Totally, 4 real roots.

來源: BeLe 2024-01-25 17:42:55 [] [博客] [舊帖] [給我悄悄話] 本文已被閱讀: 次 (633 bytes)
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回答: 哪位大神幫忙解個題風吹柳花2024-01-25 16:01:36

Let f(x) =  (x^2 - 6x +5)(x^2 + 3x + 3) - (x^2 - 2x - 3)(2 -3x) = x^4 - 18x^2 - 8x + 21

 

f(-4) = 21, f(-3) = -36, so there is a root in the interval (-4, -3)

 

f(-2)= -21, f(-1) = 12, so there is a root in (-2, -1)

 

f(0) = 21, f(1) = -4, so there is a root in (0, 1)

 

f(4) = -43, f(5) = 164, so there is a root in (4, 5)

 

Totally, 4 real roots. These are all the possible roots since f(x) is a polynomial of degree 4.

 

 

 

 

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