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If ax2+bx+g with a not 0 and abg belong to real numbers(R) and dont have solutions in R can you prove that for xy(R)
(ax2+bx+g)(ay2=by+g)>0

2007-09-30 04:40:18 · 1 answers · asked by baby 3 in Science & Mathematics Mathematics

1 answers

ax^2+bx +g will have imaginary roots if b^2-4ag <0
ay^2 +by+g will have imaginary roots if b^2-4ag <0
Say roots are (q+iu),(q-iu), (w+iz), and w-iz),respectively.
Then product = (q+iu)(q-iu) (w+iz)(w-iz)
= (q^2+u^2)(w^2+z^2) = positive since all terms are squared.
Thus (ax2+bx+g)(ay2+by+g)>0

2007-09-30 04:59:06 · answer #1 · answered by ironduke8159 7 · 1 0

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