# Thread: Why can't we solve a quadratic inequality using the multiplication rule?

1. ## Why can't we solve a quadratic inequality using the multiplication rule?

This is a "why" question. For my example I will use:
$x^2-4x-5$
Where we are trying to determine the values of x for which the polynomial is entirely positive and those for which it is entirely negative.

I understand that this equation describes a parabola that intersects the x-axis at (-1, 5). I can graph the curve and determine the answers to the question: The curve is positive for $(-\infty,-1)\cup(5,\infty)$
and negative for $(-1,5)$ I could also have tested interval values (as the textbook suggests) to arrive at the same conclusion.

However, I'm interested in the following algebraic logic:

$x^2\ -\ 4x\ -\ 5\ =\ (x-5)(x+1)$

Therefore, for positive parts of the curve, why can't we say:

$(x-5)(x+1) > 0$ therefore $(x-5)>0$, and $(x+1)>0$, therefore $x>5$ and $x>-1$?

While it's true that the curve is positive for $x>5$ it is not true that the curve is positive for $x>-1$. Why did the multiplicative rule of inequalities not work for this?

Is there another algebraic technique for answering the above question without graphing nor interval testing?

2. ## Re: Why can't we solve a quadratic inequality using the multiplication rule?

Originally Posted by B9766
This is a "why" question. For my example I will use:
$x^2-4x-5$
Where we are trying to determine the values of x for which the polynomial is entirely positive and those for which it is entirely negative.

I understand that this equation describes a parabola that intersects the x-axis at (-1, 5). I can graph the curve and determine the answers to the question: The curve is positive for $(-\infty,-1)\cup(5,\infty)$
and negative for $(-1,5)$ I could also have tested interval values (as the textbook suggests) to arrive at the same conclusion.

However, I'm interested in the following algebraic logic:

$x^2\ -\ 4x\ -\ 5\ =\ (x-5)(x+1)$

Therefore, for positive parts of the curve, why can't we say:

$(x-5)(x+1) > 0$ therefore $(x-5)>0$, and $(x+1)>0$, therefore $x>5$ and $x>-1$?
Because this is not true! If the product of two numbers is positive, it does NOT follow that the two numbers are positive, only that they have the same sign.

If $(x- 5)(x+ 5)> 0$ then either
1) x- 5> 0 and x+ 5> 0. That gives x> 5 and x> -5. Both are true for x> 5
or
2) x- 5< 0 and x+ 5< 0. That gives x< 5 and x< -5. Both are true for x< -5.

While it's true that the curve is positive for $x>5$ it is not true that the curve is positive for $x>-1$. Why did the multiplicative rule of inequalities not work for this?

Is there another algebraic technique for answering the above question without graphing nor interval testing?

3. ## Re: Why can't we solve a quadratic inequality using the multiplication rule?

I agree. I think I made the same point in the next sentence:
"While it's true that the curve is positive for x>5 it is not true that the curve is positive for x>−1."

Upon reflection, the multiplication rule states that, if a > c and b > c then ab > c but the converse isn't true. If ab > c then a > c OR b > c or both. But, a could be less than c OR b could be less than c or both could be less than c. That answers the original question I posed.

The secondary question still stands. Is there any way other than graphing or testing interval values to arrive at the correct answer?

4. ## Re: Why can't we solve a quadratic inequality using the multiplication rule?

?? What I just did did not "involve graphing or testing interval values".

5. ## Re: Why can't we solve a quadratic inequality using the multiplication rule?

I didn't mean to offend. I always appreciate your help. If you recall, sometimes it just takes me a bit more effort than others to comprehend the logic.

I DID see that your example did not use graphing nor interval testing. But, I also don't think the answer you provided satisfied the question, "we are trying to determine the values of x for which the polynomial is entirely positive and those for which it is entirely negative." Put another way, "what are the domains of the positive range and the negative range?"

If we graph the equation you provided, "the positive parts of the curve" would be those lying above he x-axis. Therefore, the true answer would be values of x < -5 and values of x > 5. But like you, when I originally tried to use inequality logic, I came up with x > -5 and x > 5. That's not true.

Note that for some examples like perfect squares, the positive domain for the function $(x^2\ +\ 10x\ +\ 25)\ >\ 0\ \ or\ \ (x\ +\ 5)(x\ +\ 5)\ >\ 0$ is "all values of x" because the function never crosses the x-axis (never goes negative).

So, some examples of the correct answers where the factors are different and the curve crosses the x-axis:
$x^2+6x+5 = (x+5)(x+1)$ Positive domain is x < -5 and x > -1
$x^2-4x-5 = (x-5)(x+1)$ Positive domain is x < -1 and x > 5
$x^2+4x-5 = (x+5)(x-1)$ Positive domain is x < -5 and x > 1
$x^2-6x+5 = (x-5)(x-1)$ Positive domain is x < 1 and x > 5

But everything gets inverted if the parabola is inverted (eg. $-x^2+6x+5$)

Working through this, I don't see a simple general rule to apply to 2 factors of a polynomial to determine the domain of a positive (or negative) range - other than graphing or interval testing. The rule would be so complex that it's easier just to graph the darned thing on a calculator and observe the answer.

Do you agree?

6. ## Re: Why can't we solve a quadratic inequality using the multiplication rule?

The "rule" is that the product of two terms is positive if and only if both terms are of the same sign- both positive or both negative.

If the problem were to find values of x such that $\displaystyle x^2+ 6x+ 5= (x+ 1)(x+ 5)> 0$ then either
(1) x+ 1> 0 and x+ 5> 0
or
(2) x+ 1< 0 and x+ 5< 0.

Then x+ 1> 0 is equivalent to x> -1 and x+ 5> 0 is equivalent to x> -5. Those are both true when x> -1.

x+ 1< 0 is equivalent to x< -1 and x+ 5< 0 is equivalent to x< -5. Those are both true when x< -5.

$\displaystyle x^2+ 6x+ 5> 0$ for all x< -5 and all x> -1.

More generally, a product of any number of terms is positive if and only if the number of negative terms is even.
Now, I will agree that usually the simplest way to visualize this is to, if not graph, at least think about intervals between zeros.

For example, to solve (x+ 5)(x+ 2)(x- 1)(x- 4)(x- 7)> 0, I would note that the zeros are at -5, -2, 1, 4, and 7. There are 5 factors. If x< -5 then, because "x minus a larger number is negative", there are 5 negative factors, an odd number so the product is negative. But for x between -5 and -2, x+ 5 has become positive and we now have an even number of factors so the product is now positive. And then it is clear that the sign alternates with each interval. The product is positive for -5< x< -2, 1< x< 4, and x> 7.

You have to be a bit careful if you have powers of factors. If, for example, the problem were to solve $\displaystyle (x+ 5)^2(x+ 2)^3(x- 1)^4(x- 7)^5> 0$ then the simplest thing to do is just throw out even powers because either positive or negative to an even power is positive. Throwing out even powers here reduces this to $\displaystyle (x+ 2)(x- 7)> 0$ and it is easy to see that this is true for x< -2 (both factors are negative) and for x> 7 (both factors are positive) (except, of course, that the original product is equal to 0 at x= -5. You have to be careful here when you have "$\displaystyle >$" rather than "$\displaystyle \ge$".