document.write( "Question 964656: the product of two number is 800. the sum of two number is 204. then find the two number. \n" ); document.write( "
Algebra.Com's Answer #589499 by addingup(3677) You can put this solution on YOUR website! a*b= 800 And: \n" ); document.write( "a+b= 204 So we can say that: \n" ); document.write( "b= 204-a Substitute in the first equation: \n" ); document.write( "a(204-a)= 800 Multiply: \n" ); document.write( "204a-a^2= 800 Let's write in standard quadratic binomial (ax^2+bx+c): \n" ); document.write( "-a^2+204a-800= 0 Now we factor this equation: \n" ); document.write( "-(a-200)(a-4)= 0 solve each separately: \n" ); document.write( "-(a-200)= 0 or a-4= 0 \n" ); document.write( "-a+200= 0 or a-4= 0 \n" ); document.write( "-a= -200 or a= 4 on the left multiply both times -1 \n" ); document.write( "a= 200 or a= 4 As you see we found our 2 numbers. 200+4=204 and 200*4=800\r \n" ); document.write( "\n" ); document.write( "Note: To factor an equation, follow FOIL and remember that the Outer and Inner have to add up to the middle number, in this case 204a, and Last have to multiply to the last number, 800. You have to break down this number into its prime components. This is how you do it. Factor the last number, 800: \n" ); document.write( "800/2=400/2=200/2=100/2=50/2=25/5=5/5=1 \n" ); document.write( "..2....2.....2.....2.....2....5....5 Now we play with the numbers. We need a combination that when added will result in 204. Since they are the prime numbers of 800, the associative property of multiplication tells us that any combination will multiply to 800, so we only worry about 204. \n" ); document.write( "2*2=4 and multiply the rest: \n" ); document.write( "2*2*2*5*5= 200 Now we have it, 200+4=204 and 200*4=800\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |