Assignment: Solve Radical Equations with Extraneous Solutions
Choose any four (4) of the five radical equations below. Solve each equation, showing all steps
leading to your answer. Then check your answer to determine if any of the solutions are
extraneous.
1.
x
1 5
x
√(x + 1) + 5 = x
√(x + 1) + 5 - 5 = x - 5
√(x + 1) = x - 5
(√(x + 1))^2 = (x - 5)^2
x + 1 =5(x
2.
x - 5)(x
x -05)
x + 1 = x^2 - 10x +25
0 - 10x + 25 - x
x√(5x)
+ 1 -- xx == x^2
√(5x)
x
+
x
+x
1 = x^2 - 11x=+025
√(5x)
=
x
1 - 1 = x^2 - 11x + 25 - 1
(√(5x))^2
= x^2
0
= x^2 - 11x
+ 24
5x
=
x^2
(x
8)(x
3)
=
3.
2 x 1 01 x
xx^2
- 8 -=5x
0=0
√(2x
+
(x)(x
5)
x - 8 + 81)=+01+=8x
x√(2x
=8
0 + 1) + 1 - 1 = x - 1
x√(2x
-3
5 =+01) = x - 1
x(√(2x
-3
5 ++3
51))^2
= 0 +=3
5 (x - 1)^2
2x
+
1
=
(x
1)(x
x
=
5
4. 3 3 x 2 3- 1)4 x
2x + 1 = x^2 - 2x + 1
Checking:
√(3x
2) +
3= =
4x
2x ++
√(5
* 1-0)
√(8
1)- -2x
+05=
=0x^2
8 - 2x + 1 - 2x
√(3x
2)
+
3
3
= 4x - 3
1 = x^2
-0
4x
√(0)
- 05==
√(9)
+
8 +1
√(3x
2)
=
4x
3
==0x^2
01 +
-- 015=
3
8 - 4x + 1 - 1
(√3x
-TRUE
2))^2
=
(4xsolution
- 3)^2
0
=
x^2
0
=
0
-real
8 8 - 4x-real
solution
3x
2
=
(4x
3)(4x
x^2*-25)
4x4
√(5
- =5 0=25x 2- 3)x
5.
3x
2
=
(x)(x+--1)
4)
√(25)
516x^2
=50
5= 3- 24x + 9
√(3
+=
3x
2
3x
=
5x -=50+= 55 = 3 16x^2 - 24x + 9 - 3x
√(4)
+9
x+
- =4516x^2
==
03 - 27x
0-2
=
FALSE
-extraneous
solution
2
-2
+
2
=
16x^2
27x
+9+2
x
4
+
4
=
0
+
4
7 = 3 FALSE -extraneous solution
0
= 416x^2 - 27x + 11
x=
16x^2
- 27x + 11 = 0
Checking:
(x
1)(16x
11)
= 00
√(2 * 0 + 1)- +
1=
x√(0
- 1+=1)0 + 1 = 0
x√(1)
- 1+
+ 11 =
= 00 + 1
x1 =
1
+1=0
16x
= 0 -extraneous solution
2 = 0- 11
FALSE
16x
11
0+
√(2 * 4 ++1)11
+=
1=
4 11
16x
=
11
√(8 + 1) + 1 = 4
16x/16
√(9) + 1==11/16
4
x3 =
+ 0.6875
1=4
Checking:
4 = 4 TRUE -real solution
√(3 * 1 - 2) + 3 = 4 * 1
√(3 - 2) + 3 = 4 * 1
√(1) + 3 = 4 * 1
1+3=4*1
4=4*1
4 = 4 TRUE -real solution
√(3 * 0.6875 - 2) + 3 = 4 * 0.6875
√(2.0625 - 2) + 3 = 4 * 0.6875
√(0.0625) + 3 = 4 * 0.6875
0.25 + 3 = 4 * 0.6875
3.25 = 4 * 0.6875
3.25 = 2.75 FALSE -extraneous solution
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