h = -0.5x² + 0.5x + 15 where x = distance from stump and h = height
We want to find the value of x when h = 10.92.
10.92 = -0.5x² + 0.5x + 15
There are many ways to solve a quadratic like this, let's start by subtracting 10.92.
0 = -0.5x² + 0.5x + 4.08
Now we have everything on one side. Let's solve for x.
Since these numbers aren't whole, I wouldn't try to factor/split the middle.
You could either complete the square or use the quadratic formula.
Completing the square
-0.5x² + 0.5x + 4.08 = 0
We need our x² coefficeint to be 1, so let's multiply by -2.
x² - x - 8.16 = 0
Move the constant to the other side.
x² - x = 8.16
Halve the x coefficient -1 to get -0.5, then square it to get 0.25.
Add this to each side.
x² - x + 0.25 = 8.41
Now we have a perfect square trinomial on the left we can factor...
(x-0.5)² = 8.41
Take the square root of each side...
x-0.5 = ±2.9
Add that 0.5 to each side...
x = 0.5±2.9
Of course, the frog can't be a negative distance x from the stump, so we have to take the positive value x = 3.4.
Using the quadratic formula
-0.5x² + 0.5x + 4.08 = 0
I'm going to multiply everything by 100 just to get rid of any decimal points.
-50x² + 50x + 408 = 0
To use the quadratic formula:
The solution to any quadratic ax² + bx + c is \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\).
In this case, a = -50, b = 50, and c = 408. Let's input these into our formula.
\(x=\frac{-50\pm\sqrt{50^2-4(-50)(408)}}{2(-50)}=\frac{-50\pm\sqrt{2500+81600}}{-100}=\frac{-50\pm{84100}}{-100}\)
\(=\frac{-50\pm290}{-100}=\frac{240}{-100}\ or\ \frac{-340}{-100}=-0.24\ or\ 0.34\)
Of course, the frog can't be a negative distance x from the stump, so we have to take the positive value x = 0.34.
