# Math Help - find x when y = 0

1. ## find x when y = 0

Hi. I have a problem which I think should be pretty easy to solve, but I'm not sure how to go about solving it. If, when x = 1990, y = 25,165,824 when x = 2000, y = 393,216 and when x = 2010, y = 12,288 what is the value of x when y = 0? I want to say this is an example of exponential decay, but I haven't solved a problem like this in quite a while, and I'm just not sure. Thanks for the help,
Risingstar64
Also, if this is the wrong section or title, please feel free to change it, or inform me and I will change it (if I can).

2. I don't think this is an exponential decay.
First, x is incremented by same amount, 10 years, each step. However, y is not reduced by the same factor.
25165824 / 393216 = 64 but
393216 / 12288 = 32
showing that y is not exponential.

Also, if it was some kind of multiplicative reduction, y would never reach 0.

3. Unfortunately, no. The only other information I can provide is that think you are supposed to use excel and a trendline to solve the equation. I appologize, but I have to go somewhere for a few hours, so I won't be able to respond again until around 3:00 pm.

4. If all you are given is three values of (x, y), there is no way to determine x in (x, 0). There are an infinite number of functions, even polynomials that go through three given points- any you can choose one that has any value of x you want for y= 0.

The simplest thing to do is to find the unique quadratic polynomial that goes through those given points.

If $y= ax^2+ bx+ c$ then (1990, 25,165,824) gives $25164824= a(1990)^2+ b(1990)+ c$ when (2000, 393,216) gives $393216= a(2000)^2+ b(2000)+ c$ and (2010, 12,288) gives $12288= a(2010)^2+ b(2010)+ c$. solve those three equations for a, b, and c and solve the resulting equation $ax^2+ bx+ c= 0$.

5. Hi. I just got back, and that seems like a fine solution. I'm not asking for the answer, but how do you solve a quadratic equation for a, b, and c? I only know how to solve for x.

6. Notice the 3 equations written by HallsOfIvy with only a, b, and c.

How do we solve 3 equations with 3 unknowns? Forget about x being quadratic.

7. I'm sorry, but I don't know how to solve 3 equations with 3 unknowns (at least, not when they are that huge). Anyway, I have to go for tonight, but I'll try to figure it out on my own from here. thanks everyone for all the help.

8. Originally Posted by risingstar64
Hi. I just got back, and that seems like a fine solution. I'm not asking for the answer, but how do you solve a quadratic equation for a, b, and c? I only know how to solve for x.
You don't. You solve three linear equations for a, b, and c.

Notice that all of them have just "+ c". Subtracting one of the equations from another eliminates c. Subtracting a different pair also eliminates c and leaves two equations in two unknowns. Solve one equation for either a or b and put it into the other to get a single equation in a single unknown.

9. $\displaystyle
\begin{bmatrix}
1990^2 & 1990 & 1 & 25164824\\
2000^2 & 2000 & 1 & 393216\\
2010^2 & 2010 & 1 & 12288
\end{bmatrix}$

Use elementary row operations--adding/subtracting two rows, multiplying a row by scalar, and multiplying by scalar and adding or subtracting--until you reach the reduced row echelon form.

Spoiler:

$\displaystyle
\begin{bmatrix}
1 & 0 & 0 & \frac{609767}{5}\\
0 & 1 & 0 & \frac{-2445356134}{5}\\
0 & 0 & 1 & 490329246816
\end{bmatrix}$

10. If you don't like those numbers, shift everything by 2000- that is, write your function as $a(x- 2000)^2+ b(x- 2000)+ c$. Tnen you have $a(-10)^2+ b(-10)+ c= 25165824$, $c= 393216$, and $a(10)^2+ b(10)+ c= 12288$. You know c from the middle equation so just subtract it from each side of the other two equations. Then one has "10b" while the other has "-10b". Just add the two equations to get a single equation in a. Since $(-10)^2= 10^2$ you can also subtract the two equations to get a single equation for b.

Of course, you can use an exponential if you want- you just can't use " $y= ae^{bx}$ because then y is never 0. You could use, instead, $y= ae^{bx}+ c$ or, better, $y= ae^{b(x- 2000)}+ c$. Then your equations become
$ae^{-10b}+ c= 25165824$
$ae^{0}+ c= a+ c= 393216$
and $ae^{10b}+ c= 12288$

You can write those as
$ae^{-10b}= 25165824- c$
$ae^{0}= a= 393216- c$
and $ae^{10b}= 12288- c$

Then, say, divide the first equation by the second to get $e^{-10b}= \frac{25165824- c}{393215- c}$ and divide the third by the second to get $e^{10}b= \frac{12288- c}{393215- c}$.
Since $e^{-10b}= \frac{1}{e^{10b}}$ that is the same as
$\frac{25165824- c}{393215- c}= \frac{393215- c}{12288- c}$
which reduces to a quadratic equation for c.