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OK, so I get the quadratic formula. it ain't such a tough nut rather it is more like a gentle baby. but to save upsetting it on this day of romance I'd like

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OK, so I get the quadratic formula. it ain't such a tough nut rather it is more like a gentle baby. but to save upsetting it on this day of romance I'd like to resolve one thing about square rooting in maths/this formula.


after half of the formula I have, -3 - Sqrt +9-4 /2*1.. my question is simple and two part.

1) to Sqrt the 9 I must also Sqrt the 4 or calculate the sum of 9-4 then Sqrt before manipulating change from outside the Sqrt right?

2) please insert parenthesis to solve including signs.
 Feb 14, 2014

15+0 Answers

 #1
avatar+6248 
0
Quote:

after half of the formula I have, -3 - Sqrt +9-4 /2*1.. my question is simple and two part.

1) to Sqrt the 9 I must also Sqrt the 4 or calculate the sum of 9-4 then Sqrt before manipulating change from outside the Sqrt right?

2) please insert parenthesis to solve including signs.



if this is the quadratic formula evaluated for a=1, b=3, and c=4

then notated properly it is

[size=200]([/size]-3 +/- sqrt[size=150]([/size]9 - 4[size=150])[/size][size=200])/[/size](2*1)

so to answer (1) you subtract the 4 from 9 to get 5 and then take the square root

(2) is answered by the formula above.
 Feb 14, 2014
 #2
avatar+118608 
0
Stu:

:D

OK, so I get the quadratic formula. it ain't such a tough nut rather it is more like a gentle baby. but to save upsetting it on this day of romance I'd like to resolve one thing about square rooting in maths/this formula.

assuming you are solving x2+3x+1=0

after half of the formula I have [-3 - Sqrt (+9-4)] / (2*1).. my question is simple and two part.

1) to Sqrt the 9 I must also Sqrt the 4 or calculate the sum of 9-4 then Sqrt before manipulating change from outside the Sqrt right?
You have to find 9-4 =5 and THEN take the square root.
2) please insert parenthesis to solve including signs.

 Feb 14, 2014
 #3
avatar+1313 
0
in fact please calculate in steps the following,

-6 + Sqrt 6 2-4*2*3 /2*3 include parenthesis please.
I got x1=-0.634

if m answer is right don't worry to solve. thanks

thanks melody, I see they will not always be nice simple numbers do I round of at 3 decimal placings as a rule?
 Feb 14, 2014
 #4
avatar+1313 
0
side question ot, when removing Sqrt in transposing given t= sqrt2mg, solve for g. it is easiest to remove the Sqrt first I get that, but answer given is g = t 2/2m. I was about to give this answer, but isnt everything under the Sqrt squared to remove it, so.. when transposing /2m shouldn't they be squared. I get that it isn't so I would just like to know why as I'm sure it is a simple answe.r. thanks.
 Feb 14, 2014
 #5
avatar+1313 
0
thanks ROM and melody just saw or post ROM. good help. I wonder in asking to solve the second equation at post 4 that I can do [-b+-(Sqrt b 2(-4*a*c))/(2*a)]
which I didn't ask very well, since I had resolved it to a further point but via this manner. it's OK right?
 Feb 14, 2014
 #6
avatar+6248 
0
Stu if you don't start using parentheses for sqrt for example, I'm gonna sic a dropbear on you.

Here is the quadratic formula properly notated

0
 Feb 14, 2014
 #7
avatar+1313 
0
@Rom, what parenthesis for Sqrt. I am on my tablet and it isn't easy to go Google the symbol and paste it nor do I know how to make the image. My question was can I work the -4ac then deduct the answer from b 2 and then solve the rest, which wasn't answered. Not sure what you meant. As well, if I'm not using it it could be either because I don't know how or cannot at this time or because in most cases it is all new to me and I'm only two weeks into any type of maths (about 10 contact hours or less.) Byut in general love your responses. Hope this helps you. Though I'm pretty certain I do need to solve for -4ac and deduct it to get the right answer and not b 2-4 then *a*c which would be wacky I think.


Lastly, few questions raised in my last few posts if anyone cares to assist.
 Feb 14, 2014
 #8
avatar+6248 
0
all I mean is to write sqrt(x) instead of sqrt x or sqrtx where x might be some complicated expression.

no fancy chars or images needed.
 Feb 14, 2014
 #9
avatar+118608 
0
Stu:

in fact please calculate in steps the following,

-6 + Sqrt 62-4*2*3 /2*3 include parenthesis please.
I got x1=-0.634

if m answer is right don't worry to solve. thanks

thanks melody, I see they will not always be nice simple numbers do I round of at 3 decimal placings as a rule?



3x 2 + 6x +2 = 0
x = [-6 ± Sqrt (6 2-4*3*2)] /[2*3]
x = [-6 ± sqrt (36-24)] / 6
x = [-6 ± sqrt(12)] / 6 (if you are asked for estimate answers you can do it on your calc from here)
x = [-6 ± sqrt(4*3)] / 6
x = [-6 ± sqrt(4)sqrt(3)]/6
x= [-6 ± 2sqrt(3)] / 6
x = 2[ -3 ± sqrt(3)] / 6
x = [ -3 ± sqrt(3)] / 3

x = [ -3 + sqrt(3)] / 3 or x = [ -3 - sqrt(3)] / 3 (These are the exact roots so they are the best answers.)

x= -0.4226 or x= -1.5774 ( both correct to 4 decimal places)

You are better off to leave the rounding until the very end because rounding errors can get exaggerated as you continue working the problem.

Now Stu, is there any other specific problems that we can help with.
perhaps you should make sure you have digested the last 2 posts that Rom and I have made first
 Feb 14, 2014
 #10
avatar+1313 
0
Rom:

all I mean is to write sqrt(x) instead of sqrt x or sqrtx where x might be some complicated expression.

no fancy chars or images needed.



Still don't get it, is the Sqrt which encloses everything under it the same as a parenthesis in and of it's self.. and the std of working is to apply anything under the tick first. And when I wrote all my other parenthesis mean to work them first? And that in the question it was kind of on topic, whether it had a punctuation error or not? So I should have wrote -->[-b+-(Sqrt (b2(-4*a*c)))/(2*a)] as opposed to [-b+-(Sqrt b2(-4*a*c))/(2*a)] My questions are not Sqrt parenthesis, I am just putting them for purpose of seeing an order to solving only, relating to the question.


Melody:

Now Stu, is there any other specific problems that we can help with.
perhaps you should make sure you have digested the last 2 posts that Rom and I have made first...



Of course. I am just starting with quadratics and graphing for the week ahead and conic sections if able. So I will do for sure.

And yes, I just had this question regarding transposing a square root posted in this thread.
side question ot, when removing Sqrt in transposing given t= sqrt2mg, solve for g. it is easiest to remove the Sqrt first I get that, but answer given is g = t 2/2m. I was about to give this answer, but isnt everything under the Sqrt squared to remove it, so.. when transposing /2m shouldn't they be squared. I get that it isn't so I would just like to know why as I'm sure it is a simple answe.r. thanks. ie can't move the Sqrt with out moving the 2m to the left and squaring it also?
 Feb 14, 2014
 #11
avatar+6248 
0
Quote:

is the Sqrt which encloses everything under it the same as a parenthesis in and of it's self.



Yes!

0

but you haven't been doing this. you've been writing

Sqrt complicatedexpression

with no parens at all.
 Feb 14, 2014
 #12
avatar+1313 
0
Ok Rom makes sense. I'l put in the effort.
 Feb 14, 2014
 #13
avatar+118608 
0
Stu:


And yes, I just had this question regarding transposing a square root posted in this thread.
side question ot, when removing Sqrt in transposing given t= sqrt2mg, solve for g. it is easiest to remove the Sqrt first I get that, but answer given is g = t2/2m. I was about to give this answer, but isnt everything under the Sqrt squared to remove it, so.. when transposing /2m shouldn't they be squared. I get that it isn't so I would just like to know why as I'm sure it is a simple answe.r. thanks. ie can't move the Sqrt with out moving the 2m to the left and squaring it also?



In your question 2mg was all under the square root sign.

So when you write it in here, you have to write sqrt(2mg)
your question is
t = sqrt(2mg)
square both sides and you get
t 2 = 2mg
divide both sides by 2m and you get
t 2/(2m) = g (NOTE: you need the brackets here too otherwise you would be dividing by 2 and multiplying the whole answer by m)
g = t 2/(2m)

Now I know that you are still confused. I will look at this with you
t = sqrt(2mg)
t = sqrt(2) * sqrt(m) * sqrt(g)
t 2 = [ sqrt(2) * sqrt(m) * sqrt(g) ] 2
t 2 = sqrt(2) 2 * sqrt(m) 2 * sqrt(g) 2
t 2 = 2 * m * g
t 2 = 2mg
See, all the terms were squared.

Think about it for a while and then tell me if you understand.
warm regards,
Melody.
 Feb 15, 2014
 #14
avatar+1313 
0
makes sense. I got it in the first bit. Just rationalising it in an improper fashion previously. Thanks.
 Feb 15, 2014
 #15
avatar+118608 
0
Stu:

makes sense. I got it in the first bit. Just rationalising it in an improper fashion previously. Thanks.



You are welcome.
 Feb 15, 2014

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