SOLUTION: Show EACH step in solving each of the following equations. State if any solution is extraneous. a. sqrt(3x + 1) = 4 b. 2x - sqrt(3x - 2) = 8 c. sqrt(2x - 1) = x - 8

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Question 39246: Show EACH step in solving each of the following equations. State if any solution is extraneous.
a. sqrt(3x + 1) = 4
b. 2x - sqrt(3x - 2) = 8
c. sqrt(2x - 1) = x - 8

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
a. sqrt(3x + 1) = 4
Square both sides to get:
3x+1=16
3x=15
x=5
b. 2x - sqrt(3x - 2) = 8
Rewrite as sqrt(3x-2)=2x-8
Square both sides to get:
3x-2=4x^2-32x+64
4x^2-35x+66=0
Solve for "x" as follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=169 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 6, 2.75. Here's your graph:

The solutions x=6 or x=2.75 are both valid.
c. sqrt(2x - 1) = x - 8
Square both sides to get:
2x-1=x^2-16x+64
x^2-18x+65=0
Solve for "x" as follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=64 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 13, 5. Here's your graph:

The solutions x=13 or x=5 are both valid.
Cheers,
Stan H.
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