If the roots of the equation $6x^2-7x+K=0$ are rational, then is equal to–
-1
$begingroup$
If the roots of the equation $6x^2-7x+K=0$ are rational, then is equal to:
$1)$ $-2$
$2)$ $-1,-2$
$3)$ $-2$
$4)$ $1,2$
quadratics
edited Jan 20 at 19:46
user289143
903313
asked Jan 20 at 18:47
ANKUR SHUKLAANKUR SHUKLA
1
$endgroup$
add a comment |
-1
$begingroup$
If the roots of the equation $6x^2-7x+K=0$ are rational, then is equal to:
$1)$ $-2$
$2)$ $-1,-2$
$3)$ $-2$
$4)$ $1,2$
quadratics
edited Jan 20 at 19:46
user289143
903313
asked Jan 20 at 18:47
ANKUR SHUKLAANKUR SHUKLA
1
$endgroup$
1
$begingroup$
The discriminant is a perfect square.
$endgroup$
– hamam_Abdallah
Jan 20 at 18:48
$begingroup$
Welcome to MSE. Please don't post what look like homework questions -- provide some context around the question and tell us what attempts you've made and where you got stuck so we can help you better. This kind of naked question tends to attract downvotes and can get closed.
$endgroup$
– postmortes
Jan 20 at 18:53
$begingroup$
"then is equal to" --- Then what is equal to?
$endgroup$
– steven gregory
Jan 20 at 19:02
add a comment |
-1
-1
-1
$begingroup$
If the roots of the equation $6x^2-7x+K=0$ are rational, then is equal to:
$1)$ $-2$
$2)$ $-1,-2$
$3)$ $-2$
$4)$ $1,2$
quadratics
edited Jan 20 at 19:46
user289143
903313
asked Jan 20 at 18:47
ANKUR SHUKLAANKUR SHUKLA
1
$endgroup$
If the roots of the equation $6x^2-7x+K=0$ are rational, then is equal to:
$1)$ $-2$
$2)$ $-1,-2$
$3)$ $-2$
$4)$ $1,2$
quadratics
quadratics
edited Jan 20 at 19:46
user289143
903313
asked Jan 20 at 18:47
ANKUR SHUKLAANKUR SHUKLA
1
edited Jan 20 at 19:46
user289143
903313
asked Jan 20 at 18:47
ANKUR SHUKLAANKUR SHUKLA
1
edited Jan 20 at 19:46
user289143
903313
edited Jan 20 at 19:46
user289143
903313
edited Jan 20 at 19:46
user289143
903313
903313
asked Jan 20 at 18:47
ANKUR SHUKLAANKUR SHUKLA
1
asked Jan 20 at 18:47
ANKUR SHUKLAANKUR SHUKLA
1
asked Jan 20 at 18:47
ANKUR SHUKLAANKUR SHUKLA
1
1
1
$begingroup$
The discriminant is a perfect square.
$endgroup$
– hamam_Abdallah
Jan 20 at 18:48
$begingroup$
Welcome to MSE. Please don't post what look like homework questions -- provide some context around the question and tell us what attempts you've made and where you got stuck so we can help you better. This kind of naked question tends to attract downvotes and can get closed.
$endgroup$
– postmortes
Jan 20 at 18:53
$begingroup$
"then is equal to" --- Then what is equal to?
$endgroup$
– steven gregory
Jan 20 at 19:02
add a comment |
1
$begingroup$
The discriminant is a perfect square.
$endgroup$
– hamam_Abdallah
Jan 20 at 18:48
$begingroup$
Welcome to MSE. Please don't post what look like homework questions -- provide some context around the question and tell us what attempts you've made and where you got stuck so we can help you better. This kind of naked question tends to attract downvotes and can get closed.
$endgroup$
– postmortes
Jan 20 at 18:53
$begingroup$
"then is equal to" --- Then what is equal to?
$endgroup$
– steven gregory
Jan 20 at 19:02
1
1
$begingroup$
The discriminant is a perfect square.
$endgroup$
– hamam_Abdallah
Jan 20 at 18:48
$begingroup$
The discriminant is a perfect square.
$endgroup$
– hamam_Abdallah
Jan 20 at 18:48
$begingroup$
Welcome to MSE. Please don't post what look like homework questions -- provide some context around the question and tell us what attempts you've made and where you got stuck so we can help you better. This kind of naked question tends to attract downvotes and can get closed.
$endgroup$
– postmortes
Jan 20 at 18:53
$begingroup$
Welcome to MSE. Please don't post what look like homework questions -- provide some context around the question and tell us what attempts you've made and where you got stuck so we can help you better. This kind of naked question tends to attract downvotes and can get closed.
$endgroup$
– postmortes
Jan 20 at 18:53
$begingroup$
"then is equal to" --- Then what is equal to?
$endgroup$
– steven gregory
Jan 20 at 19:02
$begingroup$
"then is equal to" --- Then what is equal to?
$endgroup$
– steven gregory
Jan 20 at 19:02
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
$sqrt{7^2-4times6times K}$ is rational when $K=1$ or $2$.
answered Jan 20 at 18:55
J. W. TannerJ. W. Tanner
2,1471117
$endgroup$
$begingroup$
Or $K=0$ I think ... and I think there is at least one solution for negative $K$
$endgroup$
– Mark Bennet
Jan 20 at 19:50
$begingroup$
Yes, but $K=0$ wasn't one of the choices
$endgroup$
– J. W. Tanner
Jan 20 at 19:51
$begingroup$
Well the question is somewhat unclear at the moment and is inadequately specified - negative values are admitted. I can choose an arbitrary integer value of $x$ and then fix $K$ so that this is a root. The other root will then necessarily be rational (sum of roots is rational). (Or I could choose an arbitrary rational value of $x$, though $k$ would likely not be an integer).
$endgroup$
– Mark Bennet
Jan 20 at 19:53
add a comment |
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1 Answer
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1 Answer
1
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oldest
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0
$begingroup$
$sqrt{7^2-4times6times K}$ is rational when $K=1$ or $2$.
answered Jan 20 at 18:55
J. W. TannerJ. W. Tanner
2,1471117
$endgroup$
$begingroup$
Or $K=0$ I think ... and I think there is at least one solution for negative $K$
$endgroup$
– Mark Bennet
Jan 20 at 19:50
$begingroup$
Yes, but $K=0$ wasn't one of the choices
$endgroup$
– J. W. Tanner
Jan 20 at 19:51
$begingroup$
Well the question is somewhat unclear at the moment and is inadequately specified - negative values are admitted. I can choose an arbitrary integer value of $x$ and then fix $K$ so that this is a root. The other root will then necessarily be rational (sum of roots is rational). (Or I could choose an arbitrary rational value of $x$, though $k$ would likely not be an integer).
$endgroup$
– Mark Bennet
Jan 20 at 19:53
add a comment |
0
$begingroup$
$sqrt{7^2-4times6times K}$ is rational when $K=1$ or $2$.
answered Jan 20 at 18:55
J. W. TannerJ. W. Tanner
2,1471117
$endgroup$
$begingroup$
Or $K=0$ I think ... and I think there is at least one solution for negative $K$
$endgroup$
– Mark Bennet
Jan 20 at 19:50
$begingroup$
Yes, but $K=0$ wasn't one of the choices
$endgroup$
– J. W. Tanner
Jan 20 at 19:51
$begingroup$
Well the question is somewhat unclear at the moment and is inadequately specified - negative values are admitted. I can choose an arbitrary integer value of $x$ and then fix $K$ so that this is a root. The other root will then necessarily be rational (sum of roots is rational). (Or I could choose an arbitrary rational value of $x$, though $k$ would likely not be an integer).
$endgroup$
– Mark Bennet
Jan 20 at 19:53
add a comment |
0
0
0
$begingroup$
$sqrt{7^2-4times6times K}$ is rational when $K=1$ or $2$.
answered Jan 20 at 18:55
J. W. TannerJ. W. Tanner
2,1471117
$endgroup$
$sqrt{7^2-4times6times K}$ is rational when $K=1$ or $2$.
answered Jan 20 at 18:55
J. W. TannerJ. W. Tanner
2,1471117
answered Jan 20 at 18:55
J. W. TannerJ. W. Tanner
2,1471117
answered Jan 20 at 18:55
J. W. TannerJ. W. Tanner
2,1471117
answered Jan 20 at 18:55
J. W. TannerJ. W. Tanner
2,1471117
2,1471117
$begingroup$
Or $K=0$ I think ... and I think there is at least one solution for negative $K$
$endgroup$
– Mark Bennet
Jan 20 at 19:50
$begingroup$
Yes, but $K=0$ wasn't one of the choices
$endgroup$
– J. W. Tanner
Jan 20 at 19:51
$begingroup$
Well the question is somewhat unclear at the moment and is inadequately specified - negative values are admitted. I can choose an arbitrary integer value of $x$ and then fix $K$ so that this is a root. The other root will then necessarily be rational (sum of roots is rational). (Or I could choose an arbitrary rational value of $x$, though $k$ would likely not be an integer).
$endgroup$
– Mark Bennet
Jan 20 at 19:53
add a comment |
$begingroup$
Or $K=0$ I think ... and I think there is at least one solution for negative $K$
$endgroup$
– Mark Bennet
Jan 20 at 19:50
$begingroup$
Yes, but $K=0$ wasn't one of the choices
$endgroup$
– J. W. Tanner
Jan 20 at 19:51
$begingroup$
Well the question is somewhat unclear at the moment and is inadequately specified - negative values are admitted. I can choose an arbitrary integer value of $x$ and then fix $K$ so that this is a root. The other root will then necessarily be rational (sum of roots is rational). (Or I could choose an arbitrary rational value of $x$, though $k$ would likely not be an integer).
$endgroup$
– Mark Bennet
Jan 20 at 19:53
$begingroup$
Or $K=0$ I think ... and I think there is at least one solution for negative $K$
$endgroup$
– Mark Bennet
Jan 20 at 19:50
$begingroup$
Or $K=0$ I think ... and I think there is at least one solution for negative $K$
$endgroup$
– Mark Bennet
Jan 20 at 19:50
$begingroup$
Yes, but $K=0$ wasn't one of the choices
$endgroup$
– J. W. Tanner
Jan 20 at 19:51
$begingroup$
Yes, but $K=0$ wasn't one of the choices
$endgroup$
– J. W. Tanner
Jan 20 at 19:51
$begingroup$
Well the question is somewhat unclear at the moment and is inadequately specified - negative values are admitted. I can choose an arbitrary integer value of $x$ and then fix $K$ so that this is a root. The other root will then necessarily be rational (sum of roots is rational). (Or I could choose an arbitrary rational value of $x$, though $k$ would likely not be an integer).
$endgroup$
– Mark Bennet
Jan 20 at 19:53
$begingroup$
Well the question is somewhat unclear at the moment and is inadequately specified - negative values are admitted. I can choose an arbitrary integer value of $x$ and then fix $K$ so that this is a root. The other root will then necessarily be rational (sum of roots is rational). (Or I could choose an arbitrary rational value of $x$, though $k$ would likely not be an integer).
$endgroup$
– Mark Bennet
Jan 20 at 19:53
add a comment |
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1
$begingroup$
The discriminant is a perfect square.
$endgroup$
– hamam_Abdallah
Jan 20 at 18:48
$begingroup$
Welcome to MSE. Please don't post what look like homework questions -- provide some context around the question and tell us what attempts you've made and where you got stuck so we can help you better. This kind of naked question tends to attract downvotes and can get closed.
$endgroup$
– postmortes
Jan 20 at 18:53
$begingroup$
"then is equal to" --- Then what is equal to?
$endgroup$
– steven gregory
Jan 20 at 19:02