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.
$begingroup$
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 ?
- $b+c>a$
- $c+a<2b$
- both roots of the given equation are rational
- 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
asked Mar 27 at 20:38
ThinkerThinker
356
$endgroup$
add a comment |
$begingroup$
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 ?
- $b+c>a$
- $c+a<2b$
- both roots of the given equation are rational
- 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
asked Mar 27 at 20:38
ThinkerThinker
356
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 ?
- $b+c>a$
- $c+a<2b$
- both roots of the given equation are rational
- 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
asked Mar 27 at 20:38
ThinkerThinker
356
$endgroup$
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 ?
- $b+c>a$
- $c+a<2b$
- both roots of the given equation are rational
- 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
algebra-precalculus roots quadratics
asked Mar 27 at 20:38
ThinkerThinker
356
asked Mar 27 at 20:38
ThinkerThinker
356
edited Mar 28 at 3:19
Thinker
asked Mar 27 at 20:38
ThinkerThinker
356
asked Mar 27 at 20:38
ThinkerThinker
356
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Mar 27 at 21:30
ADITYA PRAKASHADITYA PRAKASH
375110
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
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.
answered Mar 27 at 21:30
ADITYA PRAKASHADITYA PRAKASH
375110
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 27 at 21:30
ADITYA PRAKASHADITYA PRAKASH
375110
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 27 at 21:30
ADITYA PRAKASHADITYA PRAKASH
375110
$endgroup$
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.
answered Mar 27 at 21:30
ADITYA PRAKASHADITYA PRAKASH
375110
answered Mar 27 at 21:30
ADITYA PRAKASHADITYA PRAKASH
375110
answered Mar 27 at 21:30
ADITYA PRAKASHADITYA PRAKASH
375110
answered Mar 27 at 21:30
ADITYA PRAKASHADITYA PRAKASH
375110
375110
add a comment |
add a comment |
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$begingroup$
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