Verifying and understanding answer

[Oreo]

Greetings:

I want to verify my work answer for the below SSAT question.

Thank you

Cindy throws a rock in the air. The rock's height, y, in feet, with respect to time, x, in seconds can be modeled by the function y=-2x^2+10x+28. When does the rock hit the ground in seconds

My work:
y=-2x^2+10x+28
0= -2(x^2-5x-14) (While knowing that the zero is supposed to be there, I don't understand why 0= -2(x^2-5x-14) )
(x-7)(x+2)
(-7,+2)
I put 2 as the answer, but when checking it was 7?

What part did I do wrong?

 

[BigBeachBanana]

(While knowing that the zero is supposed to be there, I don't understand why 0= -2(x^2-5x-14) )

The rock hits the ground when the height is 0 feet above the ground, thus you set y=0.

(x-7)(x+2)
(-7,+2)
I put 2 as the answer, but when checking it was 7?

\(\displaystyle y=-2x^2+10x+28\\ 0=(x-7)(x+2)\)
Your work is correct up until here.
Now, for the right-hand side equal to 0, one of the factors must be 0. So either [imath](x-7)=0 \implies x=7[/imath] or [imath](x+2)=0 \implies x=-2[/imath]. Since x is time in seconds, it we omit the negative answer.

 

[Steven G]

-2x^2+10x+28 = -2(x^2 - 5x - 14)
All that was done was a -2 was factored out.

 

[blamocur]

From the domain of irrelevant comments: -2 is not completely meaningless root either. If someone wanted to throw another rock from the ground and match the trajectory of the first rock in time and space (e.g. second rock refueling the first one) they would have to have done this exactly 2 seconds earlier.