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Math Help - [SOLVED] High School Quadratic Inequalities

  1. #1
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    Cool [SOLVED] High School Quadratic Inequalities

    Q) Use the discriminant 'b^2=-4ac' to solve the following:

    i) Find the values of ' k ' for which the following equations have two separate roots.

    a) kx^2+kx+2=0

    Answer is 0>k>8

    I solved the above to get

    k(k-8)>0

    Now my question is that we know that k>8 but how do we figure k<0?

    Is it because we are 100% that if k>8 from my final step then the other value must be smaller? (k<0)

    ii) Find the values of ' k ' for which the following equations have no roots.

    a) k^2x^2+2kx+1=0

    I solved this and got:

    4k^2-4k^2<0

    0<0

    Answer is k=0

    How is the answer k=0? I mean shouldn't 0<0 mean no solution or something?

    iii) Sketch, on the same diagram, the graphs of y=1/x and y=x-3/2. Find the solution set of the inequality x-3/2>1/x

    Please tell me step by step on how to solve this question.


    Thanks in advance!
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by unstopabl3 View Post
    Q) Use the discriminant 'b^2=-4ac' to solve the following:

    i) Find the values of ' k ' for which the following equations have two separate roots.

    a) kx^2+kx+2=0

    Answer is 0>k>8

    I solved the above to get

    k(k-8)>0

    Now my question is that we know that k>8 but how do we figure k<0?

    Is it because we are 100% that if k>8 from my final step then the other value must be smaller? (k<0)

    ii) Find the values of ' k ' for which the following equations have no roots.

    a) k^2x^2+2kx+1=0

    I solved this and got:

    4k^2-4k^2<0

    0<0

    Answer is k=0

    How is the answer k=0? I mean shouldn't 0<0 mean no solution or something?

    iii) Sketch, on the same diagram, the graphs of y=1/x and y=x-3/2. Find the solution set of the inequality x-3/2>1/x

    Please tell me step by step on how to solve this question.


    Thanks in advance!
    for the first question

    k(k-8)>0 you want the values of k that make this inequality true just draw the real number line and plot 0 and 8 (zero of k(k-8)) then take a value between 0,8 and sub it in k(k-8) if the sign is negative ignore the interval between 0,8 take a number more than 8 and sub if it is positive take this interval and take a number less than 0 and sub if it is positive take the interval (-infinity , 0)

    ii) if b^2-4ac < 0 there is no solution right but here

    b^2-4ac = 4k^2 - 4k^2 = 0 for all k real number so always there is a equal roots and the root is

    \frac{-b \mp \sqrt{4k^2-4k^2}}{2a} = \frac{-b}{2a} = \frac{-2k}{2k^2}=\frac{-1}{k} repeated root so there is no value of k that make no solution

    iii) draw the two curve and take the region where x-\frac{3}{2} is above \frac{1}{x} and find the points where the two curve intersect 2,-1/2 and zero since at zero 1/x change see this pic take the intervals where x-3/2 is above 1/x

    [SOLVED] High School Quadratic Inequalities-gwe.jpg
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  3. #3
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    Thanks for your reply but I did not understand the first part at all.

    The second part I understood somewhat. I didn't get what why you included 0 as an intersection point for the line and the curve and I didn't understand how we are supposed to shade/select the required region to fulfill the inequality.
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by unstopabl3 View Post
    Thanks for your reply but I did not understand the first part at all.

    The second part I understood somewhat. I didn't get what why you included 0 as an intersection point for the line and the curve and I didn't understand how we are supposed to shade/select the required region to fulfill the inequality.
    I did not say it is a point of intersection but as you can see at zero 1/x have a disconnect point , 1/x is not continues at 0 , 1/x at 0 change it is graph as you can see .

    in general you want the interval where the line is above the curve .

    as in graph at the interval (-infinity , -1/2 ) the curve is above the line this interval dose not fulfill the inequality

    in the interval (-1/2 , 0 ) the line is above the curve and this fulfill the inequality .

    the interval (0 ,2) the curve is above the line this dose not fulfill

    (2, infinity ) line is above this fulfill .


    what I want to say in the first question , to solve this k(k-8)>0 we should study the sign of k(k-8) where this function is positive and where is negative after we determine this we take the interval where the function is positive , to determine this first we should find the point where is function equal zero and where it is dose not exist the zeros of the denominator , since usually the function change it is sign at these points , ok in our function we have 0,8 zero of it , we will take a number in each interval here we have three intervals (-infinity , 0) , (0,8) and (8, infinity ) ok take -1 in first interval and substitute it in the function is the sign is positive that mean the function is positive in all the first interval , do the same thing to the other intervals .

    I wish it is clear now .
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  5. #5
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    Cool

    Thanks much appreciated
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