what two numbers when multiplied = to 2, but the sum of is 0?
two numbers multiplied equals to 8, but added together equals to 5

i'm really confused, thanks for the help

Question

what two numbers when multiplied = to 2, but the sum of is 0?
two numbers multiplied equals to 8, but added together equals to 5

i'm really confused, thanks for the help

ganeshie8 · Answer

say the required two numbers are $x$ and $y$

ganeshie8 · Answer

you want the numbers to be multiplied to `2` ? and the sum to equal `0`
right ?

anonymous · Answer

yes

ganeshie8 · Answer

good, setup the equations accordingly

ganeshie8 · Answer

$xy = 2$

$x+y = 0$

ganeshie8 · Answer

solve them ^

anonymous · Answer

um, that's the part where i'm confused at

ganeshie8 · Answer

my display pic may help u, take a look :)

ganeshie8 · Answer

eliminate one of the variables first : solve for x from the second equation and plug it in the first equation

ganeshie8 · Answer

second equation :   
$x+ y = 0$

$x =  ?$

anonymous · Answer

are the answers decimals?

ganeshie8 · Answer

idk yet, we need to solve and see

anonymous · Answer

x=2?

ganeshie8 · Answer

not quite

$x + y =  0$

$x = -y$

ganeshie8 · Answer

plug this in first equaiton :

$xy =  2$

$(-y)y = 2$

$-y^2 = 2$

$y^2 = -2$

ganeshie8 · Answer

can the square of a `real ` number be negative ever ?

ganeshie8 · Answer

nope, so there are `no solutions` in real numbers

anonymous · Answer

so this problem doesn't have an answer?

ganeshie8 · Answer

the answer is `no solutions` in real numbers

anonymous · Answer

ok

ganeshie8 · Answer

is this ur second question ?

two numbers multiplied equals to 8, but added together equals to 5

anonymous · Answer

yes, it doesn't have a solution either right?

ganeshie8 · Answer

you cant tell it just by looking... its difficult lol

ganeshie8 · Answer

setup the equations and work it like we did the earlier problem

ganeshie8 · Answer

but you're right... there are no solutions for that also

anonymous · Answer

the only factors of 8 are 1,2,4,8, and neither of those numbers would equal 5 even if we had a negative

anonymous · Answer

ok, thanks

ganeshie8 · Answer

*no real solutions : http://www.wolframalpha.com/input/?i=solve+xy+%3D+8%2C+x%2By%3D5

ganeshie8 · Answer

very good :) thats a nice approach !

anonymous · Answer

the original problem was find the roots of each quadratic polynomial:
x^2-6x-7
(x__)(x___) and this was the way my teacher showed us

anonymous · Answer

wait, wrong problem *x^2+2*

anonymous · Answer

well, thanks for the help, and bye!