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secondary 3 | A Maths
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Please help to solve this question for me thank youu
Date Posted: 5 years ago
Views: 239
(a)
x² - 3x + 5
= (x - ³⁄₂)² - ⁹⁄₄ + 5
= (x - ³⁄₂)² + 2¾
for all real x,
(x - ³⁄₂)² is always greater than zero.
so
(x - ³⁄₂)² + 2¾ is always greater than zero.
thus, x² - 3x + 5 > 0 (shown)
x² - 3x + 5
= (x - ³⁄₂)² - ⁹⁄₄ + 5
= (x - ³⁄₂)² + 2¾
for all real x,
(x - ³⁄₂)² is always greater than zero.
so
(x - ³⁄₂)² + 2¾ is always greater than zero.
thus, x² - 3x + 5 > 0 (shown)
See 1 Answer
done
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clear
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Use completing the square method for part (a)
Date Posted:
5 years ago
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