For questions 1-2, apply the quadratic formula to find the roots of the given function, and then
graph the function.
1. f(x) = x2 + 4
2. g(x) = x2 + x + 12
0  02  4(1)(4)  16
4i
1) x 

   2i
2
2
2
2) x 
1  12  4(1)(12) 1  47
1
47

 
i
2
2
2
2
Both equations have non real roots
Graph 1)
y
f(x)=x^2+4
18
16
14
12
10
8
6
4
2
x
-5.5
-5
-4.5
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
Graph 2
y
f(x)=x^2+x+12
18
16
14
12
10
8
6
4
2
x
-5.5
-5
-4.5
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
For questions 3-4, factor the quadratic expression completely, and find the roots of the expression.
3. 135x2 - 222x + 91=0
135x2-222x+91=(15x-13)(9x-7), roots are 13/15 and 7/9
4.
6x2 - 42x + 72=0
6x2-42x+72 = 6(x2-7x+12)=6(x-3)(x-4), roots are 3 and 4
For question 5, complete the square, and find the roots of the quadratic equation.
5. x2 – 2x/3 = 0
X2-2x/3+1/9=(x-1/3)2 then x2-2x/3 = (x-1/3)2-1/9
(x-1/3)2-1/9=0 then (x-1/3)2=1/9 then x-1/3 =±(1/3) then x=±(1/3)+1/3
So roots are 0 and 2/3
In questions 6-10, use the discriminant to determine the number of solutions of the quadratic
equation, and whether the solutions are real or complex. Note: It is not necessary to find the roots;
just determine the number and types of solutions.
Let d=discriminant= b2-4ac
6.
7.
8.
9.
10.
2x2 + x - 1 = 0 d=12-4(2)(-1)=9 >0
2 real solutions
2
4/3x2 - 2x + 3/4 = 0 d= (-2) -4(4/3)(3/4)=0
1 real solution
2
2x2 + 5x + 5 = 0 d= 5 -4(2)(5)=-15<0
2 non real solutions
2
3z2 + z - 1 = 0 d= 1 -4(3)(-1)=13>0
2 real solutions
2
m2 + m + 1 = 0 d=1 -4(1)(1)=-3 <0 2 non real solutions
11. A rectangular parking lot is 50 ft longer than it is wide. Determine the dimensions of the
parking lot if it measures 250 ft diagonally.
L=50+w
Using Pythagoras theorem w2+(w+50)2=2502
2w2+100w+2500=62500 then 2w2+100w-60000=0
100  1002  4(2)(60000) 100  700
w

4
4
W is positive solution then w=150 ft, L=200 ft
12. Three consecutive even integers are such that the square of the third is 76 more than the
square of the second. Find the three integers.
Let n,n+2 and n+4 the integers
(n+4)2=76+(n+2)2 then n2+8n+16=n2+4n+80 then 8n+16=4n+80 then 4n=64
Then n=16.
Integers are 16,18 and 20
13. A photo is 3 inches longer that it is wide. A 2-inch border is placed around the photo making
the total area of the photo and border 108in2. What are the dimensions of the photo?
L=Length, W=wide
L=w+3
Dimensions of the photo with the border are
L+4 and W+4
Then area is (L+4)(W+4)=(W+7)(W+4)=108
W2+11w+28=108
W2+11w+28=108 then W2+11w-80=0
w
11  112  4(1)(80) 11  21

2
2
W is the positive solution so w=5 and L=8
Dimentions of the photo are:
Length 8 inches
Wide 5 inches