# Math Help - Pre-Cal Help

1. ## Pre-Cal Help

Consider |(x^2)-4|=c
A: Find A value of c for which this equation has four soluions
b: Find a value of c for which this equation has three soultions
c: Find a value of c for which this equation has two solution
d:Find a value of which this equation has no solutions
e are there any possble numbers of solutions of this equation?

2. What work have you done?

'd' is trivial. You should be able to get this one in under three heartbeats. It is an absolute value!

'b' is interesting. If the absoltue value were not there, could you find the minimum value? This is a really big hint.

After those two, a, c, and e shouldn't be too difficult.

Can you graph the function?

3. Hello, Rimas!

TKHunny has the best approach . . . graph it.

Consider: . $|x^2 -4| \:=\:c$

A) Find a value of c for which this equation has four soluions.
B) Find a value of c for which this equation has three soultions.
C) Find a value of c for which this equation has two solutions.
D) Find a value of which this equation has no solutions.
E) Are there any possble numbers of solutions of this equation?

$y \:=\:x^2-4$ is an up-opening parabola.
. . It has x-intercepts (±2, 0) and y-intercept (0, -4).

The absolute value says: anything below the x-axis is "reflected upward".

So the graph looks like this:
Code:
                      |
*          |          *
|
4*
*     *  |  *     *
*    |    *
* *     |     * *
|
- - - - -*- - - + - - -*- - - -
-2      |      2

Now cut the graph with a horizontal line $y \,=\,c$
. . and see where you get 4, 3, 2, or 0 intersections.

4. Originally Posted by Rimas
Consider |(x^2)-4|=c
A: Find A value of c for which this equation has four soluions
b: Find a value of c for which this equation has three soultions
c: Find a value of c for which this equation has two solution
d:Find a value of which this equation has no solutions
e are there any possble numbers of solutions of this equation?
If $c<0$ then this equation got no solutions. Because $|x^2-4|\geq 0$.

If $c=0$ then $x^2 - 4 = 0$ and so $x=\pm 2$ has two solutions.

If $c>0$ then $x^2 - 4 = \pm c \implies x^2 = 4 \pm c$. Now if $4 - c<0$ then there is no solution and the only solution we get is for $4+c$ which is $\pm \sqrt{4+c}$. So two solutions, in this case $c>4$. And if $4-c=0$ then you get three solutions: $0,\sqrt{4+c},-\sqrt{4+c}$. And finally if $4-c>0$, i.e. $c<4$ you get $\pm \sqrt{4-c},\pm \sqrt{4+c}$.

5. dor the info but i am stuck on e any help?

6. Since you have the graph, simply look at it. One can see 2, 3, 4, and 0. Can there be a 1? Can there be a 5?