# Determine m if the function f has two real roots. The roots are not equal. f = mx^2 + (m-1)*x - m + 2

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f(x) = mx^2+(m-1)x-m+2 .

Since f(x) = 0 has two distict real roots, we must have the discriminant greater than zero.

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

m^2-2m+1 +4m^2-8 > 0

5m^2 -10m -7 > 0

m = {10 +or- sqr(10^2-4*5*-7)}/2*5

m = {10 +sqrt240}/10 = 1+ 0.4sqrt15or

m = 1 - 0.4sqrt15

If a quadratic equation ax^2 + bx + c =0 has two real roots which are not equal we know that b^2 - 4ac > 0.

Therefore for mx^2 + (m-1)*x - m + 2,

a= m , b= (m-1) and c = -m +2

=> b^2 - 4ac = (m-1)^2 - 4*m*(2-m)

= m^2 + 1 - 2m - 8m + 4m^2

=> 5 m^2 - 10m +1

Now 5 m^2 - 10m +1 >0

Solving for 5 m^2 - 10m +1 , we get the roots

R1 = [10 + sqrt(100 - 20)] / 10

= (10 + 4 sqrt5) / 10

= (5 + 2sqrt 5) /5

R2 = [10 - sqrt (100 - 20)] / 10

= (10 - 4 sqrt5) / 10

= (5 - 2sqrt 5) / 5

Now 5 m^2 - 10m +1 > 0 when m is either less than (5 - 2sqrt 5)/ 5 or m is greater than (5 + 2sqrt 5) /5.

**f = mx^2 + (m-1)*x - m + 2 has two unequal real roots when m is either less than (5 - 2 sqrt 5) / 5 or m is greater than (5 + 2 sqrt 5) / 5.**

** **

The function f(x) has 2 distinct real roots if and only if the discriminant delta is strictly positive.

delta = b^2 - 4ac

We'll identify a, b and c.

a = m

b = m - 1

c = 2 - m

delta = (m-1)^2 - 4*m*(2 - m)

We'll impose the constraint: delta > 0

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

We'll expand the square and remove the brackets:

m^2 - 2m + 1 - 8m + 4m^2 > 0

We'll combine like terms:

5m^2 - 10m + 1>0

Now, we'll determine the roots of the expression 5m^2 - 10m + 1.

5m^2 - 10m + 1 = 0

We'll apply the quadratic formula:

m1 = [10+sqrt(100-20)]/10

m1 = (10+4sqrt5)/10

m1 = (5+2sqrt5)/5

m2 = (5-2sqrt5)/5

**The expression is positive outside the roots, namely when m belongs to the intervals:**

**(-infinite , (5-2sqrt5)/5) U ((5+2sqrt5)/5 , +infinite).**