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If the roots of x^2-mx+4=0 are real and different and lies in [1, 5]. Then m

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Given the quadratic equation:

x^2-mx+4=0

As we know the roots of a quadratic equation are real and different when the discriminant is greater than zero. So, 

D>0

b^2-4ac>0

(-m)^2-4(1)(4)>0

m^2-16>0

m^2>16

m>4\:\:or\:\:m<-4..........(1)

Now, it is given that the roots lie in [1, 5]. 

Since both the roots are in between 1 and 5 our equation will be positive on both points 1 and 5.So

At x = 1

x^2-mx+4=0

1^2-m(1)+4>0

m<5........(2)

Hence From (1) and (2)

m \:\epsilon \:(4, 5).

 

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Pankaj Sanodiya

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