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5. Show that: b>0 and a <0 such that a² = b⇒a= -√b. 6. Prove that there are real numbers a and b, such that (a + b)² = a² + b²

5. Show that: b>0 and a <0 such that a² = b⇒a= -√b. 6. Prove that there are real numbers a and b, such that (a + b)² = a² + b²

Advanced Engineering Mathematics
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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段階的に解決し、人工知能を使用せず、 優れた仕事を行います ご支援ありがとうございました SOLVE STEP BY STEP IN DIGITAL FORMAT DON'T USE AI | DON'T USE AI | DON'T USE AI | DON'T USE AI | 5. Show that: b>0 and a < 0 such that a² = b ⇒a=-√b. 6. Prove that there are real numbers a and b, such that (a+b)2 = a²+62
Transcribed Image Text:段階的に解決し、人工知能を使用せず、 優れた仕事を行います ご支援ありがとうございました SOLVE STEP BY STEP IN DIGITAL FORMAT DON'T USE AI | DON'T USE AI | DON'T USE AI | DON'T USE AI | 5. Show that: b>0 and a < 0 such that a² = b ⇒a=-√b. 6. Prove that there are real numbers a and b, such that (a+b)2 = a²+62
Expert Solution
Step 1: Solution of 5

We are given that b>0 and a<0

Now, let a2 = b

    Taking square root on both sides, we get, 

             a = ±√b

  As a<0, we take the negative sign. 

  So, a = -√b

  Conversely, if a = -√b, then squaring both sides, we get

        a2 = (-√b)(-√b) = +(√b)2 = b

 Hence, proved. 

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