Online Course Discussion Forum
Math Challenge IC Handout 7
How do I do: (Including my steps and the step i got stuck on)
Question 7.22: 3x‾‾√−9x‾‾√=−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: 2x‾‾√−2x−10‾‾‾‾‾‾‾√=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 rightQuestion 7.28: x−4‾‾‾‾‾√+|x2−16|=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:
2×4x‾‾‾‾‾‾√y=2×4x= 2^(x+0.5)
2*4^x=2^(2x+1)
2x+
2x+0.52x+0.5
x+0.5
Thanks so much, I really appreciate it.
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 :)
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