Answer:
The extraneous solution is m = -2
Step-by-step explanation:
This is normally solved by squaring both sides and solving the resulting quadratic.
[tex](3-m)^2=19-3m\\\\m^2-3m-10=0 \quad\text{subtract 19-3m}\\\\(m-5)(m+2)=0 \quad\text{factor}[/tex]
The values of m that make the factors zero are 5 and -2. For m=5, the equation becomes ...
5-3 = √(19-3·5) ⇒ 2 = √4 . . . . . true
For m = -2, the equation becomes ...
-2-3 = √(19-3(-2)) ⇒ -5 = √25 . . . . . not true
The extraneous solution is m = -2.
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Solving the equation graphically, it becomes readily apparent that the only solution is m=5. Extra work is required to see the extraneous solution is m=-2.
