Question

two gymnasts are running toward each other in a floor routine, and they plan to precisely time a flip to synchronize for the audience. the path of the gymnasts is parabolic and modeled by the following equations, where y is the height of the flip and x is the time in seconds:
( y = -(x - 5)^2 + 3 )
( y = -3(x - 6)^2 + x + 1 )
to make sure the flip is in unison, after how many seconds should the flip occur?
2 seconds
3 seconds
4 seconds
5 seconds

Explanation:

Step1: Set equations equal

$$-(x-5)^2 + 3 = -3(x-6)^2 + x + 1$$

Step2: Expand both sides

$$-(x^2-10x+25)+3 = -3(x^2-12x+36)+x+1$$
$$-x^2+10x-25+3 = -3x^2+36x-108+x+1$$

Step3: Simplify both sides

$$-x^2+10x-22 = -3x^2+37x-107$$

Step4: Rearrange to standard quadratic form

$$2x^2-27x+85=0$$

Step5: Solve quadratic equation

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=2$, $b=-27$, $c=85$:
$$x=\frac{27\pm\sqrt{(-27)^2-4(2)(85)}}{2(2)}$$
$$x=\frac{27\pm\sqrt{729-680}}{4}$$
$$x=\frac{27\pm\sqrt{49}}{4}$$
$$x=\frac{27\pm7}{4}$$
Solutions: $x=\frac{27+7}{4}=\frac{34}{4}=8.5$ and $x=\frac{27-7}{4}=\frac{20}{4}=5$
The relevant positive solution for a routine is $x=5$, as 8.5 is likely beyond the routine timeline.

Answer:

5 seconds