J's answer to Anonymous's Secondary 3 E Maths Singapore question.
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Your method is correct. But what you should have done is to reject the solution x = 1.
x cannot be equal to 1 as that would mean x - 1 = 0.
Look back at your original equation. It would result in the denominators being 0, and division by 0 is undefined.
So we have to reject x = 1 as a possible solution, which means the only solution is x = 0
x cannot be equal to 1 as that would mean x - 1 = 0.
Look back at your original equation. It would result in the denominators being 0, and division by 0 is undefined.
So we have to reject x = 1 as a possible solution, which means the only solution is x = 0
Date Posted:
3 years ago
Oh thanks
Btw your original equation's denominator seems to be different from your factorisation on the left.
The one on the left is x² + 2x - 3 but the original equation's is x² - x + 3.
Is there some typo?
The one on the left is x² + 2x - 3 but the original equation's is x² - x + 3.
Is there some typo?
Yea the question I wrote on the right has a typo, it's supposed to be x²+2x-3, sorry
360 Tutoring Program