synthesisFR
Well-Known Member
their method
"it's intuitive"like they say "this formula should not be memorised" and proceed to do those examples using that formula
da hell bruv
U CANT JUST SAY THATuse the discriminant?
BRO IM NOT WALTER WHITE CHILLU CANT JUST SAY THAT
JUST GET CANCER OKAY
^anyways im gonna go to sleep
@carrotsss ill leave this with you then
Basically you just solve it like a normal quadratic, except what is in the root is a complex number, so you will need to find the root of the complex number to determine it. To determine this root, you can either complete the square (extremely elegant and fun, but unnecessary) or find a result by equating real and imaginary parts in (x+iy)^2=discriminant. Then you equate real and imaginary parts, and from there it’s just algebra to find x and y, after which you can substitute in (x+iy)^2, which can be rooted to finally determine the value of z. You can also convert the discriminant to exponential form to make the root easier but that’s not always an option if it’s not a round argument obviouslycan someone do those 2 examples normally so i know what to do
when you can do q16's but not solve a quadratic equation bc ur kinda messed up af like me
yeah idk what ur ona boutBasically you just solve it like a normal quadratic, except what is in the root is a complex number, so you will need to find the root of the complex number to determine it. To determine this root, you can either complete the square (extremely elegant and fun, but unnecessary) or find a result by equating real and imaginary parts in (x+iy)^2=discriminant. Then you equate real and imaginary parts, and from there it’s just algebra to find x and y, after which you can substitute in (x+iy)^2, which can be rooted to finally determine the value of z. You can also convert the discriminant to exponential form to make the root easier but that’s not always an option if it’s not a round argument obviously
but why do we use the discriminantBasically you just solve it like a normal quadratic, except what is in the root is a complex number, so you will need to find the root of the complex number to determine it. To determine this root, you can either complete the square (extremely elegant and fun, but unnecessary) or find a result by equating real and imaginary parts in (x+iy)^2=discriminant. Then you equate real and imaginary parts, and from there it’s just algebra to find x and y, after which you can substitute in (x+iy)^2, which can be rooted to finally determine the value of z. You can also convert the discriminant to exponential form to make the root easier but that’s not always an option if it’s not a round argument obviously
I’ll send you an example tomorrow I need to sleep rn sorryyeah idk what ur ona bout
the discriminant is what’s being rooted, and we need to find a way to root the complex numberbut why do we use the discriminant
I SACRIFIEFD MY LIFE FOR U AND THIS IS HO W UREPAYI’ll send you an example tomorrow I need to sleep rn sorry
I typed my solution to your original question in post #13 at #13 .do those 2 examples
Thanks !I typed my solution to your original question in post #13 at #13 .
I did the other 2 examples but using my method instead of the method in Cambridge.
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