S4 Math

2013-10-25 11:29 pm
If p is the positive real root of

2x^3 + 7x^2 - 29x - 70 = 0,

find the value of p.

回答 (1)

2013-10-25 11:59 pm
✔ 最佳答案
Factors of 70 can be ± 1,±2,±5,±7

Let f(x) = 2x³ + 7x² - 29x - 70

f(1) = 2(1)³ + 7(1)² - 29(1) - 70 = -90

f(2) = 2(2)³ + 7(2)² - 29(2) - 70 = -84

f(5) = 2(5)³ + 7(5)² - 29(5) - 70 = 210

f(7) = 2(7)³ + 7(7)² - 29(7) - 70 = 728

f(-1) = 2(-1)³ + 7(-1)² - 29(-1) - 70 = -36

f(-2) = 2(-2)³ + 7(-2)² - 29(-2) - 70 = 0

Therefore (x+2) is a factor of f(x).

p = -2





2013-10-25 16:06:57 補充:
Sorry!

It must be factorized further : f(x) = (x+2)(2x²+3x-35)

=(x+2)(2x-7)(x+5)

p = 7/2 or 3.5


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