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secondary 4 | A Maths
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need help with this
Date Posted: 3 years ago
Views: 581
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done
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clear
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(ㆁωㆁ)
Date Posted:
3 years ago
thank youu
Alternative approach :
(x² + 2x + 4)/(x² - 5x + 4) > 0
(x² - 5x + 4 + 7x)/(x² - 5x + 4) > 0
1 + 7x/(x² - 5x + 4) > 0
7x/(x² - 5x + 4) > -1
Possibilities :
①
Denominator is positive (> 0), numerator can be either sign.
7x > -(x² - 5x + 4)
7x > -x² + 5x - 4
x² + 2x + 4 > 0
(x + 1)² + 3 > 0 (true for all real x)
Based on numerator, x > 0 or x < 0
Based on denominator, x² - 5x + 4 > 0
(x - 4)(x - 1) > 0 → x > 4 or x < 1
Overall result : x > 4 or x < 1
②
Denominator is negative (< 0), numerator can be either sign
7x < -x² + 5x - 4
x² + 2x + 4 < 0
(x + 1)² + 3 < 0 (rejected since it is positive for all real x). So ② is eliminated
(x² + 2x + 4)/(x² - 5x + 4) > 0
(x² - 5x + 4 + 7x)/(x² - 5x + 4) > 0
1 + 7x/(x² - 5x + 4) > 0
7x/(x² - 5x + 4) > -1
Possibilities :
①
Denominator is positive (> 0), numerator can be either sign.
7x > -(x² - 5x + 4)
7x > -x² + 5x - 4
x² + 2x + 4 > 0
(x + 1)² + 3 > 0 (true for all real x)
Based on numerator, x > 0 or x < 0
Based on denominator, x² - 5x + 4 > 0
(x - 4)(x - 1) > 0 → x > 4 or x < 1
Overall result : x > 4 or x < 1
②
Denominator is negative (< 0), numerator can be either sign
7x < -x² + 5x - 4
x² + 2x + 4 < 0
(x + 1)² + 3 < 0 (rejected since it is positive for all real x). So ② is eliminated
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