Question

quick question ... if I have the quadratic u^2-au+b^2 and the quadratic formula is ax^2+bx+c would a = u^2 b=-a and c = b^2?

Answer 1

Well, it depends, are we saying \(x=u\) or \(x=b\)?

Answer 2

If \(x=u\), then: \[
u^2-au+b^2 = (1)x^2+(-a)x+(b^2)
\]

Answer 3

well the standard quadratic formula is \[ax^2+bx+c\]

Answer 4

\[
a'=1\\b'=-a\\c'=b^2
\]

Answer 5

so yes x = u (sorry I know a = 1 I typo)

\[u^2-au+b^2\]

is what I have , so obviously by the definition \[a = 1, b = -1 ,c =b^2\]

Answer 6

ha ha ha my professor made a mistake XD

Answer 7

oh sorry b = -a

Answer 8

\[a = 1, b = -a ,c =b^2\]

Answer 9

so
\[b^2-4ac\]

\[(-a)^2 - 4(1)(b^2) \rightarrow a^2-4b^2\]

Answer 10

\[\frac{a + \sqrt{ a^2-4b^2}}{2}, \frac{a - \sqrt{ a^2-4b^2}}{2}\]

Answer 11

so does anyone know an online calculator that can calculate these bad boys to the 4th power?

Answer 12

Calculate what?

Answer 13

I have already calculated them to the 2nd power and already they are big... calculate the last thing I wrote in Latex

Answer 14

\[\frac{a + \sqrt{ a^2-4b^2}}{2}, \frac{a - \sqrt{ a^2-4b^2}}{2} \]

I need a calculator to calulate these two to the 4th power.

Answer 15

I already did\[(\frac{a + \sqrt{ a^2-4b^2}}{2})^2, (\frac{a - \sqrt{ a^2-4b^2}}{2})^2\] byhand and it was crazy

Answer 16

Oh

Answer 17

but I need \[(\frac{a + \sqrt{ a^2-4b^2}}{2})^4, (\frac{a - \sqrt{ a^2-4b^2}}{2})^2\] so is there a code in mathematica or an online calculator that can calculate this monster?

Answer 18

Just square it a second time

Answer 19

oh yeah and distribute like hell... that takes time :(

Answer 20

\[\frac{a^2+ 2a\sqrt{a^2-4b^2}+a^2-4b^2}{4}\]

Answer 21

\[(\frac{a^2+ 2a\sqrt{a^2-4b^2}+a^2-4b^2}{4})(\frac{a^2+ 2a\sqrt{a^2-4b^2}+a^2-4b^2}{4})\]

Answer 22

that's a beast T_T

Answer 23

that's like ok distribute a^2 all over now again with the square root la la la la la

Answer 24

\[(\frac{ 2a\sqrt{a^2-4b^2}+2a^2-4b^2}{4})(\frac{+ 2a\sqrt{a^2-4b^2}+2a^2-4b^2}{4})\]

Answer 25

\(a^2+a^2=2a^2\)

Answer 26

:/ this is a monsterrrrrr

Answer 27

anyone know a mathematica widget or something that can calculate this beast

Answer 28

Also it might help to do \(2a=\sqrt{4a^2}\)

Answer 29

is that legal?

Answer 30

Well, you don't have to do it.

Answer 31

You could use this to avoid foiling:\[
(a+b+c)^2 = a^2+b^2+c^2+2ab+2ac+2bc
\]

Answer 32

o-o

Answer 33

Another thing is binomial theorem.

Answer 34

\[(\frac{ 2a\sqrt{a^2-4b^2}+2a^2-4b^2}{4})\] with a=2a \sqrt{a^2-4b^2}
b = 2a^2
and c = 4b^2

Answer 35

\[(2a^2)^2 = 4a^4...(4b^2)^2 = 16a^4\]

Answer 36

\[
(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4
\]

Answer 37

\[(2a\sqrt{a^2-4b^2})^2 = 4a^2(a^2-4b^2) = 4a^4-16a^2b^2\]

Answer 38

Is life going well?

Answer 39

\[4a^4-16a^2b^2+4a^4+16b^4... 8a^4-16a^2b^2+a16b^4\]

Answer 40

\[4a^4-16a^2b^2+4a^4+16b^4... 8a^4-16a^2b^2+16b^4\]

Answer 41

like this?!

Answer 42

oh wait forgot the 16 from the denominator