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    Find the ranges of values of k for which the equation

     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  b^2 - 4ac \geqslant 0

    which leds to  k \leq 1   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 \leq 1 and  k\geq  9

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

     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  b^2 - 4ac \geqslant 0

    which leds to  k \leq 1   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 \leq 1 and  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
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    (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) &gt;\sqrt{(k-1)(k-9)}&gt;0
     (k-3) &gt;\sqrt{(k-1)(k-9)}&gt;0

     (k-3)^2 &gt;(k-1)(k-9)&gt;0

     (k-3)^2 &gt; (k-1)(k-9)

     k^2 - 6k + 9 &gt; k^2 -10k + 9
    k &gt; 0

    then  (k-1)(k-9)&gt; 0
      k&gt;1 or k&gt;9


    for part  (k-3) &lt;  - \sqrt{(k-1)(k-9)} &lt;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 &gt; \sqrt{b^2 - 4ac} &gt; 0
     b &lt; - \sqrt{b^2 - 4ac}  &lt; 0 from ?

    thank for your help
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    (Original post by bigmansouf)
    for part  (k-3) &gt;\sqrt{(k-1)(k-9)}&gt;0
     (k-3) &gt;\sqrt{(k-1)(k-9)}&gt;0

     (k-3)^2 &gt;(k-1)(k-9)&gt;0

     (k-3)^2 &gt; (k-1)(k-9)

     k^2 - 6k + 9 &gt; k^2 -10k + 9
    k &gt; 0

    then  (k-1)(k-9)&gt; 0
      k&gt;1 or k&gt;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
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    (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?
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    (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)
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    (Original post by ValerieKR)
    quick (yes of course you can any time)
    rearranging  b \pm \sqrt{b^2 -4ac}&gt; 0
    1)  b - \sqrt{b^2 -4ac}&gt; 0 gives  b &gt; + \sqrt{b^2 -4ac}  but from this how to i get to
     b &gt; + \sqrt{b^2 -4ac} &gt; 0

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

    and same for rearranging  b \pm \sqrt{b^2 -4ac}&lt;0
    2)  b + \sqrt{b^2 -4ac}&lt; 0 gives  b &lt; _ \sqrt{b^2 -4ac}  but from this how to i get to
     b &lt; - \sqrt{b^2 -4ac} &lt;0

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

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

    and same for rearranging  b \pm \sqrt{b^2 -4ac}&lt;0
    2)  b + \sqrt{b^2 -4ac}&lt; 0 gives  b &lt; _ \sqrt{b^2 -4ac}  but from this how to i get to
     b &lt; - \sqrt{b^2 -4ac} &lt;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
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    (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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