1. Sep 2, 2009

### musicmar

1. Solve:
l2x^2+7x-15l<10

3.

I split it up into two cases and got:

2x^2+7x-25<0 and 2x^2+7x-5>0

With the quadratic formula I got:

x= [-7+/- (249)^(1/2)]/4 and x= [-7+/- (89)^(1/2)]/4

These clearly are the answers, but I am confused about the inequality from this point.

2. Sep 2, 2009

### VietDao29

Well, no, you don't split up in 2 cases like that:

Here's a general way for you to solve equation of the type:
|f(x)| < g(x)

Since |f(x)| cannot take negative value (|f(x)| >= 0), and for the inequality (0 <=) |f(x)| < g(x) to hold (we are solving for x that makes the inequality holds), we must have g(x) > 0 (1).

Note that if |A| < B, for some positive B, then -B < A < B.

Applying that here, we have: -g(x) < f(x) < g(x). (2)

From (1), and (2), we obtain:

$$|f(x)| < g(x) \Rightarrow \left\{ \begin{array}{c} g(x) > 0 \\ -g(x) < f(x) < g(x) \end{array} \right.$$

Using the same reasoning, one can obtain:

$$|f(x)| \leq g(x) \Rightarrow \left\{ \begin{array}{c} g(x) \geq 0 \\ -g(x) \leq f(x) \leq g(x) \end{array} \right.$$

In your problem, g(x) = 10 > 0, which is a constant, so, this should be easy.

$$|2x ^ 2 + 7x - 15| < 10 \Rightarrow -10 < 2x ^ 2 + 7x - 15 < 10$$.

Can you go from here? :)

Last edited: Sep 2, 2009
3. Sep 2, 2009

### musicmar

Thank you. However, this is one of the concepts that I should have learned in 8th grade, but never did. I followed your response completely, but I don't know where to go past
5 < 2x^2 + 7x < 25

Does it help to factor it originally into (x-1.5)(x+5) ?

4. Sep 2, 2009

### njama

Now just split them

$$5<2x^2+7x$$

&&

$$2x^2+7x<25$$

5. Sep 2, 2009

Thanks.

6. Sep 2, 2009

### Elucidus

These lead to the original two cases the OP arrived at earlier.

Take for example the inequality $2x^2+7x-5>0$. The roots are

$$r_1=\frac{-7-\sqrt{89}}{4},\;r_2=\frac{-7+\sqrt{89}}{4}. \text{ (As was mentioned. Note }r_1<r_2.)$$

The polynomial on the left factors into $2(x-r_1)(x-r_2)$ and this product will only be positive if both factors on the left are positive (i.e. $x > r_1 \;\&\; x>r_2$) or both negative (i.e. $x<r_1 \;\&\; x<r_2$.) A potentially easier way of doing this is to realize that r1 and r2 subdivide the real line into three intervals in which the polynomial will have constant sign. Selecting a test value (even a nebulous one) from each and testing it into the polynmial will give the solution region:

$$\begin{array}{cccc}\text{Interval} & (-\infty,r_1) & (r_1,r_2) & (r_2, \infty) \\ \text{Test} & x<r_1 & r_1<x<r_2 & x>r_2 \\ \text{Sign} & (+) & (-) & (+) \end{array}$$

We conclude that the solution region for this inequality is the intervals $(-\infty, r_1) \cup (r_2, \infty)$.

A similar approach to the other inequality will result in a solution region for it. The solution to the original absolute value inequality would be the intersection of the intervals found from the two cases.

I hope this makes sence and is helpful.

--Elucidus

7. Sep 3, 2009

### musicmar

Thank you. I realized that it just led me to the same two cases I had obtained earlier. I discussed this problem with my dad, and we looked at it graphically. This led to basically the same solution as yours above.