Quadratic with variable buried in sqrt?

  • #1

Homework Statement


0 = -250*(sqrt(x^2+1)-0.8)^2 + 98.1*x + 59.05

Homework Equations




The Attempt at a Solution


I can't figure out how to solve this equation! After expanding the squared term, I still end up with a sqrt(x^2+1) and I can't figure out how to perform a substitution or something to make it a nice clean quadratic.
 

Answers and Replies

  • #2
andrewkirk
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Get the sqrt term alone on one side of the equation and then square both sides to get rid of the sqrt. You'll then have a quartic but if you simplify it as much as possible, a way to solve it may become apparent.
 
  • #3
Ok I liked that suggestion. Now I'm here:
0 = -149809 + x(-19806.4) + x^2(-99901.4) + x^3(49050) + x^4(62500)
so now I have a quartic equation. Not sure how to proceed.
I mean it should reduce to a quadratic somehow, right?
 
  • #4
andrewkirk
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It won't necessarily reduce to a quadratic. Sometimes one can solve a quartic by writing it as a quadratic in ##y\equiv x^2##, but this doesn't immediately appear to be one of those.

Are you required to solve it analytically? If not, the easiest thing is to solve it numerically by a root-seeking algorithm.
The coefficients don't look nice and neat like those of something that one would be given to solve analytically.
 
  • #5
Ray Vickson
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Ok I liked that suggestion. Now I'm here:
0 = -149809 + x(-19806.4) + x^2(-99901.4) + x^3(49050) + x^4(62500)
so now I have a quartic equation. Not sure how to proceed.
I mean it should reduce to a quadratic somehow, right?
No, not right. A quartic is a separate type of equation from a quadratic. Yours has two real and two complex roots. The two real roots are also the two real roots of your original equation, before you did any squaring.

Even though there are closed-form formulas for the solution of quartic equations, these formulas are rarely used in practice; numerical methods are much more frequently used.
 

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