For which values of $a$ we will get two different roots?












1












$begingroup$


In given the following system of equations:
$$ |x-1| > 2x+2 $$
$$ x^2 + ax + a -1 = 0 $$

For which values of $a$ we will get two different roots?






systems-of-equations roots quadratics absolute-value







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$endgroup$












  • $begingroup$
    Use the fact that we get two different real roots when $b^2-4ac gt 0.$
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 26 at 16:31










  • $begingroup$
    What do you need the first inequality for?
    $endgroup$
    – Dr. Mathva
    Jan 26 at 16:43
















1












$begingroup$


In given the following system of equations:
$$ |x-1| > 2x+2 $$
$$ x^2 + ax + a -1 = 0 $$

For which values of $a$ we will get two different roots?






systems-of-equations roots quadratics absolute-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    Use the fact that we get two different real roots when $b^2-4ac gt 0.$
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 26 at 16:31










  • $begingroup$
    What do you need the first inequality for?
    $endgroup$
    – Dr. Mathva
    Jan 26 at 16:43














1












1








1





$begingroup$


In given the following system of equations:
$$ |x-1| > 2x+2 $$
$$ x^2 + ax + a -1 = 0 $$

For which values of $a$ we will get two different roots?






systems-of-equations roots quadratics absolute-value







share|cite|improve this question











$endgroup$




In given the following system of equations:
$$ |x-1| > 2x+2 $$
$$ x^2 + ax + a -1 = 0 $$

For which values of $a$ we will get two different roots?






systems-of-equations roots quadratics absolute-value




systems-of-equations roots quadratics absolute-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 26 at 16:49









Michael Rozenberg

108k1895200




108k1895200










asked Jan 26 at 16:29









StackUserStackUser

61




61












  • $begingroup$
    Use the fact that we get two different real roots when $b^2-4ac gt 0.$
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 26 at 16:31










  • $begingroup$
    What do you need the first inequality for?
    $endgroup$
    – Dr. Mathva
    Jan 26 at 16:43


















  • $begingroup$
    Use the fact that we get two different real roots when $b^2-4ac gt 0.$
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 26 at 16:31










  • $begingroup$
    What do you need the first inequality for?
    $endgroup$
    – Dr. Mathva
    Jan 26 at 16:43
















$begingroup$
Use the fact that we get two different real roots when $b^2-4ac gt 0.$
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 16:31




$begingroup$
Use the fact that we get two different real roots when $b^2-4ac gt 0.$
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 16:31












$begingroup$
What do you need the first inequality for?
$endgroup$
– Dr. Mathva
Jan 26 at 16:43




$begingroup$
What do you need the first inequality for?
$endgroup$
– Dr. Mathva
Jan 26 at 16:43










2 Answers
2






active

oldest

votes


















1












$begingroup$

The first gives
$$x-1>2x+2$$ or $$x-1<-2x-2,$$ which gives $$x<-frac{1}{3}.$$
Now, let $f(x)=x^2+ax+a-1$ and solve the following system:
$$fleft(-frac{1}{3}right)>0$$
$$-frac{a}{2}<-frac{1}{3}$$ and
$$a^2-4(a-1)>0.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why you want that $f(-frac{1}{3}) > 0 $ ?
    $endgroup$
    – StackUser
    Jan 26 at 21:31










  • $begingroup$
    @StackUser Because we need that roots of the quadratic equation would be less that $-frac{1}{3}.$ Draw the graph of $f$.
    $endgroup$
    – Michael Rozenberg
    Jan 26 at 21:38





















0












$begingroup$

The equation has two roots $x_1=-1$ and $x_2=1-a$. It is easy to see $x_1neq x_2$ if $aneq 2$.



Clearly if $x=x_1$, then the inequality $|x-1|>2x+2$ holds. For $x=x_2$, then the inequality $|x-1|>2x+2$ becomes
$$ |a|>4-2a. tag{1}$$
If $age2$, (1) holds. If $0le a<2$, then the solution of (1) is $frac43<a<2$. If $a<0$, then (1) does not hold.



So if $frac43<a<2$, the equation has two different roots satisfying the inequality.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    The first gives
    $$x-1>2x+2$$ or $$x-1<-2x-2,$$ which gives $$x<-frac{1}{3}.$$
    Now, let $f(x)=x^2+ax+a-1$ and solve the following system:
    $$fleft(-frac{1}{3}right)>0$$
    $$-frac{a}{2}<-frac{1}{3}$$ and
    $$a^2-4(a-1)>0.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Why you want that $f(-frac{1}{3}) > 0 $ ?
      $endgroup$
      – StackUser
      Jan 26 at 21:31










    • $begingroup$
      @StackUser Because we need that roots of the quadratic equation would be less that $-frac{1}{3}.$ Draw the graph of $f$.
      $endgroup$
      – Michael Rozenberg
      Jan 26 at 21:38


















    1












    $begingroup$

    The first gives
    $$x-1>2x+2$$ or $$x-1<-2x-2,$$ which gives $$x<-frac{1}{3}.$$
    Now, let $f(x)=x^2+ax+a-1$ and solve the following system:
    $$fleft(-frac{1}{3}right)>0$$
    $$-frac{a}{2}<-frac{1}{3}$$ and
    $$a^2-4(a-1)>0.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Why you want that $f(-frac{1}{3}) > 0 $ ?
      $endgroup$
      – StackUser
      Jan 26 at 21:31










    • $begingroup$
      @StackUser Because we need that roots of the quadratic equation would be less that $-frac{1}{3}.$ Draw the graph of $f$.
      $endgroup$
      – Michael Rozenberg
      Jan 26 at 21:38
















    1












    1








    1





    $begingroup$

    The first gives
    $$x-1>2x+2$$ or $$x-1<-2x-2,$$ which gives $$x<-frac{1}{3}.$$
    Now, let $f(x)=x^2+ax+a-1$ and solve the following system:
    $$fleft(-frac{1}{3}right)>0$$
    $$-frac{a}{2}<-frac{1}{3}$$ and
    $$a^2-4(a-1)>0.$$






    share|cite|improve this answer









    $endgroup$



    The first gives
    $$x-1>2x+2$$ or $$x-1<-2x-2,$$ which gives $$x<-frac{1}{3}.$$
    Now, let $f(x)=x^2+ax+a-1$ and solve the following system:
    $$fleft(-frac{1}{3}right)>0$$
    $$-frac{a}{2}<-frac{1}{3}$$ and
    $$a^2-4(a-1)>0.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 26 at 16:45









    Michael RozenbergMichael Rozenberg

    108k1895200




    108k1895200












    • $begingroup$
      Why you want that $f(-frac{1}{3}) > 0 $ ?
      $endgroup$
      – StackUser
      Jan 26 at 21:31










    • $begingroup$
      @StackUser Because we need that roots of the quadratic equation would be less that $-frac{1}{3}.$ Draw the graph of $f$.
      $endgroup$
      – Michael Rozenberg
      Jan 26 at 21:38




















    • $begingroup$
      Why you want that $f(-frac{1}{3}) > 0 $ ?
      $endgroup$
      – StackUser
      Jan 26 at 21:31










    • $begingroup$
      @StackUser Because we need that roots of the quadratic equation would be less that $-frac{1}{3}.$ Draw the graph of $f$.
      $endgroup$
      – Michael Rozenberg
      Jan 26 at 21:38


















    $begingroup$
    Why you want that $f(-frac{1}{3}) > 0 $ ?
    $endgroup$
    – StackUser
    Jan 26 at 21:31




    $begingroup$
    Why you want that $f(-frac{1}{3}) > 0 $ ?
    $endgroup$
    – StackUser
    Jan 26 at 21:31












    $begingroup$
    @StackUser Because we need that roots of the quadratic equation would be less that $-frac{1}{3}.$ Draw the graph of $f$.
    $endgroup$
    – Michael Rozenberg
    Jan 26 at 21:38






    $begingroup$
    @StackUser Because we need that roots of the quadratic equation would be less that $-frac{1}{3}.$ Draw the graph of $f$.
    $endgroup$
    – Michael Rozenberg
    Jan 26 at 21:38













    0












    $begingroup$

    The equation has two roots $x_1=-1$ and $x_2=1-a$. It is easy to see $x_1neq x_2$ if $aneq 2$.



    Clearly if $x=x_1$, then the inequality $|x-1|>2x+2$ holds. For $x=x_2$, then the inequality $|x-1|>2x+2$ becomes
    $$ |a|>4-2a. tag{1}$$
    If $age2$, (1) holds. If $0le a<2$, then the solution of (1) is $frac43<a<2$. If $a<0$, then (1) does not hold.



    So if $frac43<a<2$, the equation has two different roots satisfying the inequality.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The equation has two roots $x_1=-1$ and $x_2=1-a$. It is easy to see $x_1neq x_2$ if $aneq 2$.



      Clearly if $x=x_1$, then the inequality $|x-1|>2x+2$ holds. For $x=x_2$, then the inequality $|x-1|>2x+2$ becomes
      $$ |a|>4-2a. tag{1}$$
      If $age2$, (1) holds. If $0le a<2$, then the solution of (1) is $frac43<a<2$. If $a<0$, then (1) does not hold.



      So if $frac43<a<2$, the equation has two different roots satisfying the inequality.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The equation has two roots $x_1=-1$ and $x_2=1-a$. It is easy to see $x_1neq x_2$ if $aneq 2$.



        Clearly if $x=x_1$, then the inequality $|x-1|>2x+2$ holds. For $x=x_2$, then the inequality $|x-1|>2x+2$ becomes
        $$ |a|>4-2a. tag{1}$$
        If $age2$, (1) holds. If $0le a<2$, then the solution of (1) is $frac43<a<2$. If $a<0$, then (1) does not hold.



        So if $frac43<a<2$, the equation has two different roots satisfying the inequality.






        share|cite|improve this answer









        $endgroup$



        The equation has two roots $x_1=-1$ and $x_2=1-a$. It is easy to see $x_1neq x_2$ if $aneq 2$.



        Clearly if $x=x_1$, then the inequality $|x-1|>2x+2$ holds. For $x=x_2$, then the inequality $|x-1|>2x+2$ becomes
        $$ |a|>4-2a. tag{1}$$
        If $age2$, (1) holds. If $0le a<2$, then the solution of (1) is $frac43<a<2$. If $a<0$, then (1) does not hold.



        So if $frac43<a<2$, the equation has two different roots satisfying the inequality.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 26 at 16:52









        xpaulxpaul

        23.3k24655




        23.3k24655






























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