let:
n=the number
m=the other number
n=8+m-----eqtn 1
nXm=20-----eqtn.2
substitute eqtn1 to eqtn2
(8+m)(m)=20
8m+m^2=20
arrange
m^2+8m-20=0
it forms into quadratic eqtn.
factor by grouping
(m-2)(m+10)=0
by zero property of equality we have:
m=2 and m=-10
-10 is an extraneous root since it cannot satisfy the equation.
there should'nt be negative
use m=2
substitute to eqtn.1
n=8+2
n=10
therefore the numbers are: 2 and 10
check:
the product of the numbers is 20:
2X10 = 20 check!
the number is 8 more than the other number:
n=8+2= 10 check!
A number is 8 more than another number. The product of these two numbers is 20. Find the numbers.
3 answers
Don't rule out m = -10 as extraneous, since they do satisfy the given conditionn.
of m = -10, m+8 = -2
their product = (-10)(-1) = 20
-2 is 8 more than -10
the numbers could also be -2 and -10
of m = -10, m+8 = -2
their product = (-10)(-1) = 20
-2 is 8 more than -10
the numbers could also be -2 and -10
For the following two numbers, find two factors of the first number such that their product is the first number and their sum is the second number.
44,15
44,15