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junior college 2 | H1 Maths
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QN
QN

junior college 2 chevron_right H1 Maths chevron_right Singapore

Find the sets of value for k in which x^2 + (k-2)x + 2k + 1, for all positive values of x

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Date Posted: 2 years ago
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Xing juan
Xing Juan's answer
2 answers (A Helpful Person)
1st
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QN
QN
2 years ago
Why is b^2-4ac > 0?
Eric Nicholas K
Eric Nicholas K
2 years ago
b2 - 4ac is, in fact, less than zero since “always positive” translates to no intersection of a graph with a horizontal axis.

QN, you should get 0 < k < 12.
QN
QN
2 years ago
So it’s 0
Eric Nicholas K
Eric Nicholas K
2 years ago
Yes
QN
QN
2 years ago
Thanks!!!
J
J
2 years ago
What's the original question? The posted photo had no words


'x² + (k-2)x + 2k + 1 is always positive'?


Please take note that 'for all positive values of x' ≠ 'always positive'. The phrasing in the caption is a bit odd.
Eric Nicholas K
Eric Nicholas K
2 years ago
But either way, whether it’s always positive or always negative for all real values of x, the discriminant < 0 (the coefficient of x^2 takes care of the rest).

It also can’t be the case that the expression equals zero for all real values of x.

I would, however, not think that the expression, for example, is greater than say 2 for all real values of x.
J
J
2 years ago
For x² + (k-2)x + 2k + 1 to be always positive for all real values of x,


There is no intersection with y = 0 (also where the x-axis lies) .

There will be no real roots for x² + (k-2)x + 2k + 1 = 0

Discriminant < 0

(k-2)² - 4(1)(2k+1) < 0

k² - 4k + 4 - 8k - 4 < 0

k² - 12k < 0

k(k - 12) < 0


Either sketch the curve (upward sloping, U-shaped) intersecting the x-acis, and deduce that 0 < k < 12,

Or

Test for the sign by substituting values of k for k > 12, k< 0 and 0 < k < 12.

No need to test k = 0 and k = 12 since the product will be 0 ✖

When k > 12, eg. k = 14
Both k and k-12 are positive. Their product is also positive ✖

When k < 0, eg. k = -2
Both k and k-12 are negative. Their product is positive (>0) ✖

When 0 < k < 12 eg. k = 5
k is positive but k-12 is negative. Their product is negative (<0) ✔



So 0 < k < 12
J
J
2 years ago
@Eric :


But what if the phrasing was :

Find the range/set of values of k where x² + (k - 2)x + 2k + 1 is always having an intersection with the x-axis?


Then b² - 4ac ≥ 0 (taking intersection to be at least at one point)


This is why knowing the original question is very important.
J
J
2 years ago
If the phrasing was :


'Find the set/range of values of k for which x² + (k-2)x + 2k + 1 is always positive for all positive values of x'


Only k< 0 would satisfy it.
Eric Nicholas K
Eric Nicholas K
2 years ago
For the latest example, k < 0 does not seem to always fit the situation.

For example, when k = -1, the expression becomes

x2 - 3x - 1

which is not exactly positive when x = 0 (and a tiny positive value of x)
J
J
2 years ago
Typo. It was supposed to be k > 0
J
J
2 years ago
The main point is, the question has to be clearly shown, if not it can't be accurately attempted in the first place.
J
J
2 years ago
QN wrote in the caption :

'Find the sets of value for k in which x^2 + (k-2)x + 2k + 1, for all positive values of x'

This is basically an incomplete question since the condition of the expression is not stated. Even though the range of x is given, the question is not solvable.