Use completing of the square to find d:\[x^2-dx+5\]
you want to turn this into a perfect square trinomial?
Well, actually I just made that up... but I wanted to know how you would be able to find dx of a polynomial when given ax^2-bx+c, ax^2 and c
oh nice question...you know that if you are given ax^2 + bx + c then (b/2)^2 = c right?
Yes and thank you.
so this time your b is d and c is 5 so \[\large (\frac{d}{2})^2 = 5\] \[\large \frac{d}{2} = \sqrt 5\] \[\large d = 2\sqrt 5\] got it?
Yup...
But can you do it on an hard example with integers or whole numbers...
\[2x^2 - bx + 32\] try it :)
\[4\sqrt{8}\]
nope! :p you should get a whole number ;)
b is still irrational :(
remember in completing the square method it should be in the form x^2 + bx + c? that means you divide everything by the coefficient of x first
hmm I see...
tricky right ;)
yes very... and I am stuck, haha
you divide by 2 first that will give you x^2 - bx + 16 <--perfect square
I don't divide b by 2?
wait that should be x^2 - b/2 x +16 lol sorry
Okay so b is 16
yep
you got it!
cool... Thanks @lgbasallote
\[\Huge \color{maroon}{\mathtt{\text{<tips hat>}}}\]
Lol...[English accent] Why, Thank you fine sir...
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