Question

hard Calculus I Question!

Answer 1

Oh snap!

Answer 2

If \[x _{1}, x _{2}\] are the solutions of the equation \[8^{3x ^{2}}=\frac{ 1 }{ 64^{5x+3} }\] Compute the value of \[\left| x _{1}-x _{2} \right|\]

Answer 3

first determine the quadratic
which may be some waht
2x^2 =-2*(5x+3)

Answer 4

(x1 -x2)^2 =(x1+x2)^2 -4x1.x2

Answer 5

@nincompoop

Answer 6

@matricked the first thing you wrote: 2x^2=-2(5x+3)
where did you get that?

Answer 7

I understand the 5x+3, but I don't know where you got 2x^2=-2

Answer 8

see
64^(5x+3)=(8^2)^(5x+3) =8^(2(5x+3))
hence
1/( 8^(2(5x+3) ) = 8^( - 2(5x+3))
hope u get it now

Answer 9

@matricked why did you want to make the exponent negative?

Answer 10

It's in the denominator.

Did we start with 2x^2 or 3x^2?

Answer 11

@tkhunny as the letters are small it s bit confused whether its 2x^2 or 3x^2

Answer 12

Hey, can't we simplify the 8 into 2^3?

Answer 13

@pancakeslover
as 1/ (a^n) =a^(-n)

Answer 14

@Yttrium
yup we can
but it will make us calculate/simplify more

Answer 15

Asolutely NO!!!! \(2^{3}\) is NOT a simplification of \(8\). Words mean things. Deliberately making it more complicated cannot be simplified.

Answer 16

\[8^{3x ^{2}}=8^{-10x-6}\] Am I on the right track?

Answer 17

so if they both have the same base, does that mean
\[3x ^{2}=-10x-6\]

Answer 18

yes you are. :))

Answer 19

great! can you guys tell me what to do next?

Answer 20

Solve for x

Answer 21

@Yttrium stick around so I can check with you

Answer 22

@Yttrium so do I use the quadratic formula for this equation? \[3x ^{2}+10x+6\]

Answer 23

Yes you can. :)

Answer 24

so now I have \[\frac{ -10\pm \sqrt{28} }{ 6 }\]

Answer 25

@Yttrium I need to compute the value of \[\left| x _{1}-x _{2} \right|\]

Answer 26

Yes, you're doing it right. :))

Answer 27

haha thanks. what do I do next?

Answer 28

You already have the value of your \[x _{_{1}} and x _{2}\], right? then do the arithmetic. :) You're approaching the final answer.

Answer 29

so once I simplify I get \[\frac{ -5-\sqrt{7} }{ 3 }\] and \[\frac{ -5+\sqrt{7} }{ 3 }\]

Answer 30

@Yttrium I don't know what to do next? how do I know which one is \[x _{1}\]and \[x _{2}\]

Answer 31

You can use your any of them since we are dealing with absolute values. Waht ever the x1 and x2, you will arrive at the same answer.

Answer 32

ooh right!! thank you!

Answer 33

@Yttrium is the answer \[\frac{ 10\sqrt{7} }{ 3 }\] ?

Answer 34

@Yttrium wait!

Answer 35

I did that wrong..I added

Answer 36

alright..nevermind I'm still a little confused

Answer 37

What's your final answer, then?

Answer 38

@Yttrium \[\left| \frac{ -5+\sqrt{7} }{ 3 }-\frac{ -5-\sqrt{7} }{ 3 } \right|\]

Answer 39

does it all become positive? sorry I have a little trouble with absolute value

Answer 40

Yes, because you are dealing a fraction with common denominator. It's like \[\frac{ (-5+\sqrt{7}) }{ 3 } - \frac{ (-5-\sqrt{7}) }{ 3 }\]

Answer 41

@Yttrium thank you for all your help! i got the answer and it's right! you're awesome!

Answer 42

No problem. Just post question whenever you get confused again. :))