Prove that the quadratic equation x^2 + (m+5)x +5m=0 has real roots for all values of m?

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Answer 1 · 2018-06-27T10:41:46

#"See explanation"#

#"discriminant = "(m+5)^2 - 20 m = (m - 5)^2 >= 0#
#"As the discriminant is always positive, there are always (for"#
#"all values of m) real roots."#

Answer 2 · 2018-06-27T10:52:15

Please refer to Explanation.

Observe that, the discriminant #Delta# of the given quadratic eqn. is,

#Delta=(m+5)^2-4*1*5m#,

#=(m^2+10m+25)-20m#,

#=m^2-10m+25#.

#:. Delta=(m-5)^2, &, AA m in RR, Delta ge 0#.

Hence, the given quadr. eqn. must have real roots.

In fact, these roots are, #{-(m+5)+-sqrt((m-5)^2)}/2, i.e., #

# -5 and -m,# and, are indeed real #AA m in RR#.