Extraneous roots of the equation sqr root (3x+2) = 3x

Archie Deetoo

New member
Joined
Mar 11, 2018
Messages
5
The question is :

Solve the given equation:

sqr root (3x+2) = 3x

I squared both side and then set to zero and ended up with:

(3x+1)(3x-2)=0

x = -1/3, 2/3

Now because I squared both sides in solving, I need to check for extraneous roots.

I checked x = 2/3 and got:

sqr root (3 (2/3) +2)) ?=? 3(2/3)
sqr root (4) = 2 : which is correct, so that is a correct root,

Now I checked x=-1/3:

sqr root (3(-1/3) +2)) ?=? 3(-1/3)
sqr root (1) = -1. That is also correct because square root of 1 is positive 1 and negative 1. So it appears both are correct and not extraneous solutions, yet the answer is 2/3 only. Why is that?

I have graphed the equation and can see that 2/3 is the correction solution and I can see by graphing that the extraneous solution is introduced when I square both sides when solving it. It seems that square root of 1 is only positive 1 on this occasion which is why -1/3 is extraneous.

I suppose what this comes down to is why is sometimes the square root of a number a positive answer and sometimes a positive and negative answer depending on the problem?

I appreciate any help with this.

Thanks a lot.

Zen
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
3,045
The question is :

Solve the given equation:

sqr root (3x+2) = 3x

I squared both side and then set to zero and ended up with:

(3x+1)(3x-2)=0

x = -1/3, 2/3

Now because I squared both sides in solving, I need to check for extraneous roots.

I checked x = 2/3 and got:

sqr root (3 (2/3) +2)) ?=? 3(2/3)
sqr root (4) = 2 : which is correct, so that is a correct root,

Now I checked x=-1/3:

sqr root (3(-1/3) +2)) ?=? 3(-1/3)
sqr root (1) = -1. <---- THIS IS THE ERROR
That is also correct because square root of 1 is positive 1 and negative 1. So it appears both are correct and not extraneous solutions, yet the answer is 2/3 only. Why is that?

I have graphed the equation and can see that 2/3 is the correction solution and I can see by graphing that the extraneous solution is introduced when I square both sides when solving it. It seems that square root of 1 is only positive 1 on this occasion which is why -1/3 is extraneous.

I suppose what this comes down to is why is sometimes the square root of a number a positive answer and sometimes a positive and negative answer depending on the problem?

I appreciate any help with this.

Thanks a lot.

Zen
When we write a radical, like \(\displaystyle \sqrt{1}\), it refers to only the principal root, the non-negative one. (It's defined that way so that the square root is a function.) That's why, when you graph the radical, you see only positive values.

To answer your specific question, when we take a square root ourselves, in the course of solving a problem, we need both roots (positive and negative), because the square of either is the same. But when we write (or read) a radical, that refers to only the principal root. If you want to write an expression that represents both roots, you have to use a plus-or-minus sign. Recall the quadratic formula? That's why it's there. And that's why, if you solve \(\displaystyle x^2 = 3\), the solution is not just \(\displaystyle \sqrt{3}\), but \(\displaystyle \pm\sqrt{3}\).
 

Archie Deetoo

New member
Joined
Mar 11, 2018
Messages
5
Thank you for your help. Good explanation.

Zen
 

tkhunny

Moderator
Staff member
Joined
Apr 12, 2005
Messages
9,787
The question is :

Solve the given equation:

sqr root (3x+2) = 3x
It's also important to remember the Domain. In the Real Numbers, when you write \(\displaystyle \sqrt{3x+2}\), you must necessarily also mean that \(\displaystyle 3x+2 \ge 0\). With a little algebra, we see that this means \(\displaystyle x \ge -2/3\). Having said that, if we get any supposed solutions that are less then -2/3, we can just discard them. No need to check those. Just throw them out. You should still check the other values, just in case you missed some Domain consideration along the way.

Example of slightly different issue: \(\displaystyle \sqrt{3x-2} = -4\). You may be tempted to try to solve this. Nope! That square root is NEVER negative. No solution. Done. Sometimes, a little thinking about existence can save you a lot of algebra.
 

sinx

Junior Member
Joined
Mar 1, 2018
Messages
206
The question is :

Solve the given equation:

sqr root (3x+2) = 3x

I squared both side and then set to zero and ended up with:
(3x+1)(3x-2)=0
x = -1/3, 2/3
Now I checked x=-1/3:
sqr root (3(-1/3) +2)) ?=? 3(-1/3)
sqr root (1) = -1. That is also correct because square root of 1 is positive 1 and negative 1. So it appears both are correct and not extraneous solutions, yet the answer is 2/3 only. Why is that?
unless x is restricted to positive numbers only,
-1/3 is not an extraneous root.
sqr root (3x+2) = 3x;
before you factor, the eqn. is
9x2-3x-2=0
when you put x=-1/3 into this equation you get zero, so
-1/3 should not be an extraneous root.
If it is, i am also confused.
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
3,045
unless x is restricted to positive numbers only,
-1/3 is not an extraneous root.
sqr root (3x+2) = 3x;
before you factor, the eqn. is
9x2-3x-2=0
when you put x=-1/3 into this equation you get zero, so
-1/3 should not be an extraneous root.
If it is, i am also confused.
Check your answer in the original equation, not in the one you derived from it!

The left hand side is sqrt(3(-1/3)+2) = sqrt(1) = 1 [the principal root].
The right hand side is 3(-1/3) = -1.
These are not equal, so -1/3 is not a root.
Since you found it as a root of an equation derived from the original, we call it an extraneous root.

What happens is that when you squared both sides of the equation, you lost information about signs; in solving 9x2-3x-2=0, you are actually solving both sqrt(3x+2) = 3x and sqrt(3x+2) = -3x, since both yield the same quadratic equation. It is necessary to check (or use other information such as domains or ranges) to determine whether each root you find is a solution of the intended equation, or of the other.
 
Top