Mixing
The square trinomial $y=ax^2+ bx + c$ has no roots and $a + b + c > 0$. Find the sign of the coefficient...
$begingroup$
The square trinomial $y=ax^2 + bx + c$ has no roots and $a + b + c > 0$. Find the sign of the coefficient $c$.
I'm having difficulties with this problem.
What I've tried:
I realized that a quadratic equation doesn't have roots if the discriminant $b^2 - 4ac < 0$, so I've tried to combine that with the condition $a + b + c > 0 <=> a > -b -c $, but that didn't help me that much.
I would appreciate if someone could help me to understand this. I'll ask a lot of questions on this network while I'm learning, so please don't judge me for that :) .
polynomials quadratics quadratic-forms
edited Jan 17 at 22:08
MPW
30.3k12157
asked Jan 17 at 22:04
Wolf M.Wolf M.
1097
$endgroup$
add a comment |
$begingroup$
The square trinomial $y=ax^2 + bx + c$ has no roots and $a + b + c > 0$. Find the sign of the coefficient $c$.
I'm having difficulties with this problem.
What I've tried:
I realized that a quadratic equation doesn't have roots if the discriminant $b^2 - 4ac < 0$, so I've tried to combine that with the condition $a + b + c > 0 <=> a > -b -c $, but that didn't help me that much.
I would appreciate if someone could help me to understand this. I'll ask a lot of questions on this network while I'm learning, so please don't judge me for that :) .
polynomials quadratics quadratic-forms
edited Jan 17 at 22:08
MPW
30.3k12157
asked Jan 17 at 22:04
Wolf M.Wolf M.
1097
$endgroup$
1
$begingroup$
What happens for simple choices of values of $x$? (What values might you choose, and why, to help answer the question)
$endgroup$
– Mark Bennet
Jan 17 at 22:08
1
$begingroup$
You have the right idea. You know $0 ≤ b^2 < 4ac$, so neither $a$ nor $c$ is zero. $a$ and $c$ thus have the same sign...
$endgroup$
– diracdeltafunk
Jan 17 at 22:10
add a comment |
$begingroup$
The square trinomial $y=ax^2 + bx + c$ has no roots and $a + b + c > 0$. Find the sign of the coefficient $c$.
I'm having difficulties with this problem.
What I've tried:
I realized that a quadratic equation doesn't have roots if the discriminant $b^2 - 4ac < 0$, so I've tried to combine that with the condition $a + b + c > 0 <=> a > -b -c $, but that didn't help me that much.
I would appreciate if someone could help me to understand this. I'll ask a lot of questions on this network while I'm learning, so please don't judge me for that :) .
polynomials quadratics quadratic-forms
edited Jan 17 at 22:08
MPW
30.3k12157
asked Jan 17 at 22:04
Wolf M.Wolf M.
1097
$endgroup$
The square trinomial $y=ax^2 + bx + c$ has no roots and $a + b + c > 0$. Find the sign of the coefficient $c$.
I'm having difficulties with this problem.
What I've tried:
I realized that a quadratic equation doesn't have roots if the discriminant $b^2 - 4ac < 0$, so I've tried to combine that with the condition $a + b + c > 0 <=> a > -b -c $, but that didn't help me that much.
I would appreciate if someone could help me to understand this. I'll ask a lot of questions on this network while I'm learning, so please don't judge me for that :) .
polynomials quadratics quadratic-forms
polynomials quadratics quadratic-forms
edited Jan 17 at 22:08
MPW
30.3k12157
asked Jan 17 at 22:04
Wolf M.Wolf M.
1097
edited Jan 17 at 22:08
MPW
30.3k12157
asked Jan 17 at 22:04
Wolf M.Wolf M.
1097
edited Jan 17 at 22:08
MPW
30.3k12157
edited Jan 17 at 22:08
MPW
30.3k12157
edited Jan 17 at 22:08
MPW
30.3k12157
30.3k12157
asked Jan 17 at 22:04
Wolf M.Wolf M.
1097
asked Jan 17 at 22:04
Wolf M.Wolf M.
1097
asked Jan 17 at 22:04
Wolf M.Wolf M.
1097
1097
1
$begingroup$
What happens for simple choices of values of $x$? (What values might you choose, and why, to help answer the question)
$endgroup$
– Mark Bennet
Jan 17 at 22:08
1
$begingroup$
You have the right idea. You know $0 ≤ b^2 < 4ac$, so neither $a$ nor $c$ is zero. $a$ and $c$ thus have the same sign...
$endgroup$
– diracdeltafunk
Jan 17 at 22:10
add a comment |
1
$begingroup$
What happens for simple choices of values of $x$? (What values might you choose, and why, to help answer the question)
$endgroup$
– Mark Bennet
Jan 17 at 22:08
1
$begingroup$
You have the right idea. You know $0 ≤ b^2 < 4ac$, so neither $a$ nor $c$ is zero. $a$ and $c$ thus have the same sign...
$endgroup$
– diracdeltafunk
Jan 17 at 22:10
1
1
$begingroup$
What happens for simple choices of values of $x$? (What values might you choose, and why, to help answer the question)
$endgroup$
– Mark Bennet
Jan 17 at 22:08
$begingroup$
What happens for simple choices of values of $x$? (What values might you choose, and why, to help answer the question)
$endgroup$
– Mark Bennet
Jan 17 at 22:08
1
1
$begingroup$
You have the right idea. You know $0 ≤ b^2 < 4ac$, so neither $a$ nor $c$ is zero. $a$ and $c$ thus have the same sign...
$endgroup$
– diracdeltafunk
Jan 17 at 22:10
$begingroup$
You have the right idea. You know $0 ≤ b^2 < 4ac$, so neither $a$ nor $c$ is zero. $a$ and $c$ thus have the same sign...
$endgroup$
– diracdeltafunk
Jan 17 at 22:10
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Call $p(x)=ax^2+bx+c$. Then
$$p(1)=a+b+c >0$$
$$p(0)=c$$
If $c$ were negative, then there would be a root between $0$ and $1$. This contradicts our hypothesis, hence necessarily $c>0$.
answered Jan 17 at 22:09
CrostulCrostul
27.9k22352
$endgroup$
$begingroup$
+1, beautiful argument
$endgroup$
– gt6989b
Jan 17 at 22:10
$begingroup$
Thank you so much! I understand now and I really like this way of solving the problem.
$endgroup$
– Wolf M.
Jan 17 at 22:25
add a comment |
$begingroup$
That $a+b+c > 0$ gives $ax^2+bx+c$ evaluated at $x=1$ is $a+b+c$ which is positive.
Suppose that $c$ is negative. Then $ax^2+bx+c$ evaluated at $x=0$ is $c$ which is negative. Then the Intermediate Value Theorem would imply that there is a root $x in (0,1)$. So $c$ cannot be negative.
If $c$ is 0 then 0 is a root of $ax^2 +bx+c$.
So $c$ must be positive for there to be no real root.
answered Jan 17 at 22:11
MikeMike
4,171412
$endgroup$
add a comment |
$begingroup$
It's easy to see that since $b^2<4ac$, you have $c > b^2/(4a) > 0$ if $a>0$ and $c < b^2/(4a) < 0$ if $a < 0$.
But you have no roots, so it is either a parabola opening down below $x$-axis or opening up above $x$-axis, and since $a+b+c=1$ it must be all above.
Can you conclude?
answered Jan 17 at 22:10
gt6989bgt6989b
34.4k22456
$endgroup$
$begingroup$
Yes! With your and other answers, I understand even better :). Thank you!
$endgroup$
– Wolf M.
Jan 17 at 22:25
add a comment |
$begingroup$
Since the polynomial has no roots, its graph is either strictly above or below the $x$-axis. But $f(1)=a+b+c>0$, so the graph is above the $x$-axis. The parabola then intersects the positive part $y$-axis, but this intersection point is $(0,c)$, so $c>0$.
answered Jan 17 at 22:13
Teddan the TerranTeddan the Terran
1,206210
$endgroup$
add a comment |
$begingroup$
Short answer:
Perforce
$$text{sgn}(y(0))=text{sgn}(y(1)).$$
answered Jan 17 at 22:16
Yves DaoustYves Daoust
129k675227
$endgroup$
$begingroup$
If you prefer, $text{sgn}(c)=text{sgn}(a+b+c)$.
$endgroup$
– Yves Daoust
Jan 17 at 22:22
add a comment |
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Call $p(x)=ax^2+bx+c$. Then
$$p(1)=a+b+c >0$$
$$p(0)=c$$
If $c$ were negative, then there would be a root between $0$ and $1$. This contradicts our hypothesis, hence necessarily $c>0$.
answered Jan 17 at 22:09
CrostulCrostul
27.9k22352
$endgroup$
$begingroup$
+1, beautiful argument
$endgroup$
– gt6989b
Jan 17 at 22:10
$begingroup$
Thank you so much! I understand now and I really like this way of solving the problem.
$endgroup$
– Wolf M.
Jan 17 at 22:25
add a comment |
$begingroup$
Call $p(x)=ax^2+bx+c$. Then
$$p(1)=a+b+c >0$$
$$p(0)=c$$
If $c$ were negative, then there would be a root between $0$ and $1$. This contradicts our hypothesis, hence necessarily $c>0$.
answered Jan 17 at 22:09
CrostulCrostul
27.9k22352
$endgroup$
$begingroup$
+1, beautiful argument
$endgroup$
– gt6989b
Jan 17 at 22:10
$begingroup$
Thank you so much! I understand now and I really like this way of solving the problem.
$endgroup$
– Wolf M.
Jan 17 at 22:25
add a comment |
$begingroup$
Call $p(x)=ax^2+bx+c$. Then
$$p(1)=a+b+c >0$$
$$p(0)=c$$
If $c$ were negative, then there would be a root between $0$ and $1$. This contradicts our hypothesis, hence necessarily $c>0$.
answered Jan 17 at 22:09
CrostulCrostul
27.9k22352
$endgroup$
Call $p(x)=ax^2+bx+c$. Then
$$p(1)=a+b+c >0$$
$$p(0)=c$$
If $c$ were negative, then there would be a root between $0$ and $1$. This contradicts our hypothesis, hence necessarily $c>0$.
answered Jan 17 at 22:09
CrostulCrostul
27.9k22352
answered Jan 17 at 22:09
CrostulCrostul
27.9k22352
answered Jan 17 at 22:09
CrostulCrostul
27.9k22352
answered Jan 17 at 22:09
CrostulCrostul
27.9k22352
27.9k22352
$begingroup$
+1, beautiful argument
$endgroup$
– gt6989b
Jan 17 at 22:10
$begingroup$
Thank you so much! I understand now and I really like this way of solving the problem.
$endgroup$
– Wolf M.
Jan 17 at 22:25
add a comment |
$begingroup$
+1, beautiful argument
$endgroup$
– gt6989b
Jan 17 at 22:10
$begingroup$
Thank you so much! I understand now and I really like this way of solving the problem.
$endgroup$
– Wolf M.
Jan 17 at 22:25
$begingroup$
+1, beautiful argument
$endgroup$
– gt6989b
Jan 17 at 22:10
$begingroup$
+1, beautiful argument
$endgroup$
– gt6989b
Jan 17 at 22:10
$begingroup$
Thank you so much! I understand now and I really like this way of solving the problem.
$endgroup$
– Wolf M.
Jan 17 at 22:25
$begingroup$
Thank you so much! I understand now and I really like this way of solving the problem.
$endgroup$
– Wolf M.
Jan 17 at 22:25
add a comment |
$begingroup$
That $a+b+c > 0$ gives $ax^2+bx+c$ evaluated at $x=1$ is $a+b+c$ which is positive.
Suppose that $c$ is negative. Then $ax^2+bx+c$ evaluated at $x=0$ is $c$ which is negative. Then the Intermediate Value Theorem would imply that there is a root $x in (0,1)$. So $c$ cannot be negative.
If $c$ is 0 then 0 is a root of $ax^2 +bx+c$.
So $c$ must be positive for there to be no real root.
answered Jan 17 at 22:11
MikeMike
4,171412
$endgroup$
add a comment |
$begingroup$
That $a+b+c > 0$ gives $ax^2+bx+c$ evaluated at $x=1$ is $a+b+c$ which is positive.
Suppose that $c$ is negative. Then $ax^2+bx+c$ evaluated at $x=0$ is $c$ which is negative. Then the Intermediate Value Theorem would imply that there is a root $x in (0,1)$. So $c$ cannot be negative.
If $c$ is 0 then 0 is a root of $ax^2 +bx+c$.
So $c$ must be positive for there to be no real root.
answered Jan 17 at 22:11
MikeMike
4,171412
$endgroup$
add a comment |
$begingroup$
That $a+b+c > 0$ gives $ax^2+bx+c$ evaluated at $x=1$ is $a+b+c$ which is positive.
Suppose that $c$ is negative. Then $ax^2+bx+c$ evaluated at $x=0$ is $c$ which is negative. Then the Intermediate Value Theorem would imply that there is a root $x in (0,1)$. So $c$ cannot be negative.
If $c$ is 0 then 0 is a root of $ax^2 +bx+c$.
So $c$ must be positive for there to be no real root.
answered Jan 17 at 22:11
MikeMike
4,171412
$endgroup$
That $a+b+c > 0$ gives $ax^2+bx+c$ evaluated at $x=1$ is $a+b+c$ which is positive.
Suppose that $c$ is negative. Then $ax^2+bx+c$ evaluated at $x=0$ is $c$ which is negative. Then the Intermediate Value Theorem would imply that there is a root $x in (0,1)$. So $c$ cannot be negative.
If $c$ is 0 then 0 is a root of $ax^2 +bx+c$.
So $c$ must be positive for there to be no real root.
answered Jan 17 at 22:11
MikeMike
4,171412
answered Jan 17 at 22:11
MikeMike
4,171412
answered Jan 17 at 22:11
MikeMike
4,171412
answered Jan 17 at 22:11
MikeMike
4,171412
4,171412
add a comment |
add a comment |
$begingroup$
It's easy to see that since $b^2<4ac$, you have $c > b^2/(4a) > 0$ if $a>0$ and $c < b^2/(4a) < 0$ if $a < 0$.
But you have no roots, so it is either a parabola opening down below $x$-axis or opening up above $x$-axis, and since $a+b+c=1$ it must be all above.
Can you conclude?
answered Jan 17 at 22:10
gt6989bgt6989b
34.4k22456
$endgroup$
$begingroup$
Yes! With your and other answers, I understand even better :). Thank you!
$endgroup$
– Wolf M.
Jan 17 at 22:25
add a comment |
$begingroup$
It's easy to see that since $b^2<4ac$, you have $c > b^2/(4a) > 0$ if $a>0$ and $c < b^2/(4a) < 0$ if $a < 0$.
But you have no roots, so it is either a parabola opening down below $x$-axis or opening up above $x$-axis, and since $a+b+c=1$ it must be all above.
Can you conclude?
answered Jan 17 at 22:10
gt6989bgt6989b
34.4k22456
$endgroup$
$begingroup$
Yes! With your and other answers, I understand even better :). Thank you!
$endgroup$
– Wolf M.
Jan 17 at 22:25
add a comment |
$begingroup$
It's easy to see that since $b^2<4ac$, you have $c > b^2/(4a) > 0$ if $a>0$ and $c < b^2/(4a) < 0$ if $a < 0$.
But you have no roots, so it is either a parabola opening down below $x$-axis or opening up above $x$-axis, and since $a+b+c=1$ it must be all above.
Can you conclude?
answered Jan 17 at 22:10
gt6989bgt6989b
34.4k22456
$endgroup$
It's easy to see that since $b^2<4ac$, you have $c > b^2/(4a) > 0$ if $a>0$ and $c < b^2/(4a) < 0$ if $a < 0$.
But you have no roots, so it is either a parabola opening down below $x$-axis or opening up above $x$-axis, and since $a+b+c=1$ it must be all above.
Can you conclude?
answered Jan 17 at 22:10
gt6989bgt6989b
34.4k22456
answered Jan 17 at 22:10
gt6989bgt6989b
34.4k22456
answered Jan 17 at 22:10
gt6989bgt6989b
34.4k22456
answered Jan 17 at 22:10
gt6989bgt6989b
34.4k22456
34.4k22456
$begingroup$
Yes! With your and other answers, I understand even better :). Thank you!
$endgroup$
– Wolf M.
Jan 17 at 22:25
add a comment |
$begingroup$
Yes! With your and other answers, I understand even better :). Thank you!
$endgroup$
– Wolf M.
Jan 17 at 22:25
$begingroup$
Yes! With your and other answers, I understand even better :). Thank you!
$endgroup$
– Wolf M.
Jan 17 at 22:25
$begingroup$
Yes! With your and other answers, I understand even better :). Thank you!
$endgroup$
– Wolf M.
Jan 17 at 22:25
add a comment |
$begingroup$
Since the polynomial has no roots, its graph is either strictly above or below the $x$-axis. But $f(1)=a+b+c>0$, so the graph is above the $x$-axis. The parabola then intersects the positive part $y$-axis, but this intersection point is $(0,c)$, so $c>0$.
answered Jan 17 at 22:13
Teddan the TerranTeddan the Terran
1,206210
$endgroup$
add a comment |
$begingroup$
Since the polynomial has no roots, its graph is either strictly above or below the $x$-axis. But $f(1)=a+b+c>0$, so the graph is above the $x$-axis. The parabola then intersects the positive part $y$-axis, but this intersection point is $(0,c)$, so $c>0$.
answered Jan 17 at 22:13
Teddan the TerranTeddan the Terran
1,206210
$endgroup$
add a comment |
$begingroup$
Since the polynomial has no roots, its graph is either strictly above or below the $x$-axis. But $f(1)=a+b+c>0$, so the graph is above the $x$-axis. The parabola then intersects the positive part $y$-axis, but this intersection point is $(0,c)$, so $c>0$.
answered Jan 17 at 22:13
Teddan the TerranTeddan the Terran
1,206210
$endgroup$
Since the polynomial has no roots, its graph is either strictly above or below the $x$-axis. But $f(1)=a+b+c>0$, so the graph is above the $x$-axis. The parabola then intersects the positive part $y$-axis, but this intersection point is $(0,c)$, so $c>0$.
answered Jan 17 at 22:13
Teddan the TerranTeddan the Terran
1,206210
answered Jan 17 at 22:13
Teddan the TerranTeddan the Terran
1,206210
answered Jan 17 at 22:13
Teddan the TerranTeddan the Terran
1,206210
answered Jan 17 at 22:13
Teddan the TerranTeddan the Terran
1,206210
1,206210
add a comment |
add a comment |
$begingroup$
Short answer:
Perforce
$$text{sgn}(y(0))=text{sgn}(y(1)).$$
answered Jan 17 at 22:16
Yves DaoustYves Daoust
129k675227
$endgroup$
$begingroup$
If you prefer, $text{sgn}(c)=text{sgn}(a+b+c)$.
$endgroup$
– Yves Daoust
Jan 17 at 22:22
add a comment |
$begingroup$
Short answer:
Perforce
$$text{sgn}(y(0))=text{sgn}(y(1)).$$
answered Jan 17 at 22:16
Yves DaoustYves Daoust
129k675227
$endgroup$
$begingroup$
If you prefer, $text{sgn}(c)=text{sgn}(a+b+c)$.
$endgroup$
– Yves Daoust
Jan 17 at 22:22
add a comment |
$begingroup$
Short answer:
Perforce
$$text{sgn}(y(0))=text{sgn}(y(1)).$$
answered Jan 17 at 22:16
Yves DaoustYves Daoust
129k675227
$endgroup$
Short answer:
Perforce
$$text{sgn}(y(0))=text{sgn}(y(1)).$$
answered Jan 17 at 22:16
Yves DaoustYves Daoust
129k675227
edited Jan 18 at 10:40
answered Jan 17 at 22:16
Yves DaoustYves Daoust
129k675227
answered Jan 17 at 22:16
Yves DaoustYves Daoust
129k675227
answered Jan 17 at 22:16
Yves DaoustYves Daoust
129k675227
129k675227
$begingroup$
If you prefer, $text{sgn}(c)=text{sgn}(a+b+c)$.
$endgroup$
– Yves Daoust
Jan 17 at 22:22
add a comment |
$begingroup$
If you prefer, $text{sgn}(c)=text{sgn}(a+b+c)$.
$endgroup$
– Yves Daoust
Jan 17 at 22:22
$begingroup$
If you prefer, $text{sgn}(c)=text{sgn}(a+b+c)$.
$endgroup$
– Yves Daoust
Jan 17 at 22:22
$begingroup$
If you prefer, $text{sgn}(c)=text{sgn}(a+b+c)$.
$endgroup$
– Yves Daoust
Jan 17 at 22:22
add a comment |
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What happens for simple choices of values of $x$? (What values might you choose, and why, to help answer the question)
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– Mark Bennet
Jan 17 at 22:08
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You have the right idea. You know $0 ≤ b^2 < 4ac$, so neither $a$ nor $c$ is zero. $a$ and $c$ thus have the same sign...
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– diracdeltafunk
Jan 17 at 22:10