Factor Quadratics
Factor x 2 − 7x − 18
(x-9)(x+2)
Solve x2 − 11x + 19 = −5
{3, 8}
Solve m2 − 5m − 14 = 0
{7, −2}
Solve k2 = 16
{4, −4}
A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function h = −16t2 + 36t + 9. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary.
29.25
Factor p2 − 5 p − 14
(p+2)(p-7)
n2 + 7n + 15 = 5
{−5, −2}
Solve b2− 4b + 4 = 0
{2}
Solve a2 = 4
{2, −2}
A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function h = −16t2 + 36t + 9. How high off the ground did the ball start?
9
Factor 7k2 + 9k
k(7k+9)
6n2 − 18n − 18 = 6
{4, −1}
Solve 2x2 − 3x − 5 = 0
{ 5/2 , −1}
Solve −5x2 = −500
{10, −10}
An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 19.6t + 58.8, where s is in meters. What is the height above the ground when the object is launched?
58.8m
Factor 7x2 − 45x − 28
(7x+4)(x-7)
Solve n2 + 8n = −15
{−5, −3}
Solve 4b2 + 8b + 7 = 4
{− 1/2 , − 3/2}
Solve −7n2 = −448
{8, −8}
An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 19.6t + 58.8, where s is in meters. How long before the object hits the ground after launch?
6 seconds
Factor 5p2 − p − 18
(5p + 9)( p − 2)
Solve 3r2 − 16r − 7 = 5
{− 2/3 , 6}
Solve 5r2 = 80
{4, −4}
Solve n2 − 5 = −4
{1, −1}
An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 19.6t + 58.8, where s is in meters. When does the object reach its maximum height?
2 Seconds