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What makes you think it's wrong shaniab29544?  

heureka has given the correct real number root (there are two other roots, but they are complex).

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it wrong

asin(x) asks for the angle whose sine is x.  Since the sine function only takes values between ±1, asin of anything bigger than 1 doesn't exist.

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@@ End of Day Wrap   Fri 15/5/15   Sydney, Australia Time   11:59pm  ♪ ♫

Hi all,

Another busy day       Let me see, who gave fabulous answers today?   Mmm

Zacismyname, Dbg4thebest, CPhill, EinsteinJr, Cionamzuk, Alan, Geno3141, Headingnorth, Silver27, DarkBlaze347, Gibsonj338, Heureka, Fiora and JimBob.  Yes, you all gave wonderful answer.  Thank you. 

Watch this !

Non-reducible square roots (like √2 and √3) are irrational. 

If we can find a non-reducible square root between -9 and -10, we can use that for our answer (and there is an infinite number of them!).

√81 = 9

√100 = 10

So, √83 is an irrational number between 9 and 10

So, -√83 is an irrational number between -9 and -10.

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Factor the 2^x from each term on the left-hand side:

2^x(x² - 1) = 0

The only ways the left-hand side can be 0 is if x = 1 or x = -1 or x = -∞

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Add 3x to both sides

1 ≤ 7 + 3x

Subtract 7 from both sides

-6 ≤ 3x

Divide both sides by 3

-2 ≤ x

This says x must be greater than or equal to -2, so it can also be written as x ≥ -2

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If you have a quadratic equation like a*x² + b*x + c = 0 where a, b and c are constants, then the quadratic formula is x = (-b ± √(b² - 4*a*c))/(2*a)

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Consider the function y = x² 

This doesn't have an inverse (in the real number domain) because there isn't a unique value of x for a given y, but by writing x = y² before doing anything else you imply there is (to qualify as a function an expression must map an input to a unique output).

However, as you say, the important thing is being careful about the restrictions.  As long as you keep this in mind I will not say you can't do it Melody!  

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Thanks Alan,

I thought it would work anyway so long as I am careful about the restrictions.

It still confuses me.    I don't understand why you say I can't do it.   

Yes, but in this case of course it is a simple linear function defined over all the real numbers for both x and y, so it works out ok either way.

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Sat 16/5/15

Sun 17/5/15

If you would like to comment on other site issues please do so on the Lantern Thread.  Thank you.    

Interest Posts:

FTJ means 'For the juniors' 

Sorry nicolas  

Your logic and your LaTex is great MG, unfortunately   -9-5 does not equal -4    

Could you correct it please.

I am not sure if it is a careless mistake of if you have problems with directed numbers.

If you have problems tell me and I'll teach you  

This post might help you  :)

Your answer is correct Dbg but what does it mean?

12÷2(3 )×2=

I think that using brackets instead of multiply signs with numbers is very confusing but this is what i believe to be correct. :)

$$\\12\div2(3 )\times2\\
=12\div2\times3 \times2\\
=6\times3 \times2\\
=36$$

Remember Brittany that multiply and divide are equally important.  You do them from left to right :)

YES I HAD SEEN YOUR IMPROVED ATTITUDE !

Had you not noticed my different attitude to you the last couple of days!

I did not say that your answer on this thread was a garbage post - you could not be more wrong!

I was teaching you how to teach that is all.  My thanks was guenuine!

In the thank you that I gave to MathsGod1 I was also picking him up on a small error.  Didn't you notice! 

If I try to teach you something that means I think you are worth teaching!  It is a compliment!

Wait no... 

12/(6 • 2)

12/12= 1

12/6 • 2

2 • 2 = 4

I had just tried to use this forum better than before, when you said that I just post garbage...

Had you seen?

Actually Dbg4thebest I would like to follow that up on that genuine thank you.  :)

You would have been better to use any example other than $$\sqrt4$$     This is because people confuse square root with halving numbers.

I will often see    $${{\mathtt{3}}}^{{\mathtt{2}}} = {\mathtt{6}}$$   or  $${\sqrt{{\mathtt{16}}}} = {\mathtt{8}}$$   BOTH OF WHICH ARE WRONG!

So you would be better to use an example like   $${\sqrt{{\mathtt{16}}}} = {\mathtt{4}}$$    so that there is less room for confusion. :)

Sorry Dbg4thebest, A great BIG thank you to you too :))

What about me????