Quick algebra problem?
equation for height of ball h= -16x^2+38x+8. at what time (x) well the ball reach a height of 20 feet. i know u set up the equation like this 20=-16x^2+38x+8...but then what?
equation for height of ball h= -16x^2+38x+8. at what time (x) well the ball reach a height of 20 feet. i know u set up the equation like this 20=-16x^2+38x+8...but then what?
Comments
..but then what?
Then, bring everything to the LHS leaving a zero on RHS, and simplify to get
16x^2-38x+12 = 0
Now factorise the LHS
16x^2 -32x -6x + 12 = 0
16x(x-2)-6(x-2) = 0
(16x-6)(x-2) = 0
x = 3/8 or x = 2.
They are the answers.
So you can take all the right hand side terms over to the left hand side (to make the "x^2" term positive), giving:
16x^2 - 38x + 12 = 0.
Simplifying, you can divide everything by 2:
8x^2 - 19x + 6 = 0.
Now you need to find two numbers whos product is 48 ( 8 x 6) and whose sum is -19.
-16 and -3 work.
Divide 19x up into -16x and -3x:
8x^2 - 16x - 3x + 6 = 0
Grouping the terms in twos, it can be seen that they can be easily factorised:
8x(x - 2) - 3(x - 2) = 0
So, (8x - 3)(x - 2) = 0
The two solutions for x are therefore:
(x - 2) = 0, so x = 2 seconds
and
(8x - 3) = 0, so x = 3/8 seconds
One solution is for when the ball is on it's way up and the other when it's on its way down (obviously if the ball is being thrown in the air then 3/8 seconds is for up and 2 seconds is for down).
You need to set up a quadratic equation with it being
16x^2 -38x +12= 0
then use the quadratic rule
well you're trying to find x so move the '20' over the other side and solve the quadratic so...
0=16x^2+38x-12
-this isn't factorisable so you'll have to put it into the quadratic formulae
( -b +or - square root of b^2 - 4ac ) / 2a
a=16
b=38
and c=-12
just plug the numbers into the equation
find x
solve the quadratic eqn
x= -b(+or-)root(b^2-4ac)/2a
a=16
b= -38
c=12