SOLUTION: I already know correct answer to following (is provided in back of book). What I want to know is where my error is so I don't do it again (hopefully). 3x=sqrt((9x+2)/2)); a) cle

Question 862712: I already know correct answer to following (is provided in back of book).
What I want to know is where my error is so I don't do it again (hopefully).
3x=sqrt((9x+2)/2));
a) clear radical in denominator by multiplying right side by (sqrt2/sqrt2): becomes 3x=(sqrt(9x+2)2))/2;
b) distribute 2 under radical: becomes 3x=(sqrt(18x+4))/2;
c) clear fraction by multiplying both sides by (2): becomes 6x=(sqrt(18x+4);
d) 4 under radical resolves to: 6x=2(sqrt18x);
e) square both sides to clear radical: becomes 36x^2=4(18x) or 36x^2=72x;
f) set equation to zero: 0=36x^2-72x;
g) factor out 36: becomes 0=x^2-2x;
h) complete square: x^2-2x+1=1 or (x-1)^2=1
i) take square rt both sides: x-1=+or-1
j) x={0,2}
...which is incorrect. Can't spot my error.
Found 2 solutions by josgarithmetic, richwmiller:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
The best first step is SQUARE BOTH SIDES.

will immediately become,

-

-----simplified quadratic equation.

Can you go through the rest of the process yourself?

-
Note, your step d was wrong.
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
"I already know correct answer to following (is provided in back of book)."
So why don't you tell us?
Often the book answers are wrong for a variety of reasons.
The answer may being in a form that you don't recognize as the same as your answer.

What you do to one side you must do to the other .
Square the right side and square the left side too.
3x=sqrt((9x+2)/2))
9x^2=(9x+2)/2
multiply by 2
18x^2=9x+2
bring everything to the left
18x^2-9x-2=0
solve by factoring
(3x-2)*(6x+1)

Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .

Now multiply the first coefficient by the last term to get .

Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ?

To find these two numbers, we need to list all of the factors of (the previous product).

Factors of :

1,2,3,4,6,9,12,18,36

-1,-2,-3,-4,-6,-9,-12,-18,-36

Note: list the negative of each factor. This will allow us to find all possible combinations.

These factors pair up and multiply to .

1*(-36) = -36
2*(-18) = -36
3*(-12) = -36
4*(-9) = -36
6*(-6) = -36
(-1)*(36) = -36
(-2)*(18) = -36
(-3)*(12) = -36
(-4)*(9) = -36
(-6)*(6) = -36

Now let's add up each pair of factors to see if one pair adds to the middle coefficient :

First Number Second Number Sum
1 -36 1+(-36)=-35
2 -18 2+(-18)=-16
3 -12 3+(-12)=-9
4 -9 4+(-9)=-5
6 -6 6+(-6)=0
-1 36 -1+36=35
-2 18 -2+18=16
-3 12 -3+12=9
-4 9 -4+9=5
-6 6 -6+6=0

From the table, we can see that the two numbers and add to (the middle coefficient).

So the two numbers and both multiply to and add to

Now replace the middle term with . Remember, and add to . So this shows us that .

Replace the second term with .

Group the terms into two pairs.

Factor out the GCF from the first group.

Factor out from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.

Combine like terms. Or factor out the common term

===============================================================

Answer:

So factors to .

In other words, .

Note: you can check the answer by expanding to get or by graphing the original expression and the answer (the two graphs should be identical).

or use the quadratic formula

Solved by pluggable solver: Quadratic Formula

Let's use the quadratic formula to solve for x:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=18, b=-9, and c=-2

Negate -9 to get 9

Square -9 to get 81 (note: remember when you square -9, you must square the negative as well. This is because .)

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and 18 to get 36

So now the expression breaks down into two parts

or

Lets look at the first part:

Add the terms in the numerator

Divide

So one answer is

Now lets look at the second part:

Subtract the terms in the numerator

Divide

So another answer is

So our solutions are:

or

We reject x=-1/6 as extraneous since it doesn't work in the original under the radical
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