After posting all this work, in both formats, my teacher still wants to play games and is asking me what number should be paired with -5 and what number should be paired with 2. What in the world is she talking about?

1. The product of two numbers is 10. One number is 3 more than the other number. What is the quadratic equation that models these numbers?

x(x + 3) = 10
x^2 + 3x - 10 = 0

2. Solve the quadratic equation from #1, to find the pairs of numbers.

x = -(3) +_ SQRT((3)^2) - 4(1)(-10)/2(1)

= -3 +_ SQRT(9 + 40)/2 = -3 +_ SQRT(49)/ 2

= -3 +_7/2 = -3 - 7/2, -3 + 7/2

= -10/2, 4/2 = -5, 2

x = -5, x = 2

(b + 3)b = 10

b^2 + 3b = 10

b^2 + 3b -10 = 0

x = -(3) +_ SQRT((3)^2 - 4(1)(-10)/2(1)

= -3 +_ SQRT(9 + 40)/2 = -3 +_SQRT(49)/2

= -3 +_ 7/2 = -3 -7/2, -3 +7/2

= -10/2, 4/2

= -5, 2

x = -5, 2

I mean, this just seems rediculous

2. There are two solutions to x (x+3) = 10.

one is -5, one is 2.

If you take one of the x's as 2, you see that $\displaystyle 2 \times (2+3) = 2 \times 5 = 10$. The two numbers are 2 and 5.

But also, you see you have -5 as the other solution.

And indeed, you see that $\displaystyle -5 \times (-5 + 3) = -5 \times -2 = 10$.

So that's the other pair: -5 and -2.

3. Originally Posted by bobbyboy1111
After posting all this work, in both formats, my teacher still wants to play games and is asking me what number should be paired with -5 and what number should be paired with 2. What in the world is she talking about?

1. The product of two numbers is 10. One number is 3 more than the other number. What is the quadratic equation that models these numbers?

x(x + 3) = 10
x^2 + 3x - 10 = 0

If you solve the quadratic equation you find x=-5 and x=2.
So if $\displaystyle x=-5 \Rightarrow x+3 = -5+3 = -2$
and if $\displaystyle x=2 \Rightarrow x+3 = 2+3 = 5$

The pairs -5, -2 and 2, 5 give product 10

4. And you should thank this "pest of a teacher" for requiring you to think about what you are doing rather than just apply rote formulas!