Solving an Equation

[RJay]

Hi, please anyone can help me with this one?
I have below equation and I want to find x.
x^2 + 263208.9675 = 914.4 sqrt (x^2 + 69537.69)

When I solve above manually I got a wrong answer which is x = 0.222855, -0.566343, -0.222855, 0.566343

Correct answer is x =−517.863875,−203.779553,203.779553,517.863875. How this one was derived?

Please see attached sample.

 

[RJay]

PLEASE DISREGARD THIS THREAD.
Already Found the Problem.

 

[Otis]

… I got a wrong answer which is x = 0.222855, -0.566343, -0.222855, 0.566343

Hello RJ. Without seeing your work, I can't know what went wrong there.

Correct answer is x =−517.863875, −203.779553, 203.779553, 517.863875.
How [was that] derived?

Those values are not accurate to six decimal places, and you'll need to ask the source what they did. However, if the attachment is meant to show their beginning steps, then mistakes arose from round-off error (that is, the coefficients in their 4th-degree polynomial are wrong, due to rounding intermediate results).

The correct values to six places are ±203.779406, ±517.864247

The attachment also mentions the Quartic Formula. Some people find that method tedious by hand, but maybe the source made use of it (perhaps increasing round-off error along the way). Some other approaches are numerical (eg: Newton's Method) or software assisted (which is how I got them accurately). Cheers!

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[Otis]

PLEASE DISREGARD THIS THREAD.
Already Found the Problem.

Ah, we cross posted. Did you use the Quartic Formula? (That might explain why you didn't post your work, heh.)

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[HallsofIvy]

I am going to write your equation as \(\displaystyle x^2+ A= B\sqrt{x^2+ C}\). The first thing I would do is get rid of the square root by squaring both sides of the equation: \(\displaystyle x^4+ 2Ax^2+ A^2= B^2(x^2+ C)= B^2x^2+ B^2C\). And that can be written as \(\displaystyle x^4+ (2A- B^2)x^2+ A^2- B^2C= 0\).

Yes, that is a "quartic" (fourth degree) equation, but it has only even powers of x. Let \(\displaystyle y= x^2\) and we have the quadratic equation \(\displaystyle y^2+ (2A- B^2)+ (A^2- B^2C)= 0\). That can be solved for y using the quadratic formula and then \(\displaystyle x= \pm\sqrt{y}\). Typically, a quadratic equation has two roots and then the \(\displaystyle \pm\) gives a total of 4 solutions to the quartic equation.

 

[RJay]

Ah, we cross posted. Did you use the Quartic Formula? (That might explain why you didn't post your work, heh.)

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I just use the quadratic formula as suggested by Dr. Peterson from the other thread. Then took the square root from the result.
I got wrong in the quadratic equation in which I forgot to multiply 2 by a for the denominator.

Anyway, I got the correct answer now. Thanks for the help.

 

[Otis]

I just use the quadratic formula as suggested by Dr. Peterson from the other thread … Anyway, I got the correct answer now …

Okay. Very good. (I didn't realize that there's another thread associated with this one.) Cheers!