3條quadratic equation 問題 (求詳解)

1)α and β are the roots of the quadratic equation 2x² - 4x - 7 = 0

so (x-α)(x-β) = x²-(α+β)x+αβ= x² - 2x - 7/2 = 0
so α+β = 4 αβ= -7/2

(α - β)² = α² - 2αβ + β² = α² + 2αβ +β² - 4αβ
=(α + β)² - 4αβ= 2² +14 = 18

2. Given the quadratic equation ax² + bx + c = 0, where a, b and c are positive, does the equation have a positive root? Explain your answer.

Let α， βbe the roots of ax² + bx + c = 0, since a>0, so
(x-α)(x-β) = x²-(α+β)x+αβ= x² + bx/a + c/a = 0
a, b, c are positive, so b/a, c/a are also positive, now
αβ= c/a >0 positive, so
αβmust be both positive or both negative
if αβare both positive, then -(α+β) will be a negative number, but
-(α+β) = b/a which is positive, so"αβare both positive" is wrong.
so if a,b,c are positive, the equation always have 2 negative roots if the roots exist.

3. If the quadratic equation x² + bx + c = 0 has roots of opposite signs, is it possible that c is positive? Explain your answer.

Let α， βbe the roots of x² + bx + c = 0, so we have
αβ= c
Given α andβhave different signs, so their productαβmust be negative, so c must be negative.


2013-03-13 06:54:26 補充：
寫錯
so α+β = 2 αβ= -7/2

1 (α - β)² = (α + β)² - 4(αβ)

= 2^2 - 4(-7/2)

= 4 + 14

= 18

2 In this case α + β and αβ > 0, implies that α and β should be positive

3 In this case c = αβ < 0, implies that c cannot be positive