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secondary 3 | A Maths
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How to do Question 6 ?
Date Posted: 4 years ago
Views: 258
juz solve for:
8² - 4(h)(3k-6) = 0
then subst. k = 1 for One possible pair of h & k ,
.
8² - 4(h)(3k-6) = 0
then subst. k = 1 for One possible pair of h & k ,
.
See 1 Answer
done
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clear
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I hope this helps! Things to take note of:
- the conditions when the discriminant is > or < or =0.
- when the square function is involved, you will need to consider both positive and negative square roots and check back whether the final values will satisfy the main equation. In this case, if the value of h=0, it would be rejected as the main equation will not exist as a quadratic equation with 2 roots.
- the conditions when the discriminant is > or < or =0.
- when the square function is involved, you will need to consider both positive and negative square roots and check back whether the final values will satisfy the main equation. In this case, if the value of h=0, it would be rejected as the main equation will not exist as a quadratic equation with 2 roots.
Date Posted:
4 years ago
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