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?

You were correct in setting the equation equal to 0 because the problem is asking for when the rock hits the ground and in this problem, you may assume that the ground is y=0, unless the problem says otherwise.

Ok so where you went wrong is that when you factored out and got

(x-7)(x+2), which is correct, you forgot to set these equal to zero (because your actual equation is set to zero). So you should have

x-7 = 0

and

x+2 = 0

and this gives the actual answers +7 and -2. And since time cant be negative its 7 seconds.

This was a silly mistake and happens a lot more often than you think, so I wouldn't worry too much but make sure you practice more, better to lose points on harder questions than easy ones.

Good luck with your exam !

semsee45 semsee45 · Answer 2 · 2022-06-25T21:37:30+02:00

Answer:

7 seconds

Step-by-step explanation:

Given function:

[tex]y=-2x^2+10x+28[/tex]

where:

y = height above the ground (in feet)
x = time (in seconds)

To find the time when the rock hits the ground, set the equation to zero and solve for x.

First, simplify by factoring out the common term -2:

[tex]\implies -2(x^2-5x-14)=0[/tex]

Divide both sides by -2:

[tex]\implies x^2-5x-14=0[/tex]

To factor a quadratic in the form [tex]ax^2+bx+c[/tex], find two numbers that multiply to ac and sum to b.

[tex]\implies ac=1 \cdot -14=-14[/tex]

[tex]\implies b=-5[/tex]

Therefore, the two numbers that multiply to -14 and sum to -5 are: -7 and 2

Rewrite b as the sum of these two numbers:

[tex]\implies x^2-7x+2x-14=0[/tex]

Factorize the first two terms and the last two terms separately:

[tex]\implies x(x-7)+2(x-7)=0[/tex]

Factor out the common term (x - 7):

[tex]\implies (x+2)(x-7)=0[/tex]

Therefore:

[tex]\implies x+2=0 \implies x=-2[/tex]

[tex]\implies x-7=0 \implies x=7[/tex]

As time cannot be negative, the rock hits the ground in 7 seconds.