Solving a Cubic Equation: How to Find the Real Root

  • Thread starter Thread starter xzibition8612
  • Start date Start date
  • Tags Tags
    Root
AI Thread Summary
To solve the cubic equation 2t^3 - 4t - 5 = 0, the quadratic formula is not applicable as it is designed for quadratic equations. The cubic formula is complex, so alternative methods like the rational root theorem can be used. There is confusion regarding the proposed solution of sqrt(2/3), which is incorrect. It is essential to verify the equation's accuracy before proceeding with solving techniques. The discussion emphasizes the importance of correctly identifying the type of polynomial before applying the appropriate solving methods.
xzibition8612
Messages
137
Reaction score
0

Homework Statement



2t^3-4t-5=0
Find t

Homework Equations



[-b-sqrt(b^2-4ac)]/2a

The Attempt at a Solution



The answer is sqrt(2/3). No idea how this came about. I tried plugging in the equation and got it wrong. :( Please help I've been struggling over this for 5 hours.
 
Physics news on Phys.org
The quadratic formula

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Is only for quadratics. That is, equations of the form

y=ax^2+bx+c

But what you have is a cubic, so you can't use the quadratic formula. The cubic formula is way too complicated though, so we have other methods of going about solving them, such as the rational root theorem.

But now that I look at it, are you sure you wrote the cubic down correctly? The real root definitely isn't \sqrt{2/3}.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top