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12 answers

[07]
Let the numbers be x andy
By the problem,
x+y=11.......(1)
xy=5...........(2)
We know that (x-y)^2
=(x+Y)^2-4xy
=(11)^2-4*5
=121-20
=101
Therefore x-y= sqrt101
Now,we have two equations
x+y=11 and
x-y=sqrt101
By solving ,we get
x=(11+sqrt101)/2 and y=(11-sqrt101)/2
Therefore,the nos are (11+sqrt101)/2 and (11-sqrt101)/2

2007-11-29 11:00:43 · answer #1 · answered by alpha 7 · 1 1

Let's let the numbers be x and y:
x + y = 11
xy = 5

Solve the first for y:
y = 11 - x

Substitute this into the second equation:
x(11 - x) = 5

Now expand out:
11x - x² = 5

Let's get everything on one side (add x², subtract 11x):
x² - 11x + 5 = 0

You can solve this with the quadratic formula:

x = [11 ± sqrt( (-11)² - 4*5 ) ] / 2
x = [11 ± sqrt( 121 - 20 ) ] / 2
x = [11 ± sqrt( 101 ) ] / 2

The two numbers are:
(11 - √101) / 2 ≈ 0.475062189
(11 + √101) / 2 ≈ 10.5249378

Double checking our answers with the approximations:
0.475062189 + 10.5249378 = 10.99999999 --> 11

Multiplying:
0.475062189 * 10.5249378 = 4.99999999 --> 5

The slight errors at the end are because we are using an approximation of the answer.

The answers are:
(11 - √101) / 2
(11 + √101) / 2

2007-11-29 18:56:59 · answer #2 · answered by Puzzling 7 · 0 1

You could take 5 and 1. You didn't specify if a number could only be used once!

5+5+1 = 11
AND
1 x 5 = 5

2007-11-29 18:57:09 · answer #3 · answered by luv_my_rats 5 · 1 2

Set up a system of equations.
x+y=11
x*y=5
Solve
1. y=x/5
2. x + x/5 = 11 (substitution)
3. x = (11+ square root of 101)/2 or (11 - square root of 101)/2

The two numbers are simply the same as the solutions of x.

2007-11-29 19:00:37 · answer #4 · answered by whizkid66 3 · 0 1

x + y = 11 or y = 11 - x

xy = 5
now substitute y from the first equation into the second

x(11-x) = 5

-x^2 + 11x = 5 or

x^2 - 11x + 5 = 0


now solve for x, then for y

2007-11-29 19:00:42 · answer #5 · answered by GTB 7 · 0 1

given
a+b=11
ab=5

we can determine that
a(11-a)=5
a^2-11a+5=0

therefore
a = [11 +/- sqrt(101)]/2
if we let a be the result using +, then b will be the result with -, and vice versa. in other words, the two numbers are
[11 +/- sqrt(101)]/2

that is to say,
[11 + sqrt(101)]/2
and
[11 - sqrt(101)]/2

very rough approximations are 10.5 and 0.5
clearly their sum is 11, the product of these
approximations is 5.25, but if you don't use approximations
(or more accurate approximations) you'll get closer to 5 (or exactly 5 if you use the closed form expressions)

if you want, i can rattle on and on, adding more and more to this answer.

2007-11-29 18:59:20 · answer #6 · answered by Anonymous · 0 1

puzzling made a sign mistake
you get x^2 - 11x + 5 = 0
and so the two numbers are
[11 + √101]/2 and [11 - √101]/2
~~

2007-11-29 19:00:21 · answer #7 · answered by ssssh 5 · 0 1

no numbers multiply to make 5!

2007-11-29 18:54:57 · answer #8 · answered by ~~eddie m~~~ 5 · 0 1

ok you have 2 equations
(1) x+y=11 and
(2) x*y=5

i hope from there you can solve it, but if not, i will post the answer (you have to do the work) in about a minute

but i guess puzzling did the work for you... so nv

2007-11-29 18:58:04 · answer #9 · answered by Anonymous · 0 1

if u have to have the same 2 numbers, good luck and have fun trying to make the impossible possible

2007-11-29 18:59:36 · answer #10 · answered by Caitlin M 2 · 0 1

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