Test corrects for algebra test

Answer the questions of the ones I did wrong correctly for my exam. Show all work. This is due Monday morning

=
b) + = 5
25 -5 (25)
y (25)
o
y²= 125
y = 62.5
4 니
2x
10
=
(4) 2
x+5)
–
x2 -25
x2-25
2
01
xz
– 16
2
2x
2x (87250 (x425)
X²25 x²-25 X 2+25 *225
x 2 + 25
of
dhe
2x=-6
2 2
x=-3
a 뉴
삐
R-d LI-)
-d/
2) (4 points ) Solve for t
(1-2)
–
R=
1-t
1-t=-d
t
R(1-t) = R-d
– –
tt = -d-1
t = -0-2
1
3) (4 points Solve the following equation by the square root method.
니 니
( 2x + 11)2 = 121
20
4x + 121=1212
Чx2: 0
14 na
x²=0
X=0
2
4) (4 points ) Solve the following equation by completing the square
x2 + 8x = -15
b С C
8 + 164-4x²-15
X=-b I Vb²-4ac
2x
8 + 149-49
za
–
X=8VB2.4x -15
2(x)
8 + 149-2
8+847
0 a
olu
2x
847
5) (4 points ) Find the real solutions, if any, of the following equation. Use the quadratic
formula.
X=-10 + VOO-X
x2 – 10x – 1 = 0
b c
a
12x
X = -b – Vb2-4ac
X=-10 V 100-2X
x Х
za
X = -10 + V10² 4(x) (-1)
2(x)
X=-101 V100-40 L-1)
x=-10_195
3
– 10 +98
-10.98
88 &-108
12
2x
6) (4 points each ) Perform the indicated operation, and write each expression in
standard form a + bi.
a) (2 – 4 i) + (7 + 8 )
14+ 16 + 284-321
14+2414 i
M
2ltit-28-3i
b) (3-3i) (7 + i)
– 7+2-3 i
– 7-3 i
1 7+3 i
1_(7-31)
c)
7-31
1504
/ %
35
3
1(7-3i)
17+3 i
ou
7) (4 points ) Solve the equation
”
x2 – 121 = 0
+121
+121
x? 121
X
X = 11
8) ( 5 points each) Find the real solutions of the following equations. tadion Graphite
–
a)
V7x + 13 -2 = 4
2
( (3413x+2197 / 2 (4)?
0
3434+2197-2=16
–
XXH&EN FH
Χτ)
O
343×2195=16
– 2195 – 2195
343x=2179
X=6.35
b) (x + 3)2 + 4 (x + 3) + 3 = 0
V? +9+4+12 +3=0
X – +2x+24=0
a b
듸
4 ± √4²-4(X) (24)
x 2(x)
4²+ V16-4XL24
니
26X)
4Ł V 384-2X
X Х
니 I V394 12
4 t u 382
2
4# 1382 (43U 302)
382
1390 + 374
6 + 382
–
c) 14 – 4t | + 3 = 15
41+4€+3 =15
7 +4 t = 15
4 t = 8
gye
1 t=2
=2
doo
solution set. –
9) (4 points each ) Solve the inequality. Express your answer interval notation. Graph the
– ) /
a) -2 ( 2x – 3) < -10 - Anon 0 10 A AT M 0 212843)> 10 – AX-62-10
4x+6>10
-4XL-4
4x>4
XX )
-41-4
X16
(20) 1) (003.16)
t6
b) -6 |3x – 6 2 -54
t6
34-67-48
19
a
– 14
a
t6
34+62 +48
3×254
3x-62-48
+6
–
3x? -42 10,00
3 3
x < -14 X518 (00318)(00; -14) c) |11 - 2x | > 5
TV
ܩܩܠܐ
11-2275
11+2x>5
3
Il +2 x > 8
– 11
2x) -3
2
2
N/W
2 2
x
2 x>-6
X < 3 X