The sum of two numbers is 25, the sum of their squares is 313. What are the numbers?

3 Answers
May 26, 2017

#13# and #12#

Explanation:

If #x+y=25# then #y=25-x#.

Substitute:
#x^2+y^2=313#
#x^2+(25-x)^2=313#
#2x^2-50x+312=0#
#x^2-25x+156=0#
#(x-13)(x-12)=0#

#x=13,y=12#
#x=12,y=13#

May 26, 2017

#12# and #13#.

Explanation:

Suppose the two numbers are #a# and #b#

"The sum of two numbers is 25" gives us:

# a+b=25 #

"sum of their squares is 313" gives us:

# a^2+b^2=313 #

From the first equation we have:

# b=25-a#

Substituting for #b# into the second we get:

# a^2+(25-a)^2=313 #

# :. a^2+625-50a+a^2=313 #
# :. 2a^2-50a+312=0 #
# :. a^2-25a+156=0 #
# :. (a-12)(a-13)=0 #
# :. a=12,13 #

We now use the second equation to find the value of #b# corresponding to each solution.

# a=12 => b=25 - 12 = 13 #
# a=13 => b=25 - 13 = 12 #

So there is only one solution which is that the two numbers are #12# and #13#.

May 26, 2017

The numbers are #12# and #13#

Explanation:

Let te numbers be #x# and #y#

therefore we have #x+y=25# ............(1)

which gives us #x^2+y^2+2xy=625# ............(2)

and we also have #x^2+y^2=313# ............(3)

Subtracting (3) from (2), we get #2xy=312# ............(4)

and (4) from (3) we get

#x^2+y^2-2xy=1# ............(5)

i.e. #x-y=1# ............(6)

Slolving for #x# and #y# from (1) and (6), we get

#x=13# and #y=12#

Note:we can also have #x-y=-1#, which gives #x=12# and #y=13#