Checking For Extraneous Solutions of Radical Equations

This video explains how to check for extraneous solutions when solving a radical equation.
Square Roots and Cube Roots: • Square Roots and Cube ...
Simplifying Radical Expressions:
• Simplifying Radicals
Multiplying Radical Expressions:
• Multiplying Radical Ex...
Adding & Subtracting Radicals:
• Adding and Subtracting...
Simplifying Cube Root Expressions:
• How To Simplify Cube R...
__________________________________
Rationalizing The Denominator:
• Rationalize The Denomi...
Multiplying Radicals With Different Indices:
• Multiplying Radical Ex...
How To Solve Radical Equations:
• Solving Radical Equations
Solving Equations With Cube Roots:
• Solving Equations With...
Solving Complex Radical Equations:
• How To Solve Complex R...
____________________________________
How To Graph Radical Functions:
• How To Graph Radical F...
Domain of Radical Functions:
• How To Find The Domain...
How To Graph Cube Root Functions:
• Graphing Cube Root Fun...
Radical Expressions - Test Review:
• Simplifying Radical Ex...
____________________________________
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[@TheOrganicChemistryTutor]

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MR. Organic Chemistry tutor, thank you for another exceptional video/lecture on Checking for Extraneous Solutions of Radical Equations. Extraneous solutions are created by squaring both sides of an equation.

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[@plasticpaper8725]

Lets goo extraneous solutions. I needed this

[@GOGEDIT]

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[@utilizator1701]

For more extraneous solutions do the equation type a*sin(x)+b*cos(x)=c (I think this is the equation with extraneous solutions; it can be made without checking by hand, but that method requires deep understanding of trigonometry).

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[@exegetor]

My algebra is error-prone if I do not take a LONG time fussing over it... too often I have missed the existence of alternative solutions. So my habit now is to graph in this case y=sqrt(x-2) and y=x-4. Only after this convinces me to only look for one solution, do I then start cranking the algebraic gears. It's a compensating measure, I suppose

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[@JulianShagworthy]

Can anybody explain WHY extraneous solutions exist in the medium of mathematics? Specifically, at what moment does the extraneous solution become introduced? I remember reading somewhere that there are three tests to determine if a mathematical system 'works', and one of the tests is that it isn't contradictory as a language. And yet... sqrt(x+2) + 4 = x can be proven to be equal to x^2 - 9x + 14 = 0 by rearrangement. How can the same value have two different AMOUNTS of answers (1 vs 2?)? They obviously do, but my cognitive centre just can't work it out.

[@anthonyzanghi1992]

I believe it has to do with the fact that the square root of a number can be positive or negative. i.e. Sqrt(4) is equal to both -2 and 2.

[@vilsspam]

if an equation covers a domain (range of x-values) of all real numbers, then all solutions should work out. however, some equations don't cover a domain of all real numbers. for example, sqrt(2) only covers a domain of 0 to infinity, basically the domain consists only of anything greater than or equal to 0. hope this helps

[@esnesnommoc9025]

It's because of square root. Let's say sqrt(A) = B. Then that means (sqrt(A))^2 = B^2. However, (sqrt(A))^2 = B^2 does NOT necessarily mean sqrt(A) = B, it can also mean -sqrt(A) = B. When you derive your answers from (sqrt(A))^2 = B^2 instead of the original equation, you're actually deriving the solutions to both cases of sqrt(A) and -sqrt(A).
Case in point, if the original equation were instead -sqrt(x+2) + 4 = x, then x = 2 would be the actual solution and x = 7 would be the extraneous solution. Try it yourself.
Another way to eliminate extraneous solutions is to declare conditions for x. If sqrt(x+2) = x - 4, because square root is always bigger than or equal to 0, that means x - 4 must also be bigger than or equal to 0, which means x must be bigger than or equal to 4. That eliminates the extraneous solution of x = 2.

[@utilizator1701]

I think is has to do with the fact that x^2 is not an injective (one to one) function.

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[@LarrysLandFin]

You could also find out the range what the x cannot be before proceeding. What is inside the square root cannot be negative, and the result cannot be negative as well. Both conditions must satisfy. Based on the law of even roots.

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[@vedanthsampath1073]

For the first one, You can assume that whatever is inside the square root will be a whole number because if it was a decimal inside the numbers of that decimal would never be the same as it's square root, then all u have to do is try a couple of numbers and you'll find that x=7 pretty quick

[@wandererrrrrr]

would you explain isotonic solutions?

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He can't do that here, unless you join the membership program.

[@CarlyDayDay]

What if you state that x-4 > 0? Would that eliminate the extraneous solution right away?

[@utilizator1701]

Yes, but a "more" correct condition is x-4>=0.

[@abdullahahmed1673]

if you know the answer from the start , then yes , but in a normal case no because x could be 4 or any other number

[@abdullahahmed1673]

and there is somthing else ...if you are in an exam he can give you that info if he wants a certain solution

[@CarlyDayDay]

@@abdullahahmed1673 I meant because of the radical being nonnegative

[@utilizator1701]

@@abdullahahmed1673 What? Then explain for which x

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[@malekabbr]

At the equation: sqrt(2x)=x-4
You got 8 and 2 as the solutions
And when you plugged x = 8
You got:
4=4 and that's true
But when plugging x = 2
You got:
2=-2
But the square root of 4 is either 2 or - 2
So x = 2 was also a right answer

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[@pingpong3311]

For the second problem, the square root of 4 equals 2, but it can also equal -2, because (-2)×(-2) also equals 4. So would x=2 be a solution?

[@krislegends]

The cannot take the principal square root of a negative number. You can, however, take the negative square root of a negative number when solving for an unknown.

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[@RikyPerdana]

I don't agree that the 2 in the last result called extraneous. Because the root of 4 can also be -2, right?

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Why are you giving it too long solution??
And why aren't you using the formulas??
Like (a±b)² = a² + b² ± 2ab
🤔

[@thomashenderson2936]

if you take the negative outcome of the square root of 4 than x=2 is not an extraneous solution?! Where is the logic!

[@legendaryspryzen4489]

9:36
Here, 2 can also be the solution of this equation, if you take √4= -2... Bcoz 4 can be the square of both 2 and -2. In that case, LHS=RHS.
Correct me if I am wrong.

[@GOGEDIT]

Genius idea

[@utilizator1701]

No math teacher will accept that answer (I know from experience), because sqrt(x)>=0 by definition.
If that square root was multiplied with -1, x=2 would indeed be a solution.

[@naytte9286]

This is wrong. A square root is per definition positive.

[@legendaryspryzen4489]

@@naytte9286 sqr root of a no is written like ±√a, so I think it can be, but it depends, if you take the √4=2, it is not a solution, and if u take √4= -2, it can be the solution, am I right ?

[@naytte9286]

@@legendaryspryzen4489 no the square root of a number is not written that way. That' just plus minus the square root. Just the square root is strictly positive.

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