If $3.5 - {\sqrt 2}$ and $3.5 + {\sqrt 2}$ are the roots of a quadratic equation ${ax^2 + bx + c = 0}$; then which of the following is not correct?

A. a is nonzero - I ruled out this because if was 0 then it wouldn't be a quadratic equation anymore
B. discriminant is positive - There are two solutions
C. a, b, and c are all real or all complex - I'm not sure but I had a feeling this was correct to, since the roots were all real numbers. Is this correct?

How would I choose between D or E?
D. $\frac{b}{a}$ and $\frac{c}{a}$ must be both rational
E. $\frac{b}{a}$ and $\frac{c}{a}$ must be both integral

Website on solution: http://www.mathsisfun.com/algebra/polynomials-sums-products-roots.html

  • $\begingroup$ I have to say that I find C confusing - I think they must mean "all of them are real or none of them are real" since real numbers are complex numbers. $\endgroup$ – Mark Bennet Apr 26 '14 at 13:08


Using Vieta's formula $$-\frac ba=3.5-\sqrt2+3.5+\sqrt2=7$$

and $$\frac ca=(3.5-\sqrt2)(3.5+\sqrt2)=(3.5)^2-2=\frac{41}4$$

  • $\begingroup$ Thank you! I didn't know there was a specific formula for this. I'll try to remember it for next time. $\endgroup$ – stumped Apr 26 '14 at 13:12
  • $\begingroup$ @stumped, My pleasure. Please don't remember. Try to understand en.wikipedia.org/wiki/Vieta's_formulas $\endgroup$ – lab bhattacharjee Apr 26 '14 at 13:15
  • $\begingroup$ Thanks again! I found another link, and I just finished reading it. I think I understand it much better now. $\endgroup$ – stumped Apr 26 '14 at 13:21

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