ex. 6 use the discriminant to determine the number of real solutions of the quadratic equation
DESCRIPTION
Discriminant: b 2 – 4ac used to determine the number of real solutions for quad. Equation 1) if discriminant is POSITIVE, two different solutions 2) if discriminant is ZERO, one repeated solution 3) if discriminat is NEGATIVE, no real solutions, no x-int. - PowerPoint PPT PresentationTRANSCRIPT
Ex. 6 Use the discriminant to determine the number of real solutions of the quadratic equation.
a. x2 + 4x – 32 = 0
b. 2x2 + 5x + 8 = 0
c. 2x2 + 12x +18 = 0
144, so 2 solutions
-39, so no real sol.
Discriminant: b2 – 4ac
used to determine the number of real solutions for quad. Equation
1) if discriminant is POSITIVE, two different solutions
2) if discriminant is ZERO, one repeated solution
3) if discriminat is NEGATIVE, no real solutions, no x-int.
0, so 1 sol.
Ex. 7 A room is 3 ft longer than it is wide and has an area of 154 ft2. Find the dimensions of the room.
w + 3
w
w(w + 3) = 154
w2 + 3w – 154 = 0
(w – 11)(w + 14) = 0
w = 11, - 14 width = 11 ft and length = 14 ft
Ex. 8 A construction worker on the 24th floor of a building (235 ft above ground) accidentally drops a wrench and yells “Look out below!” Could a person at ground level hear this warning in time to get out of the way? (speed of sound is about 1100 ft/sec)
s = - 16t2 + vot + so
0 = -16t2 + 0t + 235
16t2 = 235
t2 = 235/16
Person hears warning less one second after wrench is dropped. Plenty-o-time.
t = 3.83 sec
Ex. 9 From 2000 to 2007, the estimated number of Internet users, I (in millions) in the US can be modeled by the quadratic equation
I = -1.163t2 + 17.19t + 125.9 0 ≤ t ≤ 7
In which year will the number of users reach or surpass 180 million?
I = -1.163t2 + 17.19t + 125.9
180 = -1.163t2 + 17.19t + 125.9
0 = -1.163t2 + 17.19t – 54.1
)163.1(2
)1.54)(163.1(419.1719.17 2
2.105.4236.2
82.4319.17or
Since it must be between 2000 and 2007, the 10.2 does not count so it must be 4.5 or during the 2004 year.
Ex. 10 An L-shaped sidewalk was constructed so that the length of one sidewalk forming the L was twice as long as the other side. The length of the diagonal sidewalk that connects the other two is 32 feet long. How many feet does a person save by walking on the diagonal sidewalk?
x2 + (2x)2= 322
2x
x
32x2 + 4x2 = 1024
5x2 = 1024
x2 = 204.8
x = 14.3 ft
so distance around the “L” is 3(14.3) = 42.9 ft
Someone saved 10 ft going diagonally.