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.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.
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)
Start with:Originally Posted by dgolverk
$\displaystyle (a+b)^2=a^2+b^2+2ab$,
so:
$\displaystyle a^2+b^2=(a+b)^2-2ab=64-24=40$.
Now:
$\displaystyle (a-b)^2=a^2+b^2-2ab=40-24=16$
so:
$\displaystyle a-b=\pm 4$,
we can leave the $\displaystyle \pm$ here as the information we have been
given is symmetric in $\displaystyle a$ and $\displaystyle b$ and so this sign
ambiguity remains.
Now:
$\displaystyle \frac{1}{a^2}+\frac{1}{b^2}=\frac{a^2+b^2}{a^2b^2} =\frac{a^2+b^2}{(ab)^2}=\frac{40}{144}=\frac{5}{18 }$.
RonL
Is that so? So no going to find the values of a and b?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?
Easy-Peasy:Originally Posted by dgolverk
$\displaystyle (a+b)^2=a^2+b^2+2ab=49$
$\displaystyle (a-b)^2=a^2+b^2-2ab=25$
So adding these gives:
$\displaystyle (a^2+b^2+2ab)+(a^2+b^2-2ab)=49+25=74$,
but simplifying the LHS this is:
$\displaystyle 2(a^2+b^2)=74$,
so:
$\displaystyle a^2+b^2=37$.
Subtracting the same two equations as before were added gives:
$\displaystyle 4ab=49-25=24$
so:
$\displaystyle ab=6$.
RonL
Umm, you're too young yet and you have many problems already?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.