Practice Problems, Test 4, MAT 102
(1) Solve using the square-root property:
(a) x 2  25  0
(b) ( x  4)2  5  15
(2) Solve by completing the square:
x 2  6 x  16
(3) Solve using the quadratic formula:
(a) x 2  6 x  9  0
(b) x 2  6 x  10
(c) x 2  8 x  15  0
(4) Find the discriminant, and use it to determine the number and types of solutions of the
quadratic equation - do not solve.
(a) 2 x 2  7 x  3  0
(b) x 2  2 x  2  0
(c) x 2  2 x  1  0
(5) Solve the equation using u-substitution:
(a) x 4  13x 2  36  0
(b) x  2 x  15  0,
 b
(6) Find the y-intercepts, vertex   ,
 2a
function, and graph it:
 b 
f     , and x-intercepts of the Quadratic
 2a  
(a) f ( x)  x 2  6 x  2
(7) Solve the quadratic inequality using a sign graph:
x 2  x  6  0,
x 2  x  6  0,
(8) A rectangular lot is 6 feet longer than it is wide. The area of the lot is 160 feet. Find
the dimensions of the lot.
(9) A projectile is thrown upward from a height of 60 feet with an initial velocity of 50
feet per second. Find the time for the projectile to hit the ground.
(10) X drives a plane to a landing strip 300 miles West and 400 miles North of his
launching point. Assuming he flew straight, find his distance traveled.
(1)(a) x 2  25  0
-25 -25
2
x = -25
x   25,
x  5i
(Isolate x2)
(Square root both sides)
(b) ( x  4)2  5  15
+5 +5
2
( x  4)  20
(Isolate (x - 4)2 )
(Square root both sides)
x  4   20
4 4
x  4  2 5.
(2)
x 2  6 x  16
2
6 6
   
2 2
2
x  6 x  9  25,
( x 2  ax 
2
2
k )
a
a
   
2
2
2
( x  3) 2  25,
x  3   25,
3 3
x  3  5,
x  {2, 8}.
(3) First, you have to memorize the quadratic formula in order to use it:
x
b  b2  4ac
,
2a
(a) x 2  6 x  9  0
(Identify a = 1, b = 6, c = 9 )
  6  (6) 2  4(1)(9) 6  0

 3,
2(1)
2
x  3,
x
is the only solution.
(b) x 2  6 x  10
+10 +10
2
x  6 x  10  0,
(first, get standard form with left = 0)
(I.D. a = 1, b = 6, c = 10)
x 2  6 x  10  0,
  6  (6) 2  4(1)(10) 6  4 6  2i 6 2i
x


  ,
2
2
2
2 2
x  3i
(c) x 2  8 x  15  0
(I.D. a = 1, b = -8, c = 15)
  8  (8)2  4(1)(15) 8  4 4  2
x


 {3,1}.
2(1)
2
2
(4) You need to know that the discriminant is d = b2 - 4ac, then identify a, b, and c, find d
and compare it to zero:
(a) 2 x 2  7 x  3  0
a = 2, b = -7, c = 3:
d = (-7)2 - 4(2)(3) = 25, so
d > 0, and we have two real solutions.
(b) x 2  2 x  2  0
a = 1, b = -2, c = 2:
d = (-2)2 - 4(1)(2) = -4 , so
d < 0, and we have two complex solutions.
(c) x 2  2 x  1  0
a = 1, b = -2, c = 1,
d = (-2)2 - 4 (1)(1) = 0, so
d = 0, and we have one (repeated) real solution.
(5) (a) x 4  13x 2  36  0
u = x2
(middle power)
2
2 2
4
u = (x ) = x ,
(sub in: )
u2 – 13u + 36 = 0,
(u – 9) (u – 4) = 0,
u = 9, u = 4
(re-sub: )
2
2
x  9, x  4,
x   9, x   4,
x  {3, 2}
(b) x  2 x  15  0,
u x
u 2  ( x ) 2  x.
(middle power)
(sub in: )
u 2  2u  15  0,
(u  5)(u  3)  0,
u  5  0, u  3  0,
u  5, u  3,
x  5, x  3,
(square:)
x = 25, x = 9.
Check them out – x = 25, works, but x = 9 doesn’t
(6) f ( x)  x 2  6 x  2
b
6


 3,
2a
2(1)
 b 
f     f (3)  (3) 2  6(3)  2  7.
 2a 
So the vertex is (3,1). Note also that a = 1 is positive which means that the parabola will
point upward. Here is a graph:
(7)
x 2  x  6  0,
x 2  x  6  0,
(solve the corresponding quadratic equation:)
x 2  x  6  0,
( x  3)( x  2)  0,
x  3  0, x  2  0,
x  3, x  2
(then, use these to construct the sign graph: )
+
-3
(x - 3)
(x + 2)
|
-2
0
-
+
+
|
3
4
+
+
so x 2  x  6  0 where the - is: on the interval (-2,3) and x 2  x  6  0 where the +'s are:
on the intervals (, 2] U[3, )
(8)
l
l+6
( l )(l + 6) = 160,
l2 + 6l - 160 = 0,
(l + 16)(l – 10) = 0,
l = -16, l = 10
(note: no negative lengths)
(9) We have to use the formula for the height of the object that we saw in class:
h  16t 2  v0t  h0 , with ho  60, v0  50 :
h  16t 2  50t  60,
16t 2  50t  60  0,
(now, set this equal to 0 – the height of the ground:)
(solve using the Quadratic formula:)
50  (50)2  4(16)(60)
 {.92, 4.05}
2(16)
We can’t have negative time, so the answer is t = 4.05 seconds.
t
(10) We have to use the Pythagorean theorem here:
x
300
400
(300)2 + (400)2 = x2,
250,000 = x2,
x = 500 miles.