Question

Quadratic formula

Answer 1

arrange it in standard form \(\large ax^2 + bx + c\)

Answer 2

I have it worked, but apparently I have something wrong.

Answer 3

I got -H plus or minus 2 sqrt 6a

Answer 4

ohk
lets see,

\(\large ax^2 + bx + c = 0\)
quadratic formula

\(\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

Answer 5

Given equation,

\(\large 6x^2 + 12xh - A = 0\)

\(a = 6\)
\(b = 12h\)
\(c = -A\)

Answer 6

substitute them in quadratic formula

Answer 7

\(\large x = \frac{-12h \pm \sqrt{(12h)^2 - 4(6)(-A)}}{2(6)}\)

Answer 8

does it look rihgt, so far ?

Answer 9

Right. I get -12H\[\pm \sqrt{144-4(-A)(6)}\] all over 12

Answer 10

\(\large x = \frac{-12h \pm \sqrt{(12h)^2 - 4(6)(-A)}}{2(6)}\)

\(\large x = \frac{-12h \pm \sqrt{(12h)^2 + 24A)}}{12}\)

Answer 11

\(\large x = \frac{-12h \pm \sqrt{(12h)^2 - 4(6)(-A)}}{2(6)}\)

\(\large x = \frac{-12h \pm \sqrt{(12h)^2 + 24A)}}{12}\)

\(\large x = -h \pm \frac{\sqrt{(12h)^2 + 24A)}}{12}\)

Answer 12

\[-12H \pm12\sqrt{24A} \]

Answer 13

over 12?

Answer 14

So -H plus or minus...???

Answer 15

yes, and the thing inside squareroot, we can t do much as both are not like terms

Answer 16

So would it be \[-H \pm 12\sqrt{24A}\]

Answer 17

Or would that 12 cancel also?

Answer 18

nopes. its a + in between. so we cant pill that 144 out
here is the thing :-
\(\sqrt{a+b} \ne \sqrt{a} \sqrt{b}\)

Answer 19

\(\large x = \frac{-12h \pm \sqrt{(12h)^2 - 4(6)(-A)}}{2(6)}\)

\(\large x = \frac{-12h \pm \sqrt{(12h)^2 + 24A}}{12}\)

\(\large x = -h \pm \frac{\sqrt{(12h)^2 + 24A}}{12}\)

\(\large x = -h \pm \frac{\sqrt{144h^2 + 24A}}{12}\)

Answer 20

it cant be simplified further

Answer 21

Oh. I forgot the H was paired with the 12. Thats what I was dong wrong. Your a grade saver!

Answer 22

np :) good if u want u can take the bottom 12 inside, it wud become 144 and simplify - but its good the way it is now.

Answer 23

Humm. Do I need to at least take the ^2 out?

Answer 24

u forgot the 12 in the bottom of radical

Answer 25

smaller value : \(\large x = -h - \frac{\sqrt{144h^2 + A}}{12}\)


larger value : \(\large x = -h + \frac{\sqrt{144h^2 + A}}{12}\)

Answer 26

Oh. The 24.

Answer 27

Why is the 24 all gone again?

Answer 28

smaller value : \(\large x = -h - \frac{\sqrt{144h^2 + 24A}}{12}\)


larger value : \(\large x = -h + \frac{\sqrt{144h^2 + 24A}}{12}\)

Answer 29

that was a typo sorry. make the A capital... looks ur system is case sensitive

Answer 30

In the help video for this problem here is what they did to a slightly different problem. Look how they simplified the discriminant.

Answer 31

yeah if u wanto simplify u can pull out 4 max thats all

Answer 32

\(\large x = -h \pm \frac{\sqrt{4 \times 36 h^2 + 4x6A}}{12}\)

Answer 33

\(\large x = -h \pm 2\frac{\sqrt{36 h^2 + 6A}}{12}\)

Answer 34

\(\large x = -h \pm \frac{\sqrt{36 h^2 + 6A}}{6}\)

Answer 35

One final ?. Why is the max 4 and not 12?

Answer 36

cuz, 4 is the perfect square, when it comes out, it becomes 2

Answer 37

Thanks.

Answer 38

is the submission successful ?

Answer 39

One sec

Answer 40

\(\large x = -h \pm \frac{\sqrt{36 h^2 + 6A}}{6}\)

is same as,

\(\large x = -h \pm \sqrt{h^2 + \frac{A}{6}}\)

Answer 41

one of them should work

Answer 42

It worked, the 1st one!!!

Answer 43

wow ! finally !! :)

Answer 44

Thanks again.

Answer 45

np :D