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secondary 3 | A Maths
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Hi, need help on this question. Many thanks!
Date Posted: 4 years ago
Views: 645
Seen. Idea is that numerator is (if you observe carefully) a perfect square which is always non negative, so the inequality sign is not dependent on the numerator at all.
The denominator can be positive or negative, so
3x2 + x - 10 > 0
And so on.
As in, no matter the choice of x, the numerator will in no way alter the inequality sign, so in the eyes of the inequalities, the numerator is redundant.
It’s like saying
Positive divided by (3x2 + x - 10) > 0
For this to hold, denominator must also be positive.
Finally, the numerator is zero for only one value of x (x = 1), so we take this into account.
(3x - 5) (x + 2) > 0
x < -2, x > 5/3
The rejection of x = 1 will not affect our solutions.
The denominator can be positive or negative, so
3x2 + x - 10 > 0
And so on.
As in, no matter the choice of x, the numerator will in no way alter the inequality sign, so in the eyes of the inequalities, the numerator is redundant.
It’s like saying
Positive divided by (3x2 + x - 10) > 0
For this to hold, denominator must also be positive.
Finally, the numerator is zero for only one value of x (x = 1), so we take this into account.
(3x - 5) (x + 2) > 0
x < -2, x > 5/3
The rejection of x = 1 will not affect our solutions.
oic, thanks a lot Mr Eric. It is an instant enlightenment to me :) I salute you for your in depth knowledge in A maths! Thanks again.
Writing this up. There is one important caution we must take note of, which I am including inside my workings.
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done
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Good evening Candice! Important to avoid is cross multiplication as we are unsure of whether the denominator is positive or negative (which eventually results in a “dilemma” on whether the inequality signage must be changed or not).
Date Posted:
4 years ago
Excellent and super clear working and thanks for the word of caution, I will certainly take note of what you have advised. Thank you so much Mr Eric :)
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