If the sum of two numbers is #4# and their product is #3#, then what is the sum of their squares?

3 Answers
Feb 26, 2017

Answer:

#10#

Explanation:

The two numbers are #1# and #3#, being the two zeros of:

#x^2-4x+3 = (x-1)(x-3)#

In general we find:

#(x-a)(x-b) = x^2-(a+b)x+ab#

Notice that the coefficient of the middle term is #-(a+b)# and the constant term is #ab#.

Given that the two numbers are #1# and #3#, the sum of their squares is:

#1^2+3^2 = 1+9 = 10#

Feb 26, 2017

Answer:

10

Explanation:

First, let's call the two numbers #m# and #n#.

We can then write:

#m + n = 4#
#mn = 3#

We can solve the first equation for #m#:

#m + n - color(red)(n) = 4 - color(red)(n)#

#m + 0 = 4 - n#

#m = 4 - n#

Next, we can substitute #4 - n# for #m# in the second equation and solve for #n#:

#(4 - n)n = 3#

#4n - n^2 - 3 = 0#

#n^2 - 4n + 3 = 0#

#(n - 1)(n - 3) = 0#

Solution 1)
#n - 1 = 0#

#n = 1#

Solution 2)
#n - 3 = 0#

#n = 3#

Substituting these back into the solution to the first equation gives:

Solution 1)

#m = 4 - 1#

#m = 3#

Solution 2)

#m = 4 - 3#

#m = 1#

The two numbers therefore are #3# and #1#.

The sum of their squares is therefore:

#3^2 + 1^2 = 9 + 1 = 10#

Mar 30, 2018

Answer:

#10#

Explanation:

Calling the two numbers #x# and #y#, we are given:

#{ (x+y=4), (xy=3) :}#

and we find:

#x^2+y^2 = x^2+2xy+y^2-2xy#

#color(white)(x^2+y^2) = (x+y)^2-2xy#

#color(white)(x^2+y^2) = 4^2-2(3)#

#color(white)(x^2+y^2) = 16-6#

#color(white)(x^2+y^2) = 10#