The sum of two numbers is 8 and the sum of their squares is 32. What is the smaller number?

2 Answers
Mar 23, 2017

Answer:

The smaller number is #3#

(The other number is therefore #5#)

Explanation:

We could use #x and y# to represent the two numbers, but it is possible to form an equation using only one variable.

If two numbers add up to 8, and we let one of them be #x#, the other one will be the difference, which is: #8-x#

The sum of their squares is #34#

#x^2 + color(red)((8-x)^2)= 34" "larrcolor(red)((a-b)^2 = a^2 -2ab +b^2)#

#x^2 + color(red)(64-16x+x^2)= 34#

#2x^2 -16x +64-34 =0#

#2x^2 -16x +30 =0" "larrdiv 2#

#x^2 -8x +15=0" "larr# find factors of 15 which add to 8.

#(x-3)(x-5)=0#

Setting factor equal to #0# gives:

#x = 3 and x=5#

Check:

#3+5 = 8#

#3^2 +5^2 = 9+25 = 34#

Mar 23, 2017

Answer:

Smallest number is #3#.

Explanation:

Let #a# and #b# be the numbers concerned.

We are told that:

#a+b= 8 -> b=8-a# (A)

#a^2+b^2 = 34# (B)

(A) in (B):#-> a^2+(8-a)^2 = 34#

#a^2 + 64 -16a +a^2 =34#

#2a^2 -16a +30 = 0#

#a^2 -8a+15=0#

#(a-3)(a-5)=0 -> a = +3 or +5#

Hence the smallest # a=3#

#a=3# in (A) #-> b= 8-3 = 5#

Hence the two numbers are 3 and 5, where 3 is the smallest.