Find the range of values of m for which X^2-mx+4>0 for all real values of x?

2016-06-07 8:25 am
Please state working clearly and explain how you get it thank you

回答 (3)

2016-06-07 8:37 am
x² - mx + 4 > 0
[x² - mx + (m/2)²] - (m/2)² + 4 > 0
[x - (m/2)]² + [4 - (m/2)²] > 0

For any real values of x, [x - (m/2)]² > 0
If x² - mx + 4 > 0, then [4 - (m/2)²] > 0

4 - (m/2)² > 0
(m/2)² - 4 < 0
(m²/4) - 4 < 0
(1/4)(m² - 16) < 0
(m + 4)(m - 4) < 0

Range of m: -4 < m <4
2016-06-07 8:32 am
We can find the nature of the roots with the discriminant:
D = b² - 4ac

D = m² - 16

There are no roots if the discriminant is negative. So, we are looking for a range when:
m² - 16 < 0
m² < 16
|m| < 4
-4 < m < 4
2016-06-07 8:31 am
Well,

the discriminant delta is :
delta = (-m)^2 - 4*1*4
= m^2 - 16
= (m - 4)(m + 4)
therefore, delta >=0 meaning the root(s) are real
is obtained for
m € (-oo, -4) U (4, +oo)

hope it' ll help !!


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