1/3 times square root of (2x) = 2 – MANY solve Radical Equations WRONG!

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[rdspam]

7:59 By definition, the √ symbol as used in a problem statement represents the principal/positive root. While x^2=4 has two solutions, the given equation X=√4 only has one solution, x=2. Do you not teach the quadratic formula with +/- the root of the discriminant? It’s because the √ of the discriminant is defined as positive and you want both roots for the quadratic formula. There is no mathematical need for this extra work for this problem.

[danpost]

In the example, at @7:42, you are "rooting" the square; while in the problem, you are squaring the "root" -- doing the operations in reverse. Your example does introduce an extraneous solution; but the problem does NOT. Point in fact is the problem has a restricted domain to non-negative values to begin with (the value inside the radical must be non-negative, providing you are not solving for complex solutions).

[derwolf]

Alternatively, you can use equivalent transformations only, so you don't have to check the result:
(1/3) sqrt(2x) = 2 | * 3/sqrt(2)
<=> sqrt(x) = sqrt(18) | - sqrt(18)
<=> sqrt(x) - sqrt(18) = 0 | * (sqrt(x) + sqrt(18)); note (sqrt(x) + sqrt(18)) > 0
<=> x - 18 = 0 | + 18
<=> x = 18

[RBMusicCH]

If we allow x = -18 the solution would be an imaginary (i) number but not 2. Hence there is only one solution however you twist it.

[johnnyragadoo]

Hmmm... There's an alternate path with my new friends, the logs.

1/3 sqrt(2x) = 2
1/3 (2x)^1/2 = 2
1/2 log(2x) = log(2)-log(1/3)
log(2x)=2 (log(2) - log(1/3))
2x=10^(2 log(2) - 2 log(1/3))
x = 10^(log2) - 2 log(1/3))
x = 18

Wasn't it more fun that way? 🙂

[SaucyCinemaFilm]

Multiply both sides by 3 and solve for the square root of (6^2). Or solve for 2√9 if you like the scenic route.

[craigsaunders]

Not quiet sure why the extra step is needed if you already know that you can't find the square root of a negative, or right from the start you know that the square root of 2x is only positive x values. Also, if you're going to be this pedantic, don't you have to substitute -18 into the equation to show that the answer isn't -18. Being penalized because you know the difference between the square root sign being over the algebraic unknown (x) as apposed to the munerical side of the equation doesn't make sense.
PS. Love your work, most of the time. 😁

[kylekatarn]

Two equations are equivalent if they have the same set of solution. There is a theorem that shows that, if you have an equation of this type: square root of [A(x)] = B(x), where a A(x) and B(x) are polynoms, then this equation is not (always) equivalent to A(x) = B(x)^2. When you find the solutions of this second equation, you have to be sure that they are the same for the first one and so you will have to accept the only solutions that verify the first equation.

[davidpatton]

This seems like overkill since in the square root you have 2x, a positive value of 2 and the equation is looking for a positive 6 on the other side. A positive solution can never have a mixed positive and negative in the square root so this seems really to be unnecessary. Anyone with reasonable math skills should be able to deduce this just by looking at the equation.

[hexbinoban]

1/3√(2x) = 2 <=> 3*1/3√(2x) = 2 * 3 <=> √(2x) = 6 <=> {√(2x)}² = (6)² <=> 2x = 36 <=> x = 36/2 = 18
ANS : x = 18
Test by substitution : 1/3√(2(18) = 1/3√(36) = 1/3 * 6 = 2

[gilbertarnold-percy]

given: 1/3sqert(2x) = 2.... square both sides yeilding : (1/9)(2x) = 4 → 2x = 4(9) → 2x = 36 → x =36/2 → x =18 and you don't have worry about negative numbers

[johnnolen]

Mr. Z, you don't need to verify the solution to this equation by plugging it back into the original equation because there is only one possible answer. If there were more than one possible answer then you would need to verify each possibility.

Also (x, +/- sqrt[x]) is not a function. (x, sqrt[x]) is a function whose domain and range are each restricted to the positive real numbers.

sqrt[x^2] = |x| is the definition of the absolute value function. Hence for example when x^2 = 4, and you square root both sides, you get |x| on the left equal to 2 on the right, |x| = 2. There are two numbers whose absolute value is 2: +2, and -2; that's why you must include both solutions when solving x^2 - 4 = 0.

Though for the problem in question: sqrt[2x]/3 = 2, we can simplify it as sqrt[x]/3 = sqrt[2]. Multiplying both sides by 3, we obtain sqrt[x] = 3·sqrt[2]. Squaring both sides gives x = 9 × 2 = 18. Again x cannot be negative; otherwise, it couldn't live under the radical. ◼

[Herlongian]

1/3 of 6 is 2. So square root of 2x is 6. 2x is 36. x is 18

[rgrif]

You checked the answer, but did you check and validate the extraneous answer had it been negative back in the earlier step?

[tomtke]

(1/3)(sqrt(2x))=2

3( (1/3)(sqrt(2x))=2 )
Sqrt(2x)=6

( Sqrt(2x)=6 )^2
2x = 36

(1/2)( 2x = 36 )
x = 18

VERIFY

x=18
(1/3)(sqrt(2x))=2
(1/3)(sqrt(2•18))=?2
(1/3)(6)=?2
2=❤2✔️

Alternately
(1/3)(sqrt(2x))=2
(sqrt((2x)/9))=2
(2x)/9=4
x = 4×(9/2)
x = 18

[tcmxiyw]

sqrt(x^2)=4 means |x|=2, or x=+/-2. Raising both sides of an equation to a power, or multiplying both sides of an equation by an expression involving the unknown may introduce solutions that don’t satisfy the original equation. It is not being “pedantic” to check each solution in the original equation—it is not just “checking your work.” His point is somewhat obscured because his example is so simple. Equations are not always so simple.

[RustyWalker]

1/3√2x=2
1/3√2x=2

1/3 1/3

√2x=6
2x=6^2
x=18
As for -18, I've read the comments but not watched the solution yet.

[johnnyragadoo]

Hmm... If it's a given the product of a positive number and a square root is a real number, not a complex number, do you really need to verify the negative root? In this case, if the right side of the equation had been 2i instead of 2, the answer would have been x=-18 and couldn't have been 18.

Or so I understand at my current limited understanding.

[bigdog]

super easy let y = (2x) ^ 1/2.
rewrite:
y/3 = 2

multiply both sides by 3 to isolate the y then substitute.
y = 6
(2x) ^ .5 = 6

Next square both sides then solve for x
2x = 36
x = 18

Now we must prove our answer by substituting x with 18
1/3(2 * 18) ^ .5 = 2
1/3 (36) ^ .5 = 2
1/3 * 6 = 2
6/3 = 2
2 = 2

This proves that x does in fact equal 18

[KennethLongcrier]

Square root of a negative is i so I discarded that answer without bothering to prove...