equation fom I have not learned yet: oot of the equaation x^(2)-6x-81=0 can be expressed in the form x=r(+)/(-)r\sqrt(s)
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BigBeachBanana
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What method(s) of solving quadratic equations have you learned?
Otis
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Hi blackz. Are you talking about the form of the solutions?
Two methods for solving x^2 – 6x – 81 = 0 are the Quadratic Formula and Completing the Square. When we use those methods by hand, we get exact solutions. So, the form r ± r*sqrt(s) is exact, whereas using a calculator may yield decimal approximations.
For example (of a different quadratic polynomial: x^2–4x–24), using a calculator to evaluate √s would lead to roots -3.2915026 and 7.2915026, but those two solutions are rounded, decimal approximations. The exact roots are 2–2√7 and 2+2√7, respectively.
The exact form factors as r*(1±√s). In other words, such roots can by viewed as r times the combination of 1 with Irrational number √s.
(r)(1 ± √s) = (2)(1 ± √7) ≈ (2)(1 ± 2.6457513)
Also, when we look at the (nonfactored) form r ± r√s, we can see that the parabolic graph has axis of symmetry x=r with the smaller root located r√s units to the left of the axis and the larger root located r√s units to the right of the axis. So, if we were to graph the example above (y=x^2–4x–24) having roots 2+2√7 and 2–2√7, then the axis of symmetry would be at x=2 with the x-intercepts located 2√7 units on either side.
If you'd tried solving the given equation, then please share how far you got. Otherwise, let us know more about your question. Thanks
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