Solving for X with Radicals

[thejabronisayz]

Hi everyone, I need help solving for X in the following equation. I keep arriving at: 576x^2 - 1824x +1369 = 0, but since there is no GCF I am assuming that this is wrong. Any help would be greatly appreciated, thanks!

5x^(1/2) - (x-1)^(1/2) = 6

In other words, "5 [x radical 2] minus [radical (x minus 1)] equals 6."

In other words again: 5 times the square root of X, minus the square root of 'X minus one', is equal to 6.

 

[Mrspi]

Hi everyone, I need help solving for X in the following equation. I keep arriving at: 576x^2 - 1824x +1369 = 0, but since there is no GCF I am assuming that this is wrong. Any help would be greatly appreciated, thanks!

5x^(1/2) - (x-1)^(1/2) = 6

In other words, "5 [x radical 2] minus [radical (x minus 1)] equals 6."

In other words again: 5 times the square root of X, minus the square root of 'X minus one', is equal to 6.

Your equation is correct....have you considered using the quadratic formula to find the value(s) of x? No one ever guaranteed that all quadratics are factorable; in fact, most of them are NOT nicely factorable over the integers. But the quadratic formula ALWAYS works, even if the values of x do not happen to be in the realm of the real numbers.

Give it a try...and remember to check your results in the original equation, since squaring both sides of an equation may well introduce extraneous roots.

 

[tkhunny]

Try a perhaps unexpected substitution, x = y^2, but it's not a whole lot more straight-forward. Just try it again and be more careful. How did you go about it? I expect you isolated a radical and squared. You may have overlooked that you could stop after one application.

When you got this far, assuming you took the first step that I did: \(\displaystyle 24x - 60\sqrt{x} + 37 = 0\)

...were you tempted to isolate and square again or did you realize that you could apply the quadratic formula directly?

\(\displaystyle \sqrt{x} = \frac{60 \pm\sqrt{60^{2}-4\cdot 24\cdot 37}}{2\cdot 24}\)

By the way, your first notation was perfect. There was no need to make two more attempts to explain it. Good work.

 

[thejabronisayz]

I actually took Mrspi's advice and just used the quadratic formula for the equation I got myself. Using that I got x=1.9442 and that works in the original problem. Thanks for all the quick replies/help!

 

[lookagain]

\(\displaystyle 1) > > \)OBVIOUS \(\displaystyle < < \)way with ugly \(\displaystyle 2) > > arithmetic. < < \)



Not so \(\displaystyle 1) > > \)obvious \(\displaystyle < < \)way with easier \(\displaystyle 2) > > \)arithmetic. \(\displaystyle < < \)

1) This is a subjective word. What it is to you, it may not be to someone else.


2) What you showed is not arithmetic. It is algebra.

But that should be "obvious."

...but since there is no GCF...

There is a greatest common factor, and it is 1.

 

[lookagain]

Originally Posted by lookagain
1) This is a subjective word. What it is to you, it may not be to someone else. I'm writing it so it is my subjective state that is expressed. Moreover, if you were to give this problem to a random sample of students, I would be amazed if a substantial plurality did not attack the problem in just this way. Once you conduct an objective experiment with a large enough random sample that shows most students attack it a different way, you may have an objective basis for criticizing my subjective opinion. Until then, I stand by what I wrote.

\(\displaystyle \text{Pluralities are irrelevant here. The mathematical steps}\)
\(\displaystyle \text{are not motivated by what is popular.}\)
\(\displaystyle \text{Then you're standing by wrong thinking! The word "obvious,"}\)
\(\displaystyle \text{as well as dozens of other subjective words, do not belong in }\)
\(\displaystyle \text{mathematical discussions as they relate to the strategies of }\)
\(\displaystyle \text{working out exercises.}\)


2) What you showed is not arithmetic. It is algebra. Indeed, which leads to simpler arithmetic.
\(\displaystyle \text{That is changing the subject, as well as it is backpedaling by you.}\)

But that should be "obvious." Clearly, the implied conclusion was not obvious to you.

No, your implied conclusion is irrelevant.

\(\displaystyle \text{The issue is you owning up to calling it the wrong thing, instead of}\)
\(\displaystyle \text{making excuses and going off on tangents to support your errors.}\)

\(\displaystyle \text{"Clearly" is *another* one of those words in math that is not to be used.}\)
\(\displaystyle \text{You are not understanding how to be honest with me and to the }\)
\(\displaystyle \text{students to whom you are attempting to help here. }\)
\(\displaystyle \text{That's a big disservice done by you.}\)

\(\displaystyle \text{If you don't care about being accurate and don't care about making}\)
\(\displaystyle \text{excuses for a given wrong content, then consider not posting about it.}\)


I will be vigilant on these threads to help clear up errors, such as the type pointed
out about yours in this thread, so as to help the threads not be hijacked with silly
non-mathematical reasoning and its excuses for such.​

Reply With Quote

 

[Deleted member 4993]

Another non-obvious ugly way:

\(\displaystyle 5\sqrt{x} - \sqrt{x-1} = 6\)

let

\(\displaystyle x = sec^2(\theta)\)

\(\displaystyle 5*sec(\theta) - tan(\theta) = 6\)

\(\displaystyle 6*cos(\theta) + sin(\theta) = 5\)

6/√37*cos(Θ) + 1/√37*sin(Θ) = 5/√37

\(\displaystyle cos(.165149)*cos(\theta) + sin(.165149)*sin(\theta) = cos(\pm 0.60589)\)

0.165149 - Θ = ± 0.60589

Θ = 0.77104 or -0.44074 → cos(Θ) = 0.717186 or 0.904435 → x = 1.944177 or 1.222489

First answer checks - but the second one does not.

OOOps.... I should have said the first answer is in correct domain .... or some such thing......

 

[lookagain]

I shall use what words I want when responding to
\(\displaystyle > > \)rudeness. \(\displaystyle < < \)\(\displaystyle <--- Mispalced defensiveness
\(\displaystyle \text{Then you may use the words on yourself, lest you be a hypocrite.}\)

I did not nominate you to be my editor.
\(\displaystyle \text{*That* is an irrelevant statement. If you can't handle being corrected}\)
\(\displaystyle \text{for major gaffes in mathematics on math forum boards, then so be it.}\)

If the moderators object to what I am doing, then they can tell me so.
\(\displaystyle \text{*Responsible* math forum users have a duty to keep the information}\)
\(\displaystyle \text{accurate, as well as the forum user calling errors on himself, }\)
\(\displaystyle \text{much as in golf with penalties being called on the players by viewers,}\)
\(\displaystyle \text{other players, or the golfer himself who needs to take the penalty.}\)

Frankly I think it better to respond to requests to help students than to argue about
what words satisfy your standards of proper discourse.
\(\displaystyle > > \)Buzz off.\(\displaystyle < < \)\)

\(\displaystyle

\(\displaystyle \text{Frankly, if you're going to continue to be stubborn about passing on }\)
\(\displaystyle \text{mathematical falsities, then helping the student is a moot point.}\)
\(\displaystyle \text{Please lead, follow, or get out of the way in accurately helping}\)
\(\displaystyle \text{students with these difficult concepts.}\)

\(\displaystyle \text{You have my extreme thanks in advance.}\)\)

 

[lookagain]

When you got this far, assuming you took the first step that I did: \(\displaystyle 24x - 60\sqrt{x} + 37 = 0\)



\(\displaystyle \sqrt{x} = \dfrac{-(-60) \pm\sqrt{(-60)^{2}-4\cdot 24\cdot 37}}{2\cdot 24}\)

thejabronisayz,

make sure you properly substitute values into the Quadratic Formula, commonly
with parentheses, else you'd be fudging.

I actually took Mrspi's advice and just used the quadratic
formula for the equation I got myself. Using that I got x=1.9442
and that works in the original problem.
​

It's closer to x = 1.9441 for that particular solution.
You should have mentioned there is another x value that comes
out of using the Quadratic Formula. It is about 1.2226.
And then mention whether it checks into the original equation or not.
​