Question

The difference of two positive numbers is 3 and the sum of their squares is 225.What are the numbers?

Answer 1

Let's say one number is x and the other number is y.

If we take our first statement: "The difference of two positive numbers is 3 " we can rewrite this as:

x - y = 3 because "difference" implies we use subtraction.

The second statement tells us: "sum of their squares is 225." rewriting this, we would have:

x^2 + y^2 = 225 "sum" means we add the two numbers, and it states that they are squares.

So now we have:

x - y = 3
x^2 + y^2 = 225

We can use the method of substitution to solve one variable for the other.

Answer 2

Solving for y in the first equation, we would have :

-x + x - y = 3 - x
-y = 3 - x
(-1) (-y) = (-1)(3 - x)

y = x - 3 (distributing the negative)

Answer 3

Now that we have one equation of y in terms of x, we can substitute this into the second equation to solve for the other variable.

x^2 + (x-3)^2 = 225

x^2 + x^2 -6x +9 = 225

2x^2 - 6x + 9 = 225

2x^2 - 6x + 9 - 225 = 0

2x^2 - 6x - 216 = 0

2(x^2 -3x -108) = 0
2(x-12)(x+9) = 0
x = 12 , -9

Answer 4

thx

Answer 5

now we have to see which value of x will satisfy the two equations.

If we take x = 12, we'll see that

y = 12 - 3 = 9

Therefore y = 9, x - 12

12 - 9 = 3

(12)^2 + (9)^2 = 225

Answer 6

If we try x = -9 , we will get

y = -9 - 3 = -12

-9 - 12 =/= 3

(-9)^2 + (-12)^2 = 225

therefore x = -9 does not satisfy both equations.

i'm sure there are other easier methods to go about doing this, this was the first one I thought of! :P

Answer 7

Since I'm lazy, I will do an alternate method. To not confuse you with what she's written I'll call the numbers a and b.

So I'm saying

a>b>0 since they are both positive and

a-b=3 is our first statement

Now using the statement a^2+b^2=225 I noticed this: \[\LARGE a^2+b^2=15^2\] So using the triangle inequality we can safely say that \[\LARGE a+b \ge 15\] Now adding this inequality to our original statement and subtracting it as well we get two approximately good values for what a and b can be:

\[\LARGE (a+b)+(a-b) \ge (15)+(3) \\ \LARGE 2a \ge 18 \\ \LARGE a \ge 9\] \[\LARGE (a+b)-(a-b) \ge 15-3 \\ \LARGE 2b \ge 12 \\ \LARGE b \ge 6\] Since the first equation says \[\LARGE a=b+3\] we know the only possibilities are \[\LARGE a=9, b=6 \\ \LARGE a=10, b=7 \\ \LARGE a=11, b=8 \\ \LARGE etc...\] so for a^2+b^2=225 obviously we know that 9^2 and 6^2 aren't enough to get to 200, but they are nearby so it's a good starting place. You can now avoid solving any quadratic formulas and you should feel confident that you'll get to the right answer after a few successive guesses haha.