SOLUTION: What is the minimum product of two numbers whose difference is 4? What are the numbers?
This is a far as I got...
x(x+4)
x^2+4x+c=0
-4/2=-2
2^2+4(-2)+c=0
4-8+c=0
-4+c=
Question 31093: What is the minimum product of two numbers whose difference is 4? What are the numbers?
This is a far as I got...
x(x+4)
x^2+4x+c=0
-4/2=-2
2^2+4(-2)+c=0
4-8+c=0
-4+c=0
c=4
x^2+4x+4=0
(x+2)(x+2)
x=-2 and -2+4=2
This is where I realize that something is not quite right... Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! GOOD TO SEE YOUR SINCERE ATTEMPT.KEEP IT UP.YOU WILL DO WELL.SEE MY COMMENTS BELOW...
What is the minimum product of two numbers whose difference is 4? What are the numbers?
This is a far as I got...
x(x+4)..OK
ALWAYS TRY TO PUT IF POSSIBLE WHAT YOU WANT ...YOUR OBJECTIVE AS A SINGLE VARIABLE TO BE FOUND.HERE CALL Y =X^2+4X...SO Y HAS TO BE MINIMUM
NOW LOOK FOR CONSTRAINTS IF ANY....NOTHING GIVEN?..SO PROCEED
WRITE Y AS A PERFECT SQUARE,SINCE ITS MINIMUM WE KNOW IS ZERO.USE P THE X^2 AND X TERMS FULLY.USE (A+B)^2..OR..(A-B)^2 FORMULAE
SO Y = {X^2+2(X)(2)+2^2}-2^2=(X+2)^2-4
NOW WE KNOW....THE VARIABLE X IS PUT UNDER A PERFECT SQUARE..SO ITS MINIMUM IS ZERO.WHAT REMAINS IS ONLY A CONSTANT.
SO Y MINIMUM =0-4=-4...THIS HAPPENS WHEN X+2=0..OR X=-2...SINCE THERE IS NO CONSTRAINT ON X VALUE THIS IS THE ANSWER.
x^2+4x+c=0
-4/2=-2
2^2+4(-2)+c=0
4-8+c=0
-4+c=0
c=4
x^2+4x+4=0
(x+2)(x+2)
x=-2 and -2+4=2
This is where I realize that something is not quite right... YOU ARE A LITTLE CONFUSED ON OBJECTIVE AND PROCEDURE.HOPE THE ABOVE EXPLANATION CLEARS THE CLOUD.
