Find two real numbers that differ by 5 and have a product of 16.
I can figure it out by trial and error but I must show work to get credit on homework. Can someone solve and show work?
2006-08-03 21:50:00 · 8 answers · asked by Stepheni U 2 in Education & Reference ➔ Homework Help
x * ( x - 5 ) = 16
x^2 - 5x = 16
x^2 - 5x - 16 = 0
Use the quadratic equation: x = (-b +- sqrt (b^2 - 4ac)) / 2a
x = (5 +- sqrt(25 + 4 * 16)) / 2
So, the answers are...
(5 + sqrt(89)) / 2
(5 - sqrt(89)) / 2
2006-08-03 22:00:04 · answer #1 · answered by Betty L 2 · 1⤊ 0⤋
Are you sure you got the question right? The product is when the numbers are multiplied together, the only numbers with a product of 16 are 2 & 8 and 4 & 4, and neither of these differ by 5. Check your question
2006-08-04 04:56:02 · answer #2 · answered by Anonymous · 0⤊ 0⤋
The sum is wrong.
the factors of 16 are
2*8
1*16
4*4
as none these has a difference of 5, what ever method you try the answer will go into decimals.
therefore the difference of the two numbers should be 6,15 or 0
2006-08-04 05:02:03 · answer #3 · answered by Ark Angel 2 · 0⤊ 0⤋
x-y=5; x*y=16; threfore x=5+y;therefore we get y*(5+y)=16
5y+y^2=16;
therefore the two numbers are (-5+(89)^1/2)/2,(5+(89)^1/2)/2
2006-08-04 05:04:59 · answer #4 · answered by Anonymous · 0⤊ 0⤋
What it would be good for us answerers to know is what is the topic you are studying? If it's factoring then the three problems of yours I've looked at do not factor. If it's Quadratic Formula, then they all can be done..You will just get irrational or possibly complex answers.
2006-08-04 06:40:28 · answer #5 · answered by MollyMAM 6 · 0⤊ 0⤋
x-y=5 =>x=y+5
x*y=16
y*(y+5)=16 => y^2+5*y-16=0
should be easy from here
2006-08-04 04:54:19 · answer #6 · answered by raul 3 · 0⤊ 0⤋
x-y=5 and xy=16
x(x-5)=16
x^2-5x+16=0
this cannot have real roots and so the sum is wrong
2006-08-04 04:57:45 · answer #7 · answered by raj 7 · 0⤊ 0⤋
(5+sqrt(89))/2
(5-sqrt(89))/2
2006-08-04 04:57:20 · answer #8 · answered by Anonymous · 0⤊ 0⤋