18. 95 = -4(m – 3)² – 5 %3D | Мary 95 = -4(m - 3)² – 5 +5 +5 100%3D - 4(т - 3)° + V100 = -2(m – 3) + 10 = -2(m – 3) | | 10 = -2(m – 3) -10 = - 2(m – 3) - 3) | 10 = m - 3 -2 -10 - — т — 3 -2 | | -5 = m - 3 5%3D т — 3 | -2 = m 8 = m

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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18. 95 = -4(m – 3)² – 5
Mary
95 = -4(m - 3)2 - 5
+5
+5
100 = -4(m - 3)²
± V100 = -2(m – 3)
± 10 = -2(m - 3)
|
%3D
10%3D -2(т — 3) -103 -2(т - 3)
10
= m - 3
-2
-10
= m – 3
-2
-5 = m - 3
5 = m – 3
-2= m
8 = m
|
Transcribed Image Text:18. 95 = -4(m – 3)² – 5 Mary 95 = -4(m - 3)2 - 5 +5 +5 100 = -4(m - 3)² ± V100 = -2(m – 3) ± 10 = -2(m - 3) | %3D 10%3D -2(т — 3) -103 -2(т - 3) 10 = m - 3 -2 -10 = m – 3 -2 -5 = m - 3 5 = m – 3 -2= m 8 = m |
For Exercises 17 and 18, the students were asked to solve
the given equation using the square root property. Explain
what the student did wrong. Give the correct answer.
Transcribed Image Text:For Exercises 17 and 18, the students were asked to solve the given equation using the square root property. Explain what the student did wrong. Give the correct answer.
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