F4 A-MATHS [ INEQUALITIES' ]

2007-09-22 6:04 am
FIND THE RANGE OF REAL VALUES OF m FOR WHICH THE EXPRESSION x^2-mx+(m+3) IS ALWAYS POSITIVE FOR ALL REAL VALUES OF x


How can I reach the△ is <0 and why it is necessary for us to consider△<0?
What is the meaning of "always positive for all real values of x"?
How can I slove this inequalities by using other method(except the method of graph sketching?

回答 (2)

2007-09-22 6:43 am
✔ 最佳答案
We consider "△<0" because "△<0" can imply the expression "x^2-mx+(m+3)" must be > 0. Let's consider the meaning of the range of △.

1. △ > 0: Equation has 2 real roots
If x^2-mx+(m+3) = 0, there will be 2 answers of real number x.
2. △ = 0: Equation has 1 real root / a double real root
If x^2-mx+(m+3) = 0, there will be only 1 answer of real number x.
3. △ < 0: Equation has no real root
If x^2-mx+(m+3) = 0, there will be no answer of real number x, that is, you cannot find any real number x such that x^2-mx+(m+3) = 0.

Now, x^2-mx+(m+3) is always positive for all real values of x, that is x^2-mx+(m+3) > 0 for all real number x. This mean, it is not possible to find any real number x such that x^2-mx+(m+3) = 0. △ < 0 can logically represent this condition.
2007-09-22 6:11 am
參考: Myself~~~


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