formulas...

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January 28th 2006, 03:35 AM
dgolverk

formulas...

given :a+b=8 ab=12

how much is:
a-b
a^2+b^2

1 + 1
a^2 b^2

Thanks! I need it for a test tomorrow so please help me fast ;)
January 28th 2006, 04:05 AM
ticbol

Quote:

Originally Posted by dgolverk

given :a+b=8 ab=12

how much is:
a-b
a^2+b^2

1 + 1
a^2 b^2

Thanks! I need it for a test tomorrow so please help me fast ;)

Just by looking, (a or b) = 2, and (a or b) = 6.

Let us use some Math.
Here is one way.

a +b = 8 ----(i)
ab = 12 -----(ii)
From (ii), a = 12/b
Substitute that into (i),
12/b +b = 8
Clear the fraction, multiply both sides by b,
12 +b^2 = 8b
b^2 -8b +12 = 0
(b-2)(b-6) = 0

b-2 = 0
b = 2

b-6 = 0
b = 6

So, b = 2 or 6 --------***
Then, a = 12/b = 6 or 2 ----------***
Hence, we have two sets of answers:
a=2, and b=6 -----(1)
a=6, and b=2 -----(2)

-----------------------------------------------
When a=2 and b=6:

a -b = 2 -6 = -4 -----------------------------answer.
a^2 +b^2 = 2^2 +6^2 = 4 +36 = 40 ---------answer.

1/(a^2) +1/(b^2)
= 1/(2^2) +1/(6^2)
= 1/4 +1/36
= (9 +1)/36
= 10/36
= 5/18 ------------------answer.

--------------------------------------------------
When a=6 and b=2:

a -b = 6 -2 = 4 -----------------------------answer.
a^2 +b^2 = 6^2 +2^2 = 36 +4 = 40 ---------answer.

1/(a^2) +1/(b^2)
= 1/(6^2) +1/(2^2)
= 1/36 +1/4
= (1 +9)/36
= 10/36
= 5/18 ------------------answer.
January 28th 2006, 04:49 AM
dgolverk

Thanks

but...
I forgot to notice that you shouldn't find how much is a or b.
you should find how much is a-b etc... togther...
I very appreciate your help.
by the way...
You need use in these formulas:
(a+b)^2
(a-b)^2
(a-b)(a+b)
January 28th 2006, 05:12 AM
CaptainBlack

Quote:

Originally Posted by dgolverk

given :a+b=8 ab=12

how much is:
a-b
a^2+b^2

1 + 1
a^2 b^2

Thanks! I need it for a test tomorrow so please help me fast ;)

Start with:

$(a+b)^2=a^2+b^2+2ab$ ,

so:

$a^2+b^2=(a+b)^2-2ab=64-24=40$ .

Now:

$(a-b)^2=a^2+b^2-2ab=40-24=16$

so:

$a-b=\pm 4$ ,

we can leave the $\pm$ here as the information we have been
given is symmetric in $a$ and $b$ and so this sign
ambiguity remains.

Now:

$\frac{1}{a^2}+\frac{1}{b^2}=\frac{a^2+b^2}{a^2b^2} =\frac{a^2+b^2}{(ab)^2}=\frac{40}{144}=\frac{5}{18 }$ .

RonL
January 28th 2006, 05:17 AM
dgolverk

Thanks a lot!

but now i have another problem... :rolleyes:
given:
a+b=7

a-b=5

find how much is:
a^2+b^2

ab

Thanks again :)
January 28th 2006, 05:29 AM
ticbol

Quote:

Originally Posted by dgolverk

but...
I forgot to notice that you shouldn't find how much is a or b.
you should find how much is a-b etc... togther...
I very appreciate your help.
by the way...
You need use in these formulas:
(a+b)^2
(a-b)^2
(a-b)(a+b)

Is that so? So no going to find the values of a and b?
Okay.

Here is one way of doing that,

a+b = 8 ----(i)
ab = 12 --------(ii)
Find 1) a-b; 2) a^2 +b^2; and 3) 1/(a^2) +1/(b^2) without solving for the values of a and b.

----------------
The a^2 +b^2 first.

(a+b)^2 = 8^2 = 64
but (a+b)^2 = a^2 +2ab +b^2 also, so,
Substituting 12 for ab, and 64 for (a+b)^2,
64 = a^2 +2(12) +b^2
64 -24 = a^2 +b^2
40 = a^2 +b^2
Or,
a^2 +b^2 = 40 ---------answer.

-----------------------------------------
The a-b.

(a-b)^2 = a^2 -2ab +b^2
(a-b)^2 = (a^2 +b^2) -2ab
Substitutions,
(a-b)^2 = 40 -2(12)
(a-b)^2 = 16
Take the square roots of both sides,
a-b = +,-4 ----------------------------answer.
That is "plus or minus" 4.

---------------------------------------------
The 1/(a^2) + 1/(b^2).

1/(a^2) +1/(b^2)
Combine the two fractions into one fraction only,
= [(b^2)*1 +(a^2)*1] / [(a^2)(b^2)]
= [a^2 +b^2] / [(ab)^2]
Substitutions,
= [40] / [(12)^2]
= 40 / 144
Reduce to lowest term, divide numerator and denominator by 8,
= 5/18 ---------answer.

-----------------------------------------
Do the answers here tally with the answers in my first answer where I solved for and b first?
January 28th 2006, 05:40 AM
CaptainBlack

Quote:

Originally Posted by dgolverk

but now i have another problem... :rolleyes:
given:
a+b=7

a-b=5

find how much is:
a^2+b^2

ab

Thanks again :)

Easy-Peasy:

$(a+b)^2=a^2+b^2+2ab=49$
$(a-b)^2=a^2+b^2-2ab=25$

So adding these gives:

$(a^2+b^2+2ab)+(a^2+b^2-2ab)=49+25=74$ ,

but simplifying the LHS this is:

$2(a^2+b^2)=74$ ,

so:

$a^2+b^2=37$ .

Subtracting the same two equations as before were added gives:

$4ab=49-25=24$

so:

$ab=6$ .

RonL
January 28th 2006, 05:42 AM
dgolverk

Thanks!

You really helped me!
Thanks again! :)
January 28th 2006, 05:46 AM
ticbol

Quote:

Originally Posted by dgolverk

but now i have another problem... :rolleyes:
given:
a+b=7

a-b=5

find how much is:
a^2+b^2

ab

Thanks again :)

Umm, you're too young yet and you have many problems already?

(a+b)^2 = a^2 +2ab +b^2 = 7^2 = 49 ----(1)
(a-b)^2 = a^2 -2ab +b^2 = 5^2 = 25 --------(2)

Add (1) and (2),
2(a^2 +b^2) = 74
So,
a^2 +b^2 = 37 -----answer.

Substitute that into, say, (a-b)^2,
(a-b)^2 = a^2 -2ab +b^2
Substitutions,
(5)^2 = 37 -2ab
25 -37 = -2ab
-12 = -2ab
So,
ab = 6 ------------answer.

-----------------------------------------
Zeez, I posted my replies late. I need to rev this one-finger typer of mine. Steroids, maybe.
But gotta get some sleep first. 'Night.