Two numbers (negative or positive) that when multiplied is 4 and plus is 3

[formal fjord]

note that the sum should be larger and the difference is smaller

my equation is x(squared) + 3x - 4

[surreal spruce]

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[mighty sedge]

you need them to multiply to -4 and add to 3

completely different

try 4 and -1

(x+4)(x-1)

[formal fjord]

can you help me again please

two negative numbers that when added is - 5 and when multiplied is - 40

[mighty sedge]

send the question

i don't think i can do that

[formal fjord]

nvm its prime

[mighty sedge]

it's what?

[formal fjord]

Prime

Our teacher said to put prime if there is no possible two numbers.

[mighty sedge]

i suppose that is a fairly succinct way to say "irreducible over rationals"

[formal fjord]

Idk though

[mighty sedge]

i don't think there will be

one of the numbers has to be divisible by 5, and the other will not

but if they add to 5 then this is clearly not possible

[left pelican]

i know this may be a new concept for you, but we can use vieta’s formulae here to assist

[mighty sedge]

to be fair, they are already using it

by asking for the product to equal (-)4 and the sum to equal 3

[left pelican]

let us say that r and s are the roots of the equation

[mighty sedge]

also, overkill much?

[left pelican]

then we can put the quadratic into the form $x^2-(r+s)x+rs$

[last ledge]

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[left pelican]

?

there is a simple proof for this

[mighty sedge]

they are already asking for two numbers such that rs = (-)4 and r+s = (-)3

give or take some sign errors

vieta's won't even help here

besides the difficulty in explanation

[left pelican]

$(x-r)(x-s)=x^2-rx-sx+rs$

[mighty sedge]

i'm leaving

have fun

[last ledge]

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[left pelican]

factoring our the x we get $x^2-(r+s)x+rs$

[last ledge]

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