# 2 problems involving the real numbers

• Dec 20th 2010, 07:52 PM
sarahh
2 problems involving the real numbers
1. If $a$ and $b$ are real numbers, and $a > b$ and $b < 0$, then which of the following inequalities must be true?
A. $a > 0$
B. $a < 0$
C. $a^2 > b^2$
D. $a^2 < b^2$
E. $b^2 > 0$
(Turns out it is E but not sure why)

2. What is the smallest possible value for the product of two real numbers that differ by 6? (Without using derivatives and/or max/mins)

Why is it -9??
• Dec 20th 2010, 07:53 PM
dwsmith
Any real number squared that isn't 0 is greater than 0
• Dec 20th 2010, 08:17 PM
sarahh
Ahh yes of course, just by taking examples too :) But what about the second problem? Hmmm...
• Dec 20th 2010, 08:20 PM
Prove It
2. Two numbers differ by $\displaystyle 6$. Call them $\displaystyle x$ and $\displaystyle x + 6$.

Their product is $\displaystyle x(x + 6) = x^2 + 6x$

$\displaystyle = x^2 + 6x + 3^2 - 3^2$

$\displaystyle = (x+3)^2 - 9$.

This is a quadratic with minimum at $\displaystyle (-3, -9)$. So the smallest possible value is $\displaystyle -9$.
• Dec 20th 2010, 08:27 PM
dwsmith
Quote:

Originally Posted by Prove It
2. Two numbers differ by $\displaystyle 6$. Call them $\displaystyle x$ and $\displaystyle x + 6$.

Their product is $\displaystyle x(x + 6) = x^2 + 6x$

$\displaystyle = x^2 + 6x + 3^2 - 3^2$

$\displaystyle = (x+3)^2 - 9$.

This is a quadratic with minimum at $\displaystyle (-3, -9)$. So the smallest possible value is $\displaystyle -9$.

I was wondering about that one. I took that same approach and wasn't sure what do to with the parabola.
• Dec 20th 2010, 08:34 PM
Soroban
Hello, sarahh!

Quote:

2. What is the smallest possible value for the product of two real numbers
that differ by 6? (Without using derivatives and/or max/mins)

We have two numbers that differ by 6.

Let $\,x$ = larger number.
and $x-6$ = smaller number.

Their product is: . $P \:=\:x(x-6) \:=\:x^2-6x$

The graph is an up-opening parabola: . $\,\cup$
Its minimum point is at its vertex.
The vertex is at: . $x \:=\:\dfrac{\text{-}b}{2a} \:=\:\dfrac{\text{-}(\text{-}6)}{2(1)} \:=\:3$

Hence: . $P \:=\:3^2 - 6(3) \:=\:-9$

The vertex of the parabola is $(3,\,\text{-}9)$

Therefore, the minimum $\,P$ is $-9.$