## Simplification and solving of equation

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### Simplification and solving of equation

The problem are two:
(1) Simplify: 1/4(2n- 2n+2)

(2) Solve the equation 2(2x+1)- 9(2x) + 4 = 0
for the number one question, simplify:
1/4(2n- 2n+2)

1/4(2n- 2n+2)
= 4-1(2n-2n+2)
= 2(-2)(2n-2n+2)
= 2(-2)(2n-2n*22
I can't go further please I need asistance.

For the second question, solve the equation:
2(2x+1)- 9(2x) + 4 = 0

I don't just know how to start dealing with the question. It is really a troubsome question.
chikis
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### Re: Simplification and solving of equation

chikis,

For the first one, on your last step, you split $2^{n+2}$ into $2^{n}2^{2}$. You probably want to go in the other direction instead. Just distribute the coefficiet and combine the terms. It's not going to reduce down to just one term.

The second one's tricky 'cause you don't normally think of algebra problems with exponents involving the quadratic formula, right? But here's a hint: with a little manipulation, you can express that equation as $2{\left(2^x\right)}^2-9\left(2^x\right)+4=0$, which you can then rewrite as $2y^2-9y+4=0$ where $y{\equiv}2^x$.
Natural ChemE
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### Re: Simplification and solving of equation

Considering what you have posted so far, this what I have learnt:
(1) Simplify 1/4(2n-2n+2)

1/4(2n-2n+2)
= 2-2(2n-2n*22)
= 2n*2-2(1-22)
= 2n-2(1-4)
= 2n-2(-3)

(2) Solve the equation 2(2x+1)-9(2x)+4=0

2(2x+1)-9(2x)+4=0
---->22x*2-9(2x)+4=0
---->
2x(2)*2-9(2x) +4=0
Let 2x = y
2y2-9y+4=0
---->2y2-8y-y+4=0
---->2y(y-4)-1(y-4)=0
---->(2y-1)(y-4)=0
y= 1/2 or 4
but 2x=y
having that in mind
2x=1/2
---->2x=2-1
x = -1
2x=22
x = 2
therefore x = -1 or 2
Let me ask, I come across this type of equation next time and am ask to solve it, how do know that it is a quadratic equation without been told?
chikis
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### Re: Simplification and solving of equation

chikis,

Excellent work!

chikis wrote:Let me ask, I come across this type of equation next time and am ask to solve it, how do know that it is a quadratic equation without been told?

When you notice that $2^{2x+1}$ can be represented as $2\left(2^{x}\right)^{2}$, and that this gives you an $ax^2+bx+c=0$ form, that's probably the big red flag. Your mind's constantly doing little conversions and trying to see if the new configuration will fit some known form, so sometimes it's just a matter of being aware of what forms you can use and repeatedly trying 'til it comes to mind. While this process isn't deterministic, gaining enough practice will allow you to do it so quickly that you'll appear to have had some definite solution route to someone else observing you.

It helps if you first found, or at least estimated, the solutions numerically. In this case, you could do so by plotting $f\left(x\right)=2^{2x+1}-9\left(2^{x}\right)+4$ and seeing when it crosses the $x$-axis. If you did this, you may've found that $x{\in}\left{-1,2\right}$, and the whole $x$-having-two-solutions thing would've been a red flag to look for a quadratic form $ay^2+by+c=0$, which the equation already pretty much looked like (and probably written this way to make the question more accessible).

Since I'm trying to push the whole computational thing at my university, here are two quick ways to do this on the computer.

First, using Microsoft Excel. Excel is one of those programs that just about anyone in any technical discipline should have and know well. It's pretty much just computerized graphing paper, coming with all of the benefits that you'd expect computerized graphing paper to have. Additionally you can build entire programs on top of it, even complex ones.

It's common to search for the zeroes of a function (as mentioned above), so we plot $f\left(x\right)=2^{2x+1}-9\left(2^{x}\right)+4$:
This sort of approach is powerful and constantly used in practice.

Next, there's Wolfram Mathematica. Mathematica is more appropriate for Math-intensive fields, though it can also solve equations like the ones that you've given. In this case we'll just tell Mathematica to find the solutions:

As you can see, Mathematica just told us that $x{\in}\left{-1,2\right}$. In fact, it even showed the complex solutions $x{\in}\left{-1+\frac{2\mathrm{i}{\pi}\mathrm{C}\left[1\right]}{\mathrm{log}\left(2\right)},2+\frac{2\mathrm{i}{\pi}\mathrm{C}\left[1\right]}{\mathrm{log}\left(2\right)}\right}$ where (as it notes) $\mathrm{C}\left[1\right]$ refers to any integer. In other words, it's found two infinitely large sets of complex answers to this equation. And it barely took a moment to type in the equation.

Just to note it, while there are two infinite familes of solutions given to us by Mathematica, we can retrieve the only two real solutions by just using zero as our integer, $\mathrm{C}\left[1\right]=0$. This reduces the more complete answer $x{\in}\left{-1+\frac{2\mathrm{i}{\pi}\mathrm{C}\left[1\right]}{\mathrm{log}\left(2\right)},2+\frac{2\mathrm{i}{\pi}\mathrm{C}\left[1\right]}{\mathrm{log}\left(2\right)}\right}$ down to the one expected for class, $x{\in}\left{-1,2\right}$.
Natural ChemE
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### Re: Simplification and solving of equation

Thanks for the orientation!
chikis
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