prod. of 2 numbers is 16, difference is -4. no decimals

I have been unable to figure out: two numbers that when multiplied equals 16 and when subtracted equals -4. No decimals.

Re: Algebra

let the numbers be x and y

then:
eq 1) xy=16
eq2) x-y=-4

solve one equation for 1 variable and substitute it into second equation
I will do it one way , you do it a different way
solve eq 1 for x
x=16/y substitute into eq2
16/y -y =-4
multiply both sides by y
16-y^2=-4y
0=y^2-4y-16
by binomial equation
y=4 +/- [16+64]^1/2 all over 2
y=2+/- 1/2 sqrt[80]
y=2+/-2 sqrt5
y=2+2 sqrt5 answer
y=2-2 sqrt5 answer

solve for x

Arthur

Re: Algebra

There isn't any "no decimals" solution to this problem. Did you mean "added" instead of "multiplied"? It would maybe help us if you wrote the whole text of the problem. "No decimals" isn't a sentence, so I'm sure, substituting the solution of one of your equations into the other, we get

\(\displaystyle $x^2 + 4 x - 16 = 0$\)

and that gives us values for one of the numbers, but it's not an integer. You could go back and plug it in, but I'm not sure that's what you meant by the problem. Please try to write out the problem as you have it in the book. I have a feeling we're missing something here.

Re: Algebra

mmm....
I agree. to be more specific I should have done the following:
0=y^2-4y-16
complete the square by adding and subtracting 4,i.e. 1/2 the y coefficient squared
0=[y^2-4y+4]-4-16
0=[y-2]^2-20
[y-2]^2=20 take square root
y-2=+/- sqrt20
y=2+/- sqrt[4[5]]
y=2+/-2sqrt5 answer
Arthur