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 ;)

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- January 28th 2006, 03:35 AMdgolverkformulas...
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 AMticbolQuote:

Originally Posted by**dgolverk**

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 AMdgolverkThanks
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 AMCaptainBlackQuote:

Originally Posted by**dgolverk**

,

so:

.

Now:

so:

,

we can leave the here as the information we have been

given is symmetric in and and so this sign

ambiguity remains.

Now:

.

RonL - January 28th 2006, 05:17 AMdgolverkThanks 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 AMticbolQuote:

Originally Posted by**dgolverk**

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 AMCaptainBlackQuote:

Originally Posted by**dgolverk**

So adding these gives:

,

but simplifying the LHS this is:

,

so:

.

Subtracting the same two equations as before were added gives:

so:

.

RonL - January 28th 2006, 05:42 AMdgolverkThanks!
You really helped me!

Thanks again! :) - January 28th 2006, 05:46 AMticbolQuote:

Originally Posted by**dgolverk**

(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.