Quadratic Equation
$begingroup$
what does a non real solution suggest when using the quadratic equation? Explain how the graph of a quadratic equation/ function with non real solutions differs from those graphs with real solutions.
algebra-precalculus
edited Apr 28 '14 at 18:44
amWhy
1
asked Apr 28 '14 at 18:43
user146438user146438
91
$endgroup$
add a comment |
$begingroup$
what does a non real solution suggest when using the quadratic equation? Explain how the graph of a quadratic equation/ function with non real solutions differs from those graphs with real solutions.
algebra-precalculus
edited Apr 28 '14 at 18:44
amWhy
1
asked Apr 28 '14 at 18:43
user146438user146438
91
$endgroup$
$begingroup$
What have you tried? Have you grabbed a plotting package and looked at some examples?
$endgroup$
– vonbrand
Apr 28 '14 at 19:13
add a comment |
$begingroup$
what does a non real solution suggest when using the quadratic equation? Explain how the graph of a quadratic equation/ function with non real solutions differs from those graphs with real solutions.
algebra-precalculus
edited Apr 28 '14 at 18:44
amWhy
1
asked Apr 28 '14 at 18:43
user146438user146438
91
$endgroup$
what does a non real solution suggest when using the quadratic equation? Explain how the graph of a quadratic equation/ function with non real solutions differs from those graphs with real solutions.
algebra-precalculus
algebra-precalculus
edited Apr 28 '14 at 18:44
amWhy
1
asked Apr 28 '14 at 18:43
user146438user146438
91
edited Apr 28 '14 at 18:44
amWhy
1
asked Apr 28 '14 at 18:43
user146438user146438
91
edited Apr 28 '14 at 18:44
amWhy
1
edited Apr 28 '14 at 18:44
amWhy
1
edited Apr 28 '14 at 18:44
amWhy
1
1
asked Apr 28 '14 at 18:43
user146438user146438
91
asked Apr 28 '14 at 18:43
user146438user146438
91
asked Apr 28 '14 at 18:43
user146438user146438
91
91
$begingroup$
What have you tried? Have you grabbed a plotting package and looked at some examples?
$endgroup$
– vonbrand
Apr 28 '14 at 19:13
add a comment |
$begingroup$
What have you tried? Have you grabbed a plotting package and looked at some examples?
$endgroup$
– vonbrand
Apr 28 '14 at 19:13
$begingroup$
What have you tried? Have you grabbed a plotting package and looked at some examples?
$endgroup$
– vonbrand
Apr 28 '14 at 19:13
$begingroup$
What have you tried? Have you grabbed a plotting package and looked at some examples?
$endgroup$
– vonbrand
Apr 28 '14 at 19:13
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
First note that a non-real solution implies there is another one as they always come in pairs.
Graphs with of quadratic with real solutions will cross the $x$-axis. If it doesn't it will not cross the $x$-axis.
In the equation $f(x) = ax^2 + bx + c$, consider the determinant $Delta = b^2 - 4ac$. See the picture to see all the possibilities that the sign of $Delta$ can take.
In summary :
- $Delta > 0$ implies two real roots
- $Delta = 0$ implies one repeated real root
- $Delta < 0$ implies roots are not real
The line you are interested in the yellow on which doesn't cross the $x$-axis line.
answered Apr 28 '14 at 18:52
user88595user88595
3,88111830
$endgroup$
2
$begingroup$
An explanation on the downvote would be greatly appreciated.
$endgroup$
– user88595
Apr 28 '14 at 19:03
add a comment |
$begingroup$
If there is no real solution then the graph does not cross the x-axis.
answered Apr 28 '14 at 18:45
user121049user121049
1,362174
$endgroup$
add a comment |
$begingroup$
A non-real solution suggests that there is no real value of $x$ which will give a point that lies on your function. A real solution means that there is some real number $c$ where $=c$ and the points at which $x$ will equal to $c$ will cross the $x$- axis.
So $Delta$ $>$ $0$ means there are two real solutions exactly where the blue parabola crosses the $x$- axis(you can see the two bold points).
At $Delta$ $=$ $0$, you can see that the parabola only touches the $x$- axis so there will be two real solutions here but they will be the same value.
Finally, at $Delta$ $<$ $0$, you can see that there are no points at which the parabola intersects the $x$- axis. The parabola will go on without bound in the upward direction. So $x$ can't have any value.
Remember these graphs are assumed to lie on the real plane. Had this been a complex plane the picture would look different.So basically real solutions are $x$-values and if the function doesn't cross the $x$-axis there cannot be roots for that graph.
edited Jan 11 '18 at 19:05
Prakhar Nagpal
747318
answered Jul 20 '15 at 0:01
RedRed
1,782934
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First note that a non-real solution implies there is another one as they always come in pairs.
Graphs with of quadratic with real solutions will cross the $x$-axis. If it doesn't it will not cross the $x$-axis.
In the equation $f(x) = ax^2 + bx + c$, consider the determinant $Delta = b^2 - 4ac$. See the picture to see all the possibilities that the sign of $Delta$ can take.
In summary :
- $Delta > 0$ implies two real roots
- $Delta = 0$ implies one repeated real root
- $Delta < 0$ implies roots are not real
The line you are interested in the yellow on which doesn't cross the $x$-axis line.
answered Apr 28 '14 at 18:52
user88595user88595
3,88111830
$endgroup$
2
$begingroup$
An explanation on the downvote would be greatly appreciated.
$endgroup$
– user88595
Apr 28 '14 at 19:03
add a comment |
$begingroup$
First note that a non-real solution implies there is another one as they always come in pairs.
Graphs with of quadratic with real solutions will cross the $x$-axis. If it doesn't it will not cross the $x$-axis.
In the equation $f(x) = ax^2 + bx + c$, consider the determinant $Delta = b^2 - 4ac$. See the picture to see all the possibilities that the sign of $Delta$ can take.
In summary :
- $Delta > 0$ implies two real roots
- $Delta = 0$ implies one repeated real root
- $Delta < 0$ implies roots are not real
The line you are interested in the yellow on which doesn't cross the $x$-axis line.
answered Apr 28 '14 at 18:52
user88595user88595
3,88111830
$endgroup$
2
$begingroup$
An explanation on the downvote would be greatly appreciated.
$endgroup$
– user88595
Apr 28 '14 at 19:03
add a comment |
$begingroup$
First note that a non-real solution implies there is another one as they always come in pairs.
Graphs with of quadratic with real solutions will cross the $x$-axis. If it doesn't it will not cross the $x$-axis.
In the equation $f(x) = ax^2 + bx + c$, consider the determinant $Delta = b^2 - 4ac$. See the picture to see all the possibilities that the sign of $Delta$ can take.
In summary :
- $Delta > 0$ implies two real roots
- $Delta = 0$ implies one repeated real root
- $Delta < 0$ implies roots are not real
The line you are interested in the yellow on which doesn't cross the $x$-axis line.
answered Apr 28 '14 at 18:52
user88595user88595
3,88111830
$endgroup$
First note that a non-real solution implies there is another one as they always come in pairs.
Graphs with of quadratic with real solutions will cross the $x$-axis. If it doesn't it will not cross the $x$-axis.
In the equation $f(x) = ax^2 + bx + c$, consider the determinant $Delta = b^2 - 4ac$. See the picture to see all the possibilities that the sign of $Delta$ can take.
In summary :
- $Delta > 0$ implies two real roots
- $Delta = 0$ implies one repeated real root
- $Delta < 0$ implies roots are not real
The line you are interested in the yellow on which doesn't cross the $x$-axis line.
answered Apr 28 '14 at 18:52
user88595user88595
3,88111830
edited Apr 28 '14 at 19:03
answered Apr 28 '14 at 18:52
user88595user88595
3,88111830
answered Apr 28 '14 at 18:52
user88595user88595
3,88111830
answered Apr 28 '14 at 18:52
user88595user88595
3,88111830
3,88111830
2
$begingroup$
An explanation on the downvote would be greatly appreciated.
$endgroup$
– user88595
Apr 28 '14 at 19:03
add a comment |
2
$begingroup$
An explanation on the downvote would be greatly appreciated.
$endgroup$
– user88595
Apr 28 '14 at 19:03
2
2
$begingroup$
An explanation on the downvote would be greatly appreciated.
$endgroup$
– user88595
Apr 28 '14 at 19:03
$begingroup$
An explanation on the downvote would be greatly appreciated.
$endgroup$
– user88595
Apr 28 '14 at 19:03
add a comment |
$begingroup$
If there is no real solution then the graph does not cross the x-axis.
answered Apr 28 '14 at 18:45
user121049user121049
1,362174
$endgroup$
add a comment |
$begingroup$
If there is no real solution then the graph does not cross the x-axis.
answered Apr 28 '14 at 18:45
user121049user121049
1,362174
$endgroup$
add a comment |
$begingroup$
If there is no real solution then the graph does not cross the x-axis.
answered Apr 28 '14 at 18:45
user121049user121049
1,362174
$endgroup$
If there is no real solution then the graph does not cross the x-axis.
answered Apr 28 '14 at 18:45
user121049user121049
1,362174
answered Apr 28 '14 at 18:45
user121049user121049
1,362174
answered Apr 28 '14 at 18:45
user121049user121049
1,362174
answered Apr 28 '14 at 18:45
user121049user121049
1,362174
1,362174
add a comment |
add a comment |
$begingroup$
A non-real solution suggests that there is no real value of $x$ which will give a point that lies on your function. A real solution means that there is some real number $c$ where $=c$ and the points at which $x$ will equal to $c$ will cross the $x$- axis.
So $Delta$ $>$ $0$ means there are two real solutions exactly where the blue parabola crosses the $x$- axis(you can see the two bold points).
At $Delta$ $=$ $0$, you can see that the parabola only touches the $x$- axis so there will be two real solutions here but they will be the same value.
Finally, at $Delta$ $<$ $0$, you can see that there are no points at which the parabola intersects the $x$- axis. The parabola will go on without bound in the upward direction. So $x$ can't have any value.
Remember these graphs are assumed to lie on the real plane. Had this been a complex plane the picture would look different.So basically real solutions are $x$-values and if the function doesn't cross the $x$-axis there cannot be roots for that graph.
edited Jan 11 '18 at 19:05
Prakhar Nagpal
747318
answered Jul 20 '15 at 0:01
RedRed
1,782934
$endgroup$
add a comment |
$begingroup$
A non-real solution suggests that there is no real value of $x$ which will give a point that lies on your function. A real solution means that there is some real number $c$ where $=c$ and the points at which $x$ will equal to $c$ will cross the $x$- axis.
So $Delta$ $>$ $0$ means there are two real solutions exactly where the blue parabola crosses the $x$- axis(you can see the two bold points).
At $Delta$ $=$ $0$, you can see that the parabola only touches the $x$- axis so there will be two real solutions here but they will be the same value.
Finally, at $Delta$ $<$ $0$, you can see that there are no points at which the parabola intersects the $x$- axis. The parabola will go on without bound in the upward direction. So $x$ can't have any value.
Remember these graphs are assumed to lie on the real plane. Had this been a complex plane the picture would look different.So basically real solutions are $x$-values and if the function doesn't cross the $x$-axis there cannot be roots for that graph.
edited Jan 11 '18 at 19:05
Prakhar Nagpal
747318
answered Jul 20 '15 at 0:01
RedRed
1,782934
$endgroup$
add a comment |
$begingroup$
A non-real solution suggests that there is no real value of $x$ which will give a point that lies on your function. A real solution means that there is some real number $c$ where $=c$ and the points at which $x$ will equal to $c$ will cross the $x$- axis.
So $Delta$ $>$ $0$ means there are two real solutions exactly where the blue parabola crosses the $x$- axis(you can see the two bold points).
At $Delta$ $=$ $0$, you can see that the parabola only touches the $x$- axis so there will be two real solutions here but they will be the same value.
Finally, at $Delta$ $<$ $0$, you can see that there are no points at which the parabola intersects the $x$- axis. The parabola will go on without bound in the upward direction. So $x$ can't have any value.
Remember these graphs are assumed to lie on the real plane. Had this been a complex plane the picture would look different.So basically real solutions are $x$-values and if the function doesn't cross the $x$-axis there cannot be roots for that graph.
edited Jan 11 '18 at 19:05
Prakhar Nagpal
747318
answered Jul 20 '15 at 0:01
RedRed
1,782934
$endgroup$
A non-real solution suggests that there is no real value of $x$ which will give a point that lies on your function. A real solution means that there is some real number $c$ where $=c$ and the points at which $x$ will equal to $c$ will cross the $x$- axis.
So $Delta$ $>$ $0$ means there are two real solutions exactly where the blue parabola crosses the $x$- axis(you can see the two bold points).
At $Delta$ $=$ $0$, you can see that the parabola only touches the $x$- axis so there will be two real solutions here but they will be the same value.
Finally, at $Delta$ $<$ $0$, you can see that there are no points at which the parabola intersects the $x$- axis. The parabola will go on without bound in the upward direction. So $x$ can't have any value.
Remember these graphs are assumed to lie on the real plane. Had this been a complex plane the picture would look different.So basically real solutions are $x$-values and if the function doesn't cross the $x$-axis there cannot be roots for that graph.
edited Jan 11 '18 at 19:05
Prakhar Nagpal
747318
answered Jul 20 '15 at 0:01
RedRed
1,782934
edited Jan 11 '18 at 19:05
Prakhar Nagpal
747318
edited Jan 11 '18 at 19:05
Prakhar Nagpal
747318
edited Jan 11 '18 at 19:05
Prakhar Nagpal
747318
747318
answered Jul 20 '15 at 0:01
RedRed
1,782934
answered Jul 20 '15 at 0:01
RedRed
1,782934
answered Jul 20 '15 at 0:01
RedRed
1,782934
1,782934
add a comment |
add a comment |
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$begingroup$
What have you tried? Have you grabbed a plotting package and looked at some examples?
$endgroup$
– vonbrand
Apr 28 '14 at 19:13