# Thread: 2 problems involving the real numbers

1. ## 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??

2. Any real number squared that isn't 0 is greater than 0

3. Ahh yes of course, just by taking examples too But what about the second problem? Hmmm...

4. 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 $\displaystyle -9$.

5. 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.

6. Hello, sarahh!

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 $-9.$