Online Course Discussion Forum

Math Challenge IC Handout 7

 
 
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Math Challenge IC Handout 7
by Christina Peng - Thursday, 26 October 2017, 10:19 PM
 

How do I do: (Including my steps and the step i got stuck on)

Question 7.22: 3x9x=63x−9x=−6.

sqrt(3x)=-6+3sqrt(x)

3x=36-36(sqrt x)+9x

x=12-12(sqrt x)+3x

12(sqrt x)=12+2x

6(sqrt x)=6+x

sqrt x=1+x/6

x=1+x/3+x/18

Question 7.26: 2x2x10=22x−2x−10=2.

sqrt 2x= sqrt(2x-10)+2

2x=2x-10+4*sqrt(2x-10)+4

-4*sqrt(2x-10)=-6

sqrt(2x-10)=3/2

2x-10=9/4

2x=12/4=3

x=1.5

I checked my answer and something just isn't right

Question 7.28: x4+|x216|=0x−4+|x2−16|=0
sqrt (x-4)=-x^2+16

x-4=-x^2+16

What are you supposed to do with this absolute value?

Question 7.29:

2×4xy=2×4x= 2^(x+0.5)

2*4^x=2^(2x+1)

2
x+

2x+0.52x+0.5

x+0.5

P.S problem 7.9 along with 7.4 and many others do not display the writing for some reason. I don't know if it's just me.


Thanks so much, I really appreciate it.

 
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Re: Math Challenge IC Handout 7
by John Lensmire - Monday, 30 October 2017, 5:45 PM
 

Sorry for not getting back to you sooner Christina.

For 7.22, note we can rewrite the equation as $\sqrt{3}\times \sqrt{x} - 3 \sqrt{x} = -6$, so we can first solve for $\sqrt{x}$.

For 7.26, you are on the right track. Double check your arithmetic after $2x-10 = 9/4$.

For 7.28, step back a second and think about what values are possible for $\sqrt{x+4}$. Similarly what type of values would you get for $|x^2-16|$. Then, how could these possibilities add up to $0$?

For 7.29, the idea is to rewrite both as powers of $2$, so I think you are on the right track.  If you need to, try to graph them online and see what you get. It may be somewhat of a trick question :)

Picture of Christina Peng
Re: Math Challenge IC Handout 7
by Christina Peng - Monday, 30 October 2017, 7:09 PM
 

For 7.29 I am still stuck. I feel like i can't simplify any further.

Thanks for the explanation, the rest of the problems are clearer.