?
There are no factors with real numbers. If you want complex number answer then (x + 8i)(x - 8i)
^That is exactly what the answer should be if you're doing a test
x^2 +64 = 0 And we get P(x)=x^2 +64 Now, we can look for the roots of P(x) using various Algorithm : Solve by Factorization x^2 +64 = 0 Separate : ( x^2 -8*i*x ) + ( 8*i*x +64 ) = 0 Commutative Law : ( x^2 8*i*x ) + ( -8*i*x +64 ) = 0 Distributive Law : x*( x 8*i ) + -8*i*( x 8*i ) = 0 Factor : ( x -8*i )*( x 8*i ) So the Polynomial have 2 roots : x1 = 8*i x2 = -8*i
You're right There is no answer for that
How about x^2 - 64 Do you think this is factorable?
Look again. What changed from x^2 + 64 and x^2 - 64?
Yep. And so that means there is a possibility that this is factorable. Going back to where we left off with the other one... we had (x+8). Putting a square outside would not make x^2 - 64
What two combinations make a negative? Think + and - and multiplication
Actually it doesn't matter which has the higher value in multiplication But you're right So to get x^2 - 64 What would you have to multiply (x+8) with?
do you have to multiply it by (x+8) again or (x-8)?
to check your answer Expand (x+8)(x-8) that should give you x^2 - 64
Notice how when you're combining like terms, there seems to be some cancelling out during expansion
The negative stuff was just a lesson you might learn later on For now, you want to stick with showing that you cannot factor the two expressions Show your work like AA showed his
Do you know the value of i?
nope i actually is \[i = \sqrt{x^{-1}}\]
Can you square root that?
use the FOIL method to remember how to multiply two binomials (a+b)(c+d) First: a*c Outer a*d Inner: b*c Last: b*d now add up to get ac+ad+bc+bd if you have (a+b)(c -d) treat it like (a+b)(c+ -d) and do the same thing First: a*c Outer: a*-d Inner: b*c Last: b* -d put it all together: ac -ad +bc -bd
OOOPS so \[\ i = sqrt{-1}\]
\[i = \sqrt{-1}\]
just apply the (a+b)^2 formula (a+b)^2=a^2+b^2+2ab then you will get it....
after appling the formula you will see i^2 replace i^2 with -1 becouse "i" means imaginary number... i^2=-1
Lol don't say that i equals to what i said up there if you square i, what is its value?
Yep! So go back to your question and solve it using this knowledge
(x+9i)2 i is -1 as said by @OrthodoxMan
is it (x + 9i)2 or (x + 9i)^2?
the second one^ Let's see if she solves it Pretty sure she is done
yep shes done i guess
Apply the square to the i as well
Remember what i^2 = ...
NO DONT GIVE UP LOOL Treat i as a variable Substitute -1 into i Hence you must do x^2 + 18(-1)
WHAT ON EARTH IS THIS?
MOVE AWAY BHUMI WE BANISH YOU
that was pretty rude...oh well
Because you're smart You need to keep practising In no time, you'll be better than us
If you stop complaining and let me help you I can go to sleep :P
lol sleep then i can try to help
Haha alright thanks man Make sure she finishes all of her questions
yep i will
oh,well shes offline -.-
-.- guess we can both get off goodnight
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