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C1 quadratic inequality question

Find the ranges of values of k for which the equation

x2+(k−3)x+k=0 x^2 +(k-3)x +k = 0

has roots of the same sign

My comment:
I understand that if a> 0 then k>0 and for f(x) to have roots b2−4ac⩾0 b^2 - 4ac \geqslant 0

which leds to k≤1 k \leq 1 k≥9 k\geq 9
but from the nature of the question I assume that it also considers distinct roots.

but I dont know how to get to that particular part of the solution.
from the book the answer is 0<k≤1 0 < k \leq 1 and k≥9 k\geq 9

please help
Original post by bigmansouf
Find the ranges of values of k for which the equation

x2+(k−3)x+k=0 x^2 +(k-3)x +k = 0

has roots of the same sign

My comment:
I understand that if a> 0 then k>0 and for f(x) to have roots b2−4ac⩾0 b^2 - 4ac \geqslant 0

which leds to k≤1 k \leq 1 k≥9 k\geq 9
but from the nature of the question I assume that it also considers distinct roots.

but I dont know how to get to that particular part of the solution.
from the book the answer is 0<k≤1 0 < k \leq 1 and k≥9 k\geq 9

please help


the answers to a quadratic=0 are x=(b+-sqrt(b^2-4ac))/2a

these have the same sign when b>sqrt(b^2-4ac)>0 or b<-sqrt(b^2-4ac)<0
plug in the k things for b and c and 1 for a
(edited 7 years ago)
Reply 2
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bigmansouf
OP
20
Original post by ValerieKR
the answers to a quadratic=0 are x=(b+-sqrt(b^2-4ac))/2a

these have the same sign when b>sqrt(b^2-4ac)>0 or b<-sqrt(b^2-4ac)<0
plug in the k things for b and c and 1 for a


for part (k−3)>(k−1)(k−9)>0 (k-3) >\sqrt{(k-1)(k-9)}>0
(k−3)>(k−1)(k−9)>0 (k-3) >\sqrt{(k-1)(k-9)}>0

(k−3)2>(k−1)(k−9)>0 (k-3)^2 >(k-1)(k-9)>0

(k−3)2>(k−1)(k−9) (k-3)^2 > (k-1)(k-9)

k2−6k+9>k2−10k+9 k^2 - 6k + 9 > k^2 -10k + 9
k>0k > 0

then (k−1)(k−9)>0 (k-1)(k-9)> 0
k>1ork>9 k>1 or k>9


for part (k−3)<−(k−1)(k−9)<0 (k-3) < - \sqrt{(k-1)(k-9)} <0
which will give the same result as above

Im very sorry but i am confused on what to do next. This is because from my view if k > 0, k> 1 and k > 9 according to edexcel math the answer is K > 9 which is not the answer given in the textbook




Also, how did you get the formula of b>b2−4ac>0 b > \sqrt{b^2 - 4ac} > 0
b<−b2−4ac<0 b < - \sqrt{b^2 - 4ac} < 0 from ?

thank for your help
(edited 7 years ago)
Original post by bigmansouf
for part (k−3)>(k−1)(k−9)>0 (k-3) >\sqrt{(k-1)(k-9)}>0
(k−3)>(k−1)(k−9)>0 (k-3) >\sqrt{(k-1)(k-9)}>0

(k−3)2>(k−1)(k−9)>0 (k-3)^2 >(k-1)(k-9)>0

(k−3)2>(k−1)(k−9) (k-3)^2 > (k-1)(k-9)

k2−6k+9>k2−10k+9 k^2 - 6k + 9 > k^2 -10k + 9
k>0k > 0

then (k−1)(k−9)>0 (k-1)(k-9)> 0
k>1ork>9 k>1 or k>9
thank for your help


That k>1 should be a k<1 (sketch it and see where it's positive)
which gives you the answer

maybe you don't need the other one

they come from the fact that b+-sqrt(b^2-4ac)>0 or <0
and rearranging that in different ways
Reply 4
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bigmansouf
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20
Original post by ValerieKR
That k>1 should be a k<1 (sketch it and see where it's positive)
which gives you the answer

maybe you don't need the other one

they come from the fact that b+-sqrt(b^2-4ac)>0 or <0
and rearranging that in different ways

yes you are right about the k > 1

please can i ask one more question i know its early in the morning?
Original post by bigmansouf
yes you are right about the k > 1

please can i ask one more question i know its early in the morning?


quick (yes of course you can any time)
Reply 6
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bigmansouf
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20
Original post by ValerieKR
quick (yes of course you can any time)


rearranging b±b2−4ac>0 b \pm \sqrt{b^2 -4ac}> 0
1) b−b2−4ac>0 b - \sqrt{b^2 -4ac}> 0 gives b>+b2−4ac b > + \sqrt{b^2 -4ac} but from this how to i get to
b>+b2−4ac>0 b > + \sqrt{b^2 -4ac} > 0

do i just add > 0? is there a rule for this?

and same for rearranging b±b2−4ac<0 b \pm \sqrt{b^2 -4ac}<0
2) b+b2−4ac<0 b + \sqrt{b^2 -4ac}< 0 gives
Unparseable latex formula:

b < _ \sqrt{b^2 -4ac}

but from this how to i get to
b<−b2−4ac<0 b < - \sqrt{b^2 -4ac} <0

sorry if this may be trivial but i self studying maths
Original post by bigmansouf
rearranging b±b2−4ac>0 b \pm \sqrt{b^2 -4ac}> 0
1) b−b2−4ac>0 b - \sqrt{b^2 -4ac}> 0 gives b>+b2−4ac b > + \sqrt{b^2 -4ac} but from this how to i get to
b>+b2−4ac>0 b > + \sqrt{b^2 -4ac} > 0

do i just add > 0? is there a rule for this?

and same for rearranging b±b2−4ac<0 b \pm \sqrt{b^2 -4ac}<0
2) b+b2−4ac<0 b + \sqrt{b^2 -4ac}< 0 gives
Unparseable latex formula:

b < _ \sqrt{b^2 -4ac}

but from this how to i get to
b<−b2−4ac<0 b < - \sqrt{b^2 -4ac} <0

sorry if this may be trivial but i self studying maths


the square root sign makes whatever is inside positive so you need to add the <0 and >0 to avoid awkward scenarios
Reply 8
Avatar for bigmansouf
bigmansouf
OP
20
Original post by ValerieKR
the square root sign makes whatever is inside positive so you need to add the <0 and >0 to avoid awkward scenarios

thank you very much im truly sorry for the late reply
i cant rate you any more but thank a lot

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