Jack's answer to glittr's Secondary 3 A Maths Singapore question.
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Don't know why my answer is abit different from the question, but the method should be the same. An equation has no solution (no real roots) when b^2-4ac<0 as it cannot be square rooted. (based on quadratic equation)
Date Posted:
4 years ago
I have verified from the Desmos application that k is indeed > 0.75.
discriminant b² - 4ac = 9 - 12k
If k > 0.5,
-k < -0.5
-12k < -6
9 - 12k < 9 - 6
9 - 12k < 3
So it would be possible for real solutions to exist if k > 0.5. k needs to be > 0.75 for no real solutions to exist
If k > 0.5,
-k < -0.5
-12k < -6
9 - 12k < 9 - 6
9 - 12k < 3
So it would be possible for real solutions to exist if k > 0.5. k needs to be > 0.75 for no real solutions to exist
thanks y’all are correct , i wrote 0.5 wrongly . it should have been 0.75
360 Tutoring Program