Question

Solve. 3x2 = 33x + 24!!??

Answer 1

3x^2 = 33x + 24

3x^2 - 33x - 24 = 0

Now use the quadratic formula to solve for x

Answer 2

The quadratic formula is

\[\Large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

and in this case

a = 3
b = -33
c = -24

Answer 3

hmmm i am really bad at thiss!?? :(

Answer 4

ok I'll take it a bit further

Answer 5

thank you so much :)

Answer 6

\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]


\[\Large x = \frac{-(-33)\pm\sqrt{(-33)^2-4(3)(-24)}}{2(3)}\]


\[\Large x = \frac{33\pm\sqrt{1089-(-288)}}{6}\]


\[\Large x = \frac{33\pm\sqrt{1089+288}}{6}\]


Let me know if you can take it from here or not

Answer 7

woww your good at this!! im still stuck though if you could just take it to the end for me i would be forever grateful :))))

Answer 8

Alright,


\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]


\[\Large x = \frac{-(-33)\pm\sqrt{(-33)^2-4(3)(-24)}}{2(3)}\]


\[\Large x = \frac{33\pm\sqrt{1089-(-288)}}{6}\]


\[\Large x = \frac{33\pm\sqrt{1089+288}}{6}\]


\[\Large x = \frac{33\pm\sqrt{1377}}{6}\]


\[\Large x = \frac{33+\sqrt{1377}}{6} \ \text{or} \ x = \frac{33-\sqrt{1377}}{6}\]


\[\Large x = \frac{33+9\sqrt{17}}{6} \ \text{or} \ x = \frac{33-9\sqrt{17}}{6}\]


\[\Large x = \frac{3(11+3\sqrt{17})}{6} \ \text{or} \ x = \frac{3(11-3\sqrt{17})}{6}\]


\[\Large x = \frac{11+3\sqrt{17}}{2} \ \text{or} \ x = \frac{11-3\sqrt{17}}{2} \]

Answer 9

The last line is one way to write the fully simplified exact answers

the approximate answers (after using a calculator) are

\[\Large x \approx 11.684658 \ \text{or} \ x \approx -0.684658\]

Answer 10

wowww thank you sooooo much!!! i am forever grateful now haha :))))))

Answer 11

lol glad to be of help

Answer 12

yess you are best math teacher on here!!!!

Answer 13

thanks lol

Answer 14

:))

Answer 15

ohhh noo we have a problem jim!!

Answer 16

there is no answer like that out of the choices\ :(

Answer 17

ok if possible show me the answer choices with a screenshot

Answer 18

okk :)

Answer 19

ahh turns out you answered this question for somebody else so its A!

Answer 20

so thanks again!

Answer 21

oh glad you figured it out, I'm betting the answer is probably in a different format or something

Answer 22

exactly! :)

Answer 23

ah probably something like

\[\Large x = \frac{11}{2}+\frac{3\sqrt{17}}{2} \ \text{or} \ x = \frac{11}{2}-\frac{3\sqrt{17}}{2} \]

Answer 24

yeap :) great job!!!!

Answer 25

hmm clever lol, hate when it gets that picky though

Answer 26

mee tooo im already confused enough haha

Answer 27

yeah it should be smart enough to know that

\[\Large x = \frac{11+3\sqrt{17}}{2} \ \text{or} \ x = \frac{11-3\sqrt{17}}{2} \]

is the same as

\[\Large x = \frac{11}{2}+\frac{3\sqrt{17}}{2} \ \text{or} \ x = \frac{11}{2}-\frac{3\sqrt{17}}{2} \]