Answer
The roots are non-real when \( \frac{3}{2} < m < 5 \).
Solution
We are given that the roots of f(x) = 0 are
x = [4 ± √( (6 – 4m)/(–m + 5) )] / (2m).
For the roots to be non‐real the expression under the square root (the discriminant) must be negative. In other words, we require
(6 – 4m)/(–m + 5) < 0.
Step 1. Rewrite the inequality in a slightly simpler form. Notice that
–m + 5 = 5 – m.
So the inequality becomes
(6 – 4m)/(5 – m) < 0.
Step 2. Find the critical values by setting the numerator and denominator equal to zero.
• Numerator: 6 – 4m = 0 ⇒ 4m = 6 ⇒ m = 3/2.
• Denominator: 5 – m = 0 ⇒ m = 5.
Also note that m = 5 must be excluded since it would make the denominator zero (and would also make the expression under the square root undefined). Additionally, since m appears as a coefficient in the denominator of the entire root expression (x = …/(2m)), one might exclude m = 0 to maintain a quadratic formula—but our focus here is determining when the discriminant is negative.
Step 3. Determine the sign of the expression (6 – 4m)/(5 – m) in the intervals determined by m = 3/2 and m = 5.
The intervals to test are:
(a) m < 3/2,
(b) 3/2 < m < 5,
(c) m > 5.
Test a value in each region:
(a) For m = 0 (which is less than 3/2):
Numerator: 6 – 4(0) = 6 (positive).
Denominator: 5 – 0 = 5 (positive).
So (6 – 4m)/(5 – m) > 0.
(b) For m = 2 (which lies between 3/2 and 5):
Numerator: 6 – 4(2) = 6 – 8 = –2 (negative).
Denominator: 5 – 2 = 3 (positive).
So (6 – 4m)/(5 – m) < 0.
(c) For m = 6 (which is greater than 5):
Numerator: 6 – 4(6) = 6 – 24 = –18 (negative).
Denominator: 5 – 6 = –1 (negative).
So (6 – 4m)/(5 – m) > 0 (since a negative divided by a negative is positive).
Step 4. Conclusion
The expression (6 – 4m)/(5 – m) is negative only on the interval
3/2 < m < 5.
Thus, the roots of the equation are non‐real if and only if m is in the open interval (3/2, 5).
Answer: The roots are non‐real for all m satisfying 3/2 < m < 5.
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