Now square both sides of the equation, and move everything to the right side of the equation:
[â(3x + 1)]² = (x - 3)²
(3x + 1) = x² - 6x + 9
0 = x² - 9x + 8.
Since a = b, then b = a. That allows us to flip the equation above around:
x² - 9x + 8 = 0.
The left side easily factors:
(x - 1)(x - 8) = 0.
Set each factor equal to zero and solve for x for each factor:
x - 1 = 0 ---> x = 1
x - 8 = 0 ---> x = 8.
To check the solutions, plug them into the original equation:
Let x = 1:
3 + â(3x + 1) = x
3 + â[3(1) + 1] = 1
3 + â(3 + 1) = 1
3 + â4 = 1
3 + (-2) = 1
3 - 2 = 1
1 = 1.
x = 1 actually works, if we consider that every real number actually has both a positive and negative square root. However, if your book or instructor teaches you to only use the principal square root of a number, which is its positive value, then x = 1 will not work, because 3 + 2 = 5, not 1.
Let x = 8:
3 + â(3x + 1) = x
3 + â[3(8) + 1] = 8
3 + â(24 + 1) = 8
3 + â25 = 8
3 + 5 = 8
8 = 8.
x = 8 works when we use the principal square root of 25, which is positive. If we use the secondary square root of 25, which is -5, then x = 8 does not work, since 3 - 5 = -2, not 8.
Answers & Comments
Verified answer
3 + SQR(3x + 1) = x
subtract 3 from each side:
SQR(3x + 1) = x - 3
Square both sides:
3x + 1 = (x - 3)^2
3x + 1 = (x - 3)(x - 3)
3x + 1 = x^2 - 6x + 9
x^2 - 9x + 8 = 0
using quadratic formula:
x = [9 +/- SQR(81 - 4(1)(8))]/2(1)
x = [9 +/- SQR(49)]/2
x = (9 +/- 7)/2
x = 1, 8
- .--
First isolate the square root:
â(3x + 1) = x - 3.
Now square both sides of the equation, and move everything to the right side of the equation:
[â(3x + 1)]² = (x - 3)²
(3x + 1) = x² - 6x + 9
0 = x² - 9x + 8.
Since a = b, then b = a. That allows us to flip the equation above around:
x² - 9x + 8 = 0.
The left side easily factors:
(x - 1)(x - 8) = 0.
Set each factor equal to zero and solve for x for each factor:
x - 1 = 0 ---> x = 1
x - 8 = 0 ---> x = 8.
To check the solutions, plug them into the original equation:
Let x = 1:
3 + â(3x + 1) = x
3 + â[3(1) + 1] = 1
3 + â(3 + 1) = 1
3 + â4 = 1
3 + (-2) = 1
3 - 2 = 1
1 = 1.
x = 1 actually works, if we consider that every real number actually has both a positive and negative square root. However, if your book or instructor teaches you to only use the principal square root of a number, which is its positive value, then x = 1 will not work, because 3 + 2 = 5, not 1.
Let x = 8:
3 + â(3x + 1) = x
3 + â[3(8) + 1] = 8
3 + â(24 + 1) = 8
3 + â25 = 8
3 + 5 = 8
8 = 8.
x = 8 works when we use the principal square root of 25, which is positive. If we use the secondary square root of 25, which is -5, then x = 8 does not work, since 3 - 5 = -2, not 8.