#### allegansveritatem

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Equation:

Correct answer:

What I keep getting:

- Thread starter allegansveritatem
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Equation:

Correct answer:

What I keep getting:

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When you square both sides, you have to squareThis seems to be a very simple equation...but somehow the solution is eluding me. I have tried again and again and can't get it to come out right. lIt is easy to see what the answer should be without working it out. But when I try to work it out, Mr Right doesn't show up.

Equation:

View attachment 10826

Correct answer:

View attachment 10827

What I keep getting:

View attachment 10828

The square of \(\displaystyle \sqrt{m+4}+\sqrt{m-4}\) is not \(\displaystyle \sqrt{m+4}^2+\sqrt{m-4}^2\), but \(\displaystyle (\sqrt{m+4}+\sqrt{m-4})^2\). Do you see the difference?

Doing that will not simplify the equation. What you have to do is to isolate one radical (move the other to the right side), and then square each side.

Your textbook should have examples of this.

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\(\displaystyle (a+ b)^2= a^2+ 2ab+ b^2\), not \(\displaystyle a^2+ b^2\).

\(\displaystyle (\sqrt{m+4}+ \sqrt{m-4})^2=\)

\(\displaystyle (\sqrt{m+4})^2+ 2\sqrt{m+4}\sqrt{m-4}+ (\sqrt{m-4})^2\)

\(\displaystyle = m+ 4+ 2\sqrt{(m+4)(m-4)}+ m- 4= 2m+ 2\sqrt{m^2- 16}\).

So the equation, after squaring both sides, is \(\displaystyle 2m+ 2\sqrt{m^2- 16}= 16\). Rewrite this as \(\displaystyle \sqrt{m^2- 16}= 8- m\) and square both sides again to get an equation for m.

Caution: While any solution to the original equation must satisfy the new equation, squaring both sides of an equation can introduce "spurious solutions"- solutions to the new equation that do not satisfy the original equation. Be sure to check any solutions to your final equation in the original equation!

\(\displaystyle (\sqrt{m+4}+ \sqrt{m-4})^2=\)

\(\displaystyle (\sqrt{m+4})^2+ 2\sqrt{m+4}\sqrt{m-4}+ (\sqrt{m-4})^2\)

\(\displaystyle = m+ 4+ 2\sqrt{(m+4)(m-4)}+ m- 4= 2m+ 2\sqrt{m^2- 16}\).

So the equation, after squaring both sides, is \(\displaystyle 2m+ 2\sqrt{m^2- 16}= 16\). Rewrite this as \(\displaystyle \sqrt{m^2- 16}= 8- m\) and square both sides again to get an equation for m.

Caution: While any solution to the original equation must satisfy the new equation, squaring both sides of an equation can introduce "spurious solutions"- solutions to the new equation that do not satisfy the original equation. Be sure to check any solutions to your final equation in the original equation!

Last edited:

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Please learn to experiment with simpler examples. Maybe...This seems to be a very simple equation...but somehow the solution is eluding me. I have tried again and again and can't get it to come out right. lIt is easy to see what the answer should be without working it out. But when I try to work it out, Mr Right doesn't show up.

Equation:

View attachment 10826

Correct answer:

View attachment 10827

What I keep getting:

View attachment 10828

\(\displaystyle 4 + 3 = 7\)

\(\displaystyle 16 + 9 \ne 49\) -- Something went wrong.

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I 2nd that technique.Please learn to experiment with simpler examples. Maybe...

\(\displaystyle 4 + 3 = 7\)

\(\displaystyle 16 + 9 \ne 49\) -- Something went wrong.

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I will look at this more closely in the morning but I see what is being said..and also, I know that squaring this type of problem with all the radicals on one side is not a means to an easy life. In fact I did try to move one of the radical terms to the right but somehow things seemed to end up exactly the same. But I will go at this again tomorrow and keep what you say in mind. Thanks for pointing out the problem. I think I did it this way in the hope of avoiding a messy FOIL.When you square both sides, you have to squareeach entire side-- not justeach termon each side.

The square of \(\displaystyle \sqrt{m+4}+\sqrt{m-4}\) is not \(\displaystyle \sqrt{m+4}^2+\sqrt{m-4}^2\), but \(\displaystyle (\sqrt{m+4}+\sqrt{m-4})^2\). Do you see the difference?

Doing that will not simplify the equation. What you have to do is to isolate one radical (move the other to the right side), and then square each side.

Your textbook should have examples of this.

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Thanks for working through it...I will get back to this in the morning when my brain comes back from the dead. I do know about the fdalse solution pitfall, and have actually fallen into it more than once.\(\displaystyle (a+ b)^2= a^2+ 2ab+ b^2\), not \(\displaystyle a^2+ b^2\).

\(\displaystyle (\sqrt{m+4}+ \sqrt{m-4})^2=\)

\(\displaystyle (\sqrt{m+4})^2+ 2\sqrt{m+4}\sqrt{m-4}+ (\sqrt{m-4})^2\)

\(\displaystyle = m+ 4+ 2\sqrt{(m+4)(m-4)}+ m- 4= 2m+ 2\sqrt{m^2- 16}\).

So the equation, after squaring both sides, is \(\displaystyle 2m+ 2\sqrt{m^2- 16}= 16\). Rewrite this as \(\displaystyle \sqrt{m^2- 16}= 8- m\) and square both sides again to get an equation for m.

Caution: While any solution to the original equation must satisfy the new equation, squaring both sides of an equation can introduce "spurious solutions"- solutions to the new equation that do not satisfy the original equation. Be sure to check any solutions to your final equation in the original equation!

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I know this technique and I assure you I use it all the time. I use 2 and 3 as my dummies of choicePlease learn to experiment with simpler examples. Maybe...

\(\displaystyle 4 + 3 = 7\)

\(\displaystyle 16 + 9 \ne 49\) -- Something went wrong.

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There are no shortcuts to any place worth going. :cool:… I think I did it this way in the hope of avoiding a messy FOIL.

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Kinda sloppy, but correct...you get a pink star for your forehead :razz:Finally:

View attachment 10840

This is not correct.Finally:

View attachment 10840

\(\displaystyle a + b > c \text { DOES NOT GENERALLY ENTAIL } a > b + c.\)

Furthermore, by definition \(\displaystyle \sqrt{x} \in \mathbb R \implies 0 \le \sqrt{x}.\)

So, \(\displaystyle - 3 = \sqrt{m + 4}\) is impossible.

EDIT: You can sometimes get a correct answer through an incorrect process, but do not count on it.

Last edited:

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No, this is not correct. First when you moved sqrt(m+4) to the other side you failed to change the sign. 2nd mistake is that you got a solution to sqrt(m+4) = -3 !! The sqrt of nothing is ever negative. Go back and fix your mistakes.Finally:

View attachment 10840

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Correct??? Go to the corner for a few hours and review your algebra as you are clearly a bit rusty.Kinda sloppy, but correct...you get a pink star for your forehead :razz:

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Five is right. But along the way, two mistakes are made that cancel one another out. Getting the right answer in an invalid way doesn't count as correct.Problem posted: sqrt(M + 4) + sqrt(M - 4) = 4

All I did was substitute his solution M=5 to get sqrt(9)+sqrt(1) = 3+1 = 4

Why is M=5 wrong?

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I'm sorry but this needs to be unpacked a little so a novice like me can make it out.This is not correct.

\(\displaystyle a + b > c \text { DOES NOT GENERALLY ENTAIL } a > b + c.\)

Furthermore, by definition \(\displaystyle \sqrt{x} \in \mathbb R \implies 0 \le \sqrt{x}.\)

So, \(\displaystyle - 3 = \sqrt{m + 4}\) is impossible.

EDIT: You can sometimes get a correct answer through an incorrect process, but do not count on it.

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Yes, I see that now. I mean, I see that I forgot to change the sign when I switched the radical over to the right. That is a huge mistake right there and yet...the answer came out right. So....I must either have done something else that cancelled that error or maybe the nature of this type of problem is such that flukes of this sort are, if not likely, yet still possible. Anyway, tomorrow I will have another lunge at this beast, cuz it ain't dead yet!No, this is not correct. First when you moved sqrt(m+4) to the other side you failed to change the sign. 2nd mistake is that you got a solution to sqrt(m+4) = -3 !! The sqrt of nothing is ever negative. Go back and fix your mistakes.