If a,b,c are positive rational numbers such that a>b>c then tell which of the following statement are correct following quadratic equation Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Quadratic equation which has rational rootsIf the roots of the quadratic equation $2kx^2+(4k-1)x+2k-3=0$ are rational and k is an integer, how many values can k take which are less that 50?Find the three positive values of p for which the equation $px^2-4x+1=0$ will have rational rootsProbability that the roots of a quadratic equation are realIf coefficients of the Quadratic Equation are in AP find $alpha+beta +alphabeta$.What is the largest $a$ for which all the solutions to the equation $3x^2+ax-(a^2-1)=0$ are positive and real?Quadratic functions - using substitutionRoot of a complex number - signFind the condition that the roots of the quadratic equationForm a quadratic equation with the following details.

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If a,b,c are positive rational numbers such that a>b>c then tell which of the following statement are correct following quadratic equation



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Quadratic equation which has rational rootsIf the roots of the quadratic equation $2kx^2+(4k-1)x+2k-3=0$ are rational and k is an integer, how many values can k take which are less that 50?Find the three positive values of p for which the equation $px^2-4x+1=0$ will have rational rootsProbability that the roots of a quadratic equation are realIf coefficients of the Quadratic Equation are in AP find $alpha+beta +alphabeta$.What is the largest $a$ for which all the solutions to the equation $3x^2+ax-(a^2-1)=0$ are positive and real?Quadratic functions - using substitutionRoot of a complex number - signFind the condition that the roots of the quadratic equationForm a quadratic equation with the following details.










4












$begingroup$


I am solving following question based on quadratic equation




If $a,b,c$ are positive rational numbers such that $a>b>c$ and the quadratic equation $(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=0$ has a root in the interval $(-1,0)$ then which of the following statements are true ?



  1. $b+c>a$

  2. $c+a<2b$

  3. both roots of the given equation are rational

  4. the equation $ax^2+2bx+c=0$ has both negative real roots.



My Approach



First I calculated discriminant of the given quadratic equation which turns out to be $3(b-c)$ (This proves statement 3).



So root 1 $r_1$ is



$$r_1 = frac-b-c+2a+3b-3c2(a+b-2c)=1$$



So root 2 will be



$$fracc+a-2ba+b-2c$$



As it is mentioned that one root will in $(-1,0) $ so $fracc+a-2ba+b-2c$ will be that root.
So



$$ -1<fracc+a-2ba+b-2c<0 \
-a-b+2c<c+a-2b<0 $$



Solving first half of the above inequality i.e.
$c+a-2b<0$ will prove statement 2 to be true.



Solving another half of the inequality i.e.
$-a-b+2c < c+a-2b$ will prove statement 1 to false as our results are $b+c<2a$.



But I am not able to find the reasoning for fourth statement. My work for proving 4th statement to true:



As $a,b,c$ are all positive and sum of the root for $ax^2+2bx+c$ is $alpha+beta=-2b/a$. This proves that at least one of the root is negative. The product of the root is
$alphabeta=c/a$ as $c/a$ is positive this states that both the roots are negative.



if the discriminant of the $ax^2+2bx+c$ is > 0 only then this equation will have real roots.
How do I prove that the discriminant $D=b^2-4ac>0$?










share|cite|improve this question











$endgroup$











  • $begingroup$
    arent you missing something, option 4 has a different quadratic equation than the one you mentioned in the end
    $endgroup$
    – ADITYA PRAKASH
    Mar 27 at 21:08










  • $begingroup$
    I'll solve it for u
    $endgroup$
    – ADITYA PRAKASH
    Mar 27 at 21:13















4












$begingroup$


I am solving following question based on quadratic equation




If $a,b,c$ are positive rational numbers such that $a>b>c$ and the quadratic equation $(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=0$ has a root in the interval $(-1,0)$ then which of the following statements are true ?



  1. $b+c>a$

  2. $c+a<2b$

  3. both roots of the given equation are rational

  4. the equation $ax^2+2bx+c=0$ has both negative real roots.



My Approach



First I calculated discriminant of the given quadratic equation which turns out to be $3(b-c)$ (This proves statement 3).



So root 1 $r_1$ is



$$r_1 = frac-b-c+2a+3b-3c2(a+b-2c)=1$$



So root 2 will be



$$fracc+a-2ba+b-2c$$



As it is mentioned that one root will in $(-1,0) $ so $fracc+a-2ba+b-2c$ will be that root.
So



$$ -1<fracc+a-2ba+b-2c<0 \
-a-b+2c<c+a-2b<0 $$



Solving first half of the above inequality i.e.
$c+a-2b<0$ will prove statement 2 to be true.



Solving another half of the inequality i.e.
$-a-b+2c < c+a-2b$ will prove statement 1 to false as our results are $b+c<2a$.



But I am not able to find the reasoning for fourth statement. My work for proving 4th statement to true:



As $a,b,c$ are all positive and sum of the root for $ax^2+2bx+c$ is $alpha+beta=-2b/a$. This proves that at least one of the root is negative. The product of the root is
$alphabeta=c/a$ as $c/a$ is positive this states that both the roots are negative.



if the discriminant of the $ax^2+2bx+c$ is > 0 only then this equation will have real roots.
How do I prove that the discriminant $D=b^2-4ac>0$?










share|cite|improve this question











$endgroup$











  • $begingroup$
    arent you missing something, option 4 has a different quadratic equation than the one you mentioned in the end
    $endgroup$
    – ADITYA PRAKASH
    Mar 27 at 21:08










  • $begingroup$
    I'll solve it for u
    $endgroup$
    – ADITYA PRAKASH
    Mar 27 at 21:13













4












4








4





$begingroup$


I am solving following question based on quadratic equation




If $a,b,c$ are positive rational numbers such that $a>b>c$ and the quadratic equation $(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=0$ has a root in the interval $(-1,0)$ then which of the following statements are true ?



  1. $b+c>a$

  2. $c+a<2b$

  3. both roots of the given equation are rational

  4. the equation $ax^2+2bx+c=0$ has both negative real roots.



My Approach



First I calculated discriminant of the given quadratic equation which turns out to be $3(b-c)$ (This proves statement 3).



So root 1 $r_1$ is



$$r_1 = frac-b-c+2a+3b-3c2(a+b-2c)=1$$



So root 2 will be



$$fracc+a-2ba+b-2c$$



As it is mentioned that one root will in $(-1,0) $ so $fracc+a-2ba+b-2c$ will be that root.
So



$$ -1<fracc+a-2ba+b-2c<0 \
-a-b+2c<c+a-2b<0 $$



Solving first half of the above inequality i.e.
$c+a-2b<0$ will prove statement 2 to be true.



Solving another half of the inequality i.e.
$-a-b+2c < c+a-2b$ will prove statement 1 to false as our results are $b+c<2a$.



But I am not able to find the reasoning for fourth statement. My work for proving 4th statement to true:



As $a,b,c$ are all positive and sum of the root for $ax^2+2bx+c$ is $alpha+beta=-2b/a$. This proves that at least one of the root is negative. The product of the root is
$alphabeta=c/a$ as $c/a$ is positive this states that both the roots are negative.



if the discriminant of the $ax^2+2bx+c$ is > 0 only then this equation will have real roots.
How do I prove that the discriminant $D=b^2-4ac>0$?










share|cite|improve this question











$endgroup$




I am solving following question based on quadratic equation




If $a,b,c$ are positive rational numbers such that $a>b>c$ and the quadratic equation $(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=0$ has a root in the interval $(-1,0)$ then which of the following statements are true ?



  1. $b+c>a$

  2. $c+a<2b$

  3. both roots of the given equation are rational

  4. the equation $ax^2+2bx+c=0$ has both negative real roots.



My Approach



First I calculated discriminant of the given quadratic equation which turns out to be $3(b-c)$ (This proves statement 3).



So root 1 $r_1$ is



$$r_1 = frac-b-c+2a+3b-3c2(a+b-2c)=1$$



So root 2 will be



$$fracc+a-2ba+b-2c$$



As it is mentioned that one root will in $(-1,0) $ so $fracc+a-2ba+b-2c$ will be that root.
So



$$ -1<fracc+a-2ba+b-2c<0 \
-a-b+2c<c+a-2b<0 $$



Solving first half of the above inequality i.e.
$c+a-2b<0$ will prove statement 2 to be true.



Solving another half of the inequality i.e.
$-a-b+2c < c+a-2b$ will prove statement 1 to false as our results are $b+c<2a$.



But I am not able to find the reasoning for fourth statement. My work for proving 4th statement to true:



As $a,b,c$ are all positive and sum of the root for $ax^2+2bx+c$ is $alpha+beta=-2b/a$. This proves that at least one of the root is negative. The product of the root is
$alphabeta=c/a$ as $c/a$ is positive this states that both the roots are negative.



if the discriminant of the $ax^2+2bx+c$ is > 0 only then this equation will have real roots.
How do I prove that the discriminant $D=b^2-4ac>0$?







algebra-precalculus roots quadratics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 28 at 3:19







Thinker

















asked Mar 27 at 20:38









ThinkerThinker

356




356











  • $begingroup$
    arent you missing something, option 4 has a different quadratic equation than the one you mentioned in the end
    $endgroup$
    – ADITYA PRAKASH
    Mar 27 at 21:08










  • $begingroup$
    I'll solve it for u
    $endgroup$
    – ADITYA PRAKASH
    Mar 27 at 21:13
















  • $begingroup$
    arent you missing something, option 4 has a different quadratic equation than the one you mentioned in the end
    $endgroup$
    – ADITYA PRAKASH
    Mar 27 at 21:08










  • $begingroup$
    I'll solve it for u
    $endgroup$
    – ADITYA PRAKASH
    Mar 27 at 21:13















$begingroup$
arent you missing something, option 4 has a different quadratic equation than the one you mentioned in the end
$endgroup$
– ADITYA PRAKASH
Mar 27 at 21:08




$begingroup$
arent you missing something, option 4 has a different quadratic equation than the one you mentioned in the end
$endgroup$
– ADITYA PRAKASH
Mar 27 at 21:08












$begingroup$
I'll solve it for u
$endgroup$
– ADITYA PRAKASH
Mar 27 at 21:13




$begingroup$
I'll solve it for u
$endgroup$
– ADITYA PRAKASH
Mar 27 at 21:13










1 Answer
1






active

oldest

votes


















1












$begingroup$

Building up on your work :- You have found that $$2b>a+c... (1)$$$$2a>b+c...(2)$$Note that $ax^2+2bx+c$ has all positive coefficients , there fore the parabola of this quadratic equation will have its vertex in the negative X - direction ($-bover 2a <0$ and we will have an upturned parabola ($a>0$) and Y-intercept of the parabola will be positive ($c>0$).



Analysing the above characteristics, there are only two contending parabolas ($A$ and $B$ in the diagram)



Note that (1) can be rearranged to $a-2b+c <0$ which is nothing but f(-1) or the y-value at x=-1 , therefore case B is correct and the equation has both negative roots.
enter image description here






share|cite|improve this answer









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    active

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    1












    $begingroup$

    Building up on your work :- You have found that $$2b>a+c... (1)$$$$2a>b+c...(2)$$Note that $ax^2+2bx+c$ has all positive coefficients , there fore the parabola of this quadratic equation will have its vertex in the negative X - direction ($-bover 2a <0$ and we will have an upturned parabola ($a>0$) and Y-intercept of the parabola will be positive ($c>0$).



    Analysing the above characteristics, there are only two contending parabolas ($A$ and $B$ in the diagram)



    Note that (1) can be rearranged to $a-2b+c <0$ which is nothing but f(-1) or the y-value at x=-1 , therefore case B is correct and the equation has both negative roots.
    enter image description here






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      Building up on your work :- You have found that $$2b>a+c... (1)$$$$2a>b+c...(2)$$Note that $ax^2+2bx+c$ has all positive coefficients , there fore the parabola of this quadratic equation will have its vertex in the negative X - direction ($-bover 2a <0$ and we will have an upturned parabola ($a>0$) and Y-intercept of the parabola will be positive ($c>0$).



      Analysing the above characteristics, there are only two contending parabolas ($A$ and $B$ in the diagram)



      Note that (1) can be rearranged to $a-2b+c <0$ which is nothing but f(-1) or the y-value at x=-1 , therefore case B is correct and the equation has both negative roots.
      enter image description here






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        Building up on your work :- You have found that $$2b>a+c... (1)$$$$2a>b+c...(2)$$Note that $ax^2+2bx+c$ has all positive coefficients , there fore the parabola of this quadratic equation will have its vertex in the negative X - direction ($-bover 2a <0$ and we will have an upturned parabola ($a>0$) and Y-intercept of the parabola will be positive ($c>0$).



        Analysing the above characteristics, there are only two contending parabolas ($A$ and $B$ in the diagram)



        Note that (1) can be rearranged to $a-2b+c <0$ which is nothing but f(-1) or the y-value at x=-1 , therefore case B is correct and the equation has both negative roots.
        enter image description here






        share|cite|improve this answer









        $endgroup$



        Building up on your work :- You have found that $$2b>a+c... (1)$$$$2a>b+c...(2)$$Note that $ax^2+2bx+c$ has all positive coefficients , there fore the parabola of this quadratic equation will have its vertex in the negative X - direction ($-bover 2a <0$ and we will have an upturned parabola ($a>0$) and Y-intercept of the parabola will be positive ($c>0$).



        Analysing the above characteristics, there are only two contending parabolas ($A$ and $B$ in the diagram)



        Note that (1) can be rearranged to $a-2b+c <0$ which is nothing but f(-1) or the y-value at x=-1 , therefore case B is correct and the equation has both negative roots.
        enter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 27 at 21:30









        ADITYA PRAKASHADITYA PRAKASH

        375110




        375110



























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