AmTran_Bus
  • AmTran_Bus
Quadratic formula
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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AmTran_Bus
  • AmTran_Bus
ganeshie8
  • ganeshie8
arrange it in standard form \(\large ax^2 + bx + c\)
AmTran_Bus
  • AmTran_Bus
I have it worked, but apparently I have something wrong.

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AmTran_Bus
  • AmTran_Bus
I got -H plus or minus 2 sqrt 6a
ganeshie8
  • ganeshie8
ohk lets see, \(\large ax^2 + bx + c = 0\) quadratic formula \(\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)
ganeshie8
  • ganeshie8
Given equation, \(\large 6x^2 + 12xh - A = 0\) \(a = 6\) \(b = 12h\) \(c = -A\)
ganeshie8
  • ganeshie8
substitute them in quadratic formula
ganeshie8
  • ganeshie8
\(\large x = \frac{-12h \pm \sqrt{(12h)^2 - 4(6)(-A)}}{2(6)}\)
ganeshie8
  • ganeshie8
does it look rihgt, so far ?
AmTran_Bus
  • AmTran_Bus
Right. I get -12H\[\pm \sqrt{144-4(-A)(6)}\] all over 12
ganeshie8
  • ganeshie8
\(\large x = \frac{-12h \pm \sqrt{(12h)^2 - 4(6)(-A)}}{2(6)}\) \(\large x = \frac{-12h \pm \sqrt{(12h)^2 + 24A)}}{12}\)
ganeshie8
  • ganeshie8
\(\large x = \frac{-12h \pm \sqrt{(12h)^2 - 4(6)(-A)}}{2(6)}\) \(\large x = \frac{-12h \pm \sqrt{(12h)^2 + 24A)}}{12}\) \(\large x = -h \pm \frac{\sqrt{(12h)^2 + 24A)}}{12}\)
AmTran_Bus
  • AmTran_Bus
\[-12H \pm12\sqrt{24A} \]
AmTran_Bus
  • AmTran_Bus
over 12?
AmTran_Bus
  • AmTran_Bus
So -H plus or minus...???
ganeshie8
  • ganeshie8
yes, and the thing inside squareroot, we can t do much as both are not like terms
AmTran_Bus
  • AmTran_Bus
So would it be \[-H \pm 12\sqrt{24A}\]
AmTran_Bus
  • AmTran_Bus
Or would that 12 cancel also?
ganeshie8
  • ganeshie8
nopes. its a + in between. so we cant pill that 144 out here is the thing :- \(\sqrt{a+b} \ne \sqrt{a} \sqrt{b}\)
ganeshie8
  • ganeshie8
\(\large x = \frac{-12h \pm \sqrt{(12h)^2 - 4(6)(-A)}}{2(6)}\) \(\large x = \frac{-12h \pm \sqrt{(12h)^2 + 24A}}{12}\) \(\large x = -h \pm \frac{\sqrt{(12h)^2 + 24A}}{12}\) \(\large x = -h \pm \frac{\sqrt{144h^2 + 24A}}{12}\)
ganeshie8
  • ganeshie8
it cant be simplified further
AmTran_Bus
  • AmTran_Bus
Oh. I forgot the H was paired with the 12. Thats what I was dong wrong. Your a grade saver!
ganeshie8
  • ganeshie8
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.
AmTran_Bus
  • AmTran_Bus
Humm. Do I need to at least take the ^2 out?
ganeshie8
  • ganeshie8
u forgot the 12 in the bottom of radical
AmTran_Bus
  • AmTran_Bus
ganeshie8
  • ganeshie8
smaller value : \(\large x = -h - \frac{\sqrt{144h^2 + A}}{12}\) larger value : \(\large x = -h + \frac{\sqrt{144h^2 + A}}{12}\)
AmTran_Bus
  • AmTran_Bus
Oh. The 24.
AmTran_Bus
  • AmTran_Bus
Why is the 24 all gone again?
ganeshie8
  • ganeshie8
smaller value : \(\large x = -h - \frac{\sqrt{144h^2 + 24A}}{12}\) larger value : \(\large x = -h + \frac{\sqrt{144h^2 + 24A}}{12}\)
ganeshie8
  • ganeshie8
that was a typo sorry. make the A capital... looks ur system is case sensitive
AmTran_Bus
  • AmTran_Bus
In the help video for this problem here is what they did to a slightly different problem. Look how they simplified the discriminant.
ganeshie8
  • ganeshie8
yeah if u wanto simplify u can pull out 4 max thats all
ganeshie8
  • ganeshie8
\(\large x = -h \pm \frac{\sqrt{4 \times 36 h^2 + 4x6A}}{12}\)
ganeshie8
  • ganeshie8
\(\large x = -h \pm 2\frac{\sqrt{36 h^2 + 6A}}{12}\)
ganeshie8
  • ganeshie8
\(\large x = -h \pm \frac{\sqrt{36 h^2 + 6A}}{6}\)
AmTran_Bus
  • AmTran_Bus
One final ?. Why is the max 4 and not 12?
ganeshie8
  • ganeshie8
cuz, 4 is the perfect square, when it comes out, it becomes 2
AmTran_Bus
  • AmTran_Bus
Thanks.
ganeshie8
  • ganeshie8
is the submission successful ?
AmTran_Bus
  • AmTran_Bus
One sec
ganeshie8
  • ganeshie8
\(\large x = -h \pm \frac{\sqrt{36 h^2 + 6A}}{6}\) is same as, \(\large x = -h \pm \sqrt{h^2 + \frac{A}{6}}\)
ganeshie8
  • ganeshie8
one of them should work
AmTran_Bus
  • AmTran_Bus
It worked, the 1st one!!!
ganeshie8
  • ganeshie8
wow ! finally !! :)
AmTran_Bus
  • AmTran_Bus
Thanks again.
ganeshie8
  • ganeshie8
np :D

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