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secondary 4 | A Maths
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This question need another answer.
Pls help. Thank you
Date Posted: 3 years ago
Views: 211
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I believe this should be the correct answer. Since there is only 1 equation and 2 unknown variable, you would have to make an assumption in this question.
Hope it helps! :)
Hope it helps! :)
Date Posted:
3 years ago
when the number of unknowns is more than the number of equations, it usually means that there are infinite solutions.
from 4(h)(3k-6) = 64,
gives k = 16/(3h) + 2 .... (1)
we can substitute any real number (except 0) for h, then calculate the corresponding value for k using (1). this will always give a quadratic equation with zero discriminant which means two real and equal roots.
using the quadratic formula, the root will be x = -4/h
for example,
when h = 1, k = 22/3, and the root is x = -4
when h = 2, k = 14/3, and the root is x = -2
when h = 3, k = 34/9, and the root is x = -4/3
the question asks for "... a possible pair ... h & k",
this hints that there is more than one possible answer, and the student just needs to write down any one possible pair.
from 4(h)(3k-6) = 64,
gives k = 16/(3h) + 2 .... (1)
we can substitute any real number (except 0) for h, then calculate the corresponding value for k using (1). this will always give a quadratic equation with zero discriminant which means two real and equal roots.
using the quadratic formula, the root will be x = -4/h
for example,
when h = 1, k = 22/3, and the root is x = -4
when h = 2, k = 14/3, and the root is x = -2
when h = 3, k = 34/9, and the root is x = -4/3
the question asks for "... a possible pair ... h & k",
this hints that there is more than one possible answer, and the student just needs to write down any one possible pair.
From the eqn, if we let x=0
3k=6
k=3
We can solve for h once we know k
3k=6
k=3
We can solve for h once we know k
by putting x=0, gives 3k = 6, k=2 (not 3!)
for discriminant = 0
4(h)(3k-6) = 64
h = 16/(3k-6)
when k=2, you have division by 0
therefore, for k=2 no possible solution for h
for discriminant = 0
4(h)(3k-6) = 64
h = 16/(3k-6)
when k=2, you have division by 0
therefore, for k=2 no possible solution for h
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