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If m is a real number and 2x^2+mx+8 has two distinct real roots, then what are the possible values of m? Express your answer in interval notation. Thanks

Guest Jan 21, 2018
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2x^2  +  mx   + 8

 

If  we have  two distinct real roots, then this will be true

 

m^2  - 4 (2)(8)   >  0

 

m^2  -  64   >   0

 

add  64 to both sides

 

m^2    >  64

 

So either

 

m >  8    or     m < -8

 

So.....the solution is     (-infinity, -8) U  (8, infinity)

 

 

 

cool cool cool

CPhill  Jan 21, 2018

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