+0

# HELP QUICK

0
216
6

The sum of the squares of two nonnegative numbers is 327. The product of the two numbers is 101. What is the sum of the two numbers?

Jun 16, 2021

#1
0

The sum of the two numbers is 25.

Jun 16, 2021
#2
0

The sum of the squares of two nonnegative numbers is 327. The product of the two numbers is 101. What is the sum of the two numbers?

I don't think this will work.  The only factors of 101 are 1 and 101.  If you square 101 you're already larger than 10,000 so the "sum of the squares" part of the problem isn't satisfied.

Jun 17, 2021
#3
-1

x^2+y^2 = 327

x+y = 101

y = 101-x

x^2 + (101-x)^2 = 327

Solve for x and then substitute the value of x into the equation and solve for y

I hope this helps!!!

Jun 17, 2021
#4
0

In theory, the approach is good, but as I said above, I don't think this will work.

x2 + (10,201 – 202x + x2)  =  327

2x2 – 202x + 10,201  =  327

2x2 – 202x + 9874  =  0

x2 – 101x + 4937  =  0

4937 is prime so we can't factor this.

The alternative is to use the quadratic formula.

I'll spare you the gory details, but you end up with a negative number under the radical.

I don't know where to go from here, mnkmnk.  Have I made a mistake?

Guest Jun 17, 2021
#5
+2

This question has illusion of complexity because it’s intuitive for many mathematicians to find the values of the two (2) numbers (a) and (b), but the question only asks for their sum.

Solution:

\small \text{ }\\ \begin{align*} \small x^2 + y^2 &= 327\\ \hspace{1em} \small x + y &= 101\\ \end{align*} \;\\\;\\ \text {Note that }\\ \left(x+y\right)^2 \;\; = \quad x^2 + y^2 +2xy \hspace{3ex}| \scriptsize \text { perfect square expansion formula }\\ \text{then}\\ \left(x + y\right)^2 \;\; = \quad 327 +2(101)\\ \left(x + y\right)^2 \;\; = \quad 529\\ \text{then }\\ \sqrt{\left(x + y\right)^2} \;\; = \quad \sqrt{529}\\ \hspace{1.5em}{\left(x + y\right)} \;\; = \quad23 \; \hspace{2em} \leftarrow \small \text{sum of (x) and (y)}

GA

Jun 17, 2021
#6
+1

Nice answer Ginger Melody  Jun 17, 2021