Practice the technique with the following quadratics: (i) x? – 8x + 7 (ii) 2x2 – 5x + 1 (iii) Зx? — х — 4
Practice the technique with the following quadratics: (i) x? – 8x + 7 (ii) 2x2 – 5x + 1 (iii) Зx? — х — 4
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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Question
![All quadratics can be expressed in the form a(x – h)² + k.
For example, x? – 4x – 5 = (x – 2)² – 9
How to do it:
a(x – h)² + k = a[(x – h)(x – h)] + k = a[x² – hx – hx + h²] + k
= ax? – 2ahx + ah² + k
So, we want to express x? – 4x – 5 in form a(x – h)² + k
x² – 4x – 5 = a(x – h)² + k
x2 – 4x – 5 = ax² – 2ahx + ah² + k
Now equate the coefficients:
1 = a
-4 = -2ah - 4 = -2(1)h
Therefore, h = 2
-5 = h2 + k
- 5 = 4 + k
Therefore, k = -9
So
x² – 4x – 5 = (x – 2)² – 9](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffb6ef112-0fda-41dc-bad6-913ca56d167c%2F38a77e7c-e2a5-471d-ba63-fcc2162360d5%2F7no8to_processed.png&w=3840&q=75)
Transcribed Image Text:All quadratics can be expressed in the form a(x – h)² + k.
For example, x? – 4x – 5 = (x – 2)² – 9
How to do it:
a(x – h)² + k = a[(x – h)(x – h)] + k = a[x² – hx – hx + h²] + k
= ax? – 2ahx + ah² + k
So, we want to express x? – 4x – 5 in form a(x – h)² + k
x² – 4x – 5 = a(x – h)² + k
x2 – 4x – 5 = ax² – 2ahx + ah² + k
Now equate the coefficients:
1 = a
-4 = -2ah - 4 = -2(1)h
Therefore, h = 2
-5 = h2 + k
- 5 = 4 + k
Therefore, k = -9
So
x² – 4x – 5 = (x – 2)² – 9
Transcribed Image Text:Practice the technique with the following quadratics:
(i)
х2 — 8х + 7
(ii)
2x2 — 5х + 1
(iii)
3x2 – x – 4
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