How to graphically depict the possible solutions of a quadratic equation
up vote
0
down vote
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I have the following quadratic equation :
$$am^2 + bm + (c_1^2 +c_2^2) =0,$$
where the solution is given by
$$m = frac{-bpmsqrt{b^2-4a(c_1^2+c_2^2)}}{2a}.$$ Here, $Delta>0$.
Thus I have two real roots What I would like to do is to graphically depict possible solutions based on the values of $c_1$ and $c_2$. Here, $c_1$ and $c_2$ are two components of a vector and are real numbers.
Can anyone help me out with this? And is it possible to do that?
roots graphing-functions quadratics
asked Nov 28 at 9:17
newstudent
346
|
show 4 more comments
up vote
0
down vote
favorite
I have the following quadratic equation :
$$am^2 + bm + (c_1^2 +c_2^2) =0,$$
where the solution is given by
$$m = frac{-bpmsqrt{b^2-4a(c_1^2+c_2^2)}}{2a}.$$ Here, $Delta>0$.
Thus I have two real roots What I would like to do is to graphically depict possible solutions based on the values of $c_1$ and $c_2$. Here, $c_1$ and $c_2$ are two components of a vector and are real numbers.
Can anyone help me out with this? And is it possible to do that?
roots graphing-functions quadratics
asked Nov 28 at 9:17
newstudent
346
There is a formula for the two (positive) solutions $m_{1,2}$.
– Wuestenfux
Nov 28 at 9:26
@Wuestenfux I guess here, both the solutions might not be positive. Most likely one is positive and other one is negative.
– newstudent
Nov 28 at 9:41
Add the condition that the discriminant is positive. This should give you a condition on the coefficients.
– Wuestenfux
Nov 28 at 9:42
What do you mean by "depict possible solutions"?
– Todor Markov
Nov 28 at 9:43
@TodorMarkov I would like to plot the possible values of m as a function of $c_1$ and $c_2$ or $b$.
– newstudent
Nov 28 at 9:47
|
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the following quadratic equation :
$$am^2 + bm + (c_1^2 +c_2^2) =0,$$
where the solution is given by
$$m = frac{-bpmsqrt{b^2-4a(c_1^2+c_2^2)}}{2a}.$$ Here, $Delta>0$.
Thus I have two real roots What I would like to do is to graphically depict possible solutions based on the values of $c_1$ and $c_2$. Here, $c_1$ and $c_2$ are two components of a vector and are real numbers.
Can anyone help me out with this? And is it possible to do that?
roots graphing-functions quadratics
asked Nov 28 at 9:17
newstudent
346
I have the following quadratic equation :
$$am^2 + bm + (c_1^2 +c_2^2) =0,$$
where the solution is given by
$$m = frac{-bpmsqrt{b^2-4a(c_1^2+c_2^2)}}{2a}.$$ Here, $Delta>0$.
Thus I have two real roots What I would like to do is to graphically depict possible solutions based on the values of $c_1$ and $c_2$. Here, $c_1$ and $c_2$ are two components of a vector and are real numbers.
Can anyone help me out with this? And is it possible to do that?
roots graphing-functions quadratics
roots graphing-functions quadratics
asked Nov 28 at 9:17
newstudent
346
asked Nov 28 at 9:17
newstudent
346
edited Nov 28 at 10:17
asked Nov 28 at 9:17
newstudent
346
asked Nov 28 at 9:17
newstudent
346
asked Nov 28 at 9:17
newstudent
346
346
There is a formula for the two (positive) solutions $m_{1,2}$.
– Wuestenfux
Nov 28 at 9:26
@Wuestenfux I guess here, both the solutions might not be positive. Most likely one is positive and other one is negative.
– newstudent
Nov 28 at 9:41
Add the condition that the discriminant is positive. This should give you a condition on the coefficients.
– Wuestenfux
Nov 28 at 9:42
What do you mean by "depict possible solutions"?
– Todor Markov
Nov 28 at 9:43
@TodorMarkov I would like to plot the possible values of m as a function of $c_1$ and $c_2$ or $b$.
– newstudent
Nov 28 at 9:47
|
show 4 more comments
There is a formula for the two (positive) solutions $m_{1,2}$.
– Wuestenfux
Nov 28 at 9:26
@Wuestenfux I guess here, both the solutions might not be positive. Most likely one is positive and other one is negative.
– newstudent
Nov 28 at 9:41
Add the condition that the discriminant is positive. This should give you a condition on the coefficients.
– Wuestenfux
Nov 28 at 9:42
What do you mean by "depict possible solutions"?
– Todor Markov
Nov 28 at 9:43
@TodorMarkov I would like to plot the possible values of m as a function of $c_1$ and $c_2$ or $b$.
– newstudent
Nov 28 at 9:47
There is a formula for the two (positive) solutions $m_{1,2}$.
– Wuestenfux
Nov 28 at 9:26
There is a formula for the two (positive) solutions $m_{1,2}$.
– Wuestenfux
Nov 28 at 9:26
@Wuestenfux I guess here, both the solutions might not be positive. Most likely one is positive and other one is negative.
– newstudent
Nov 28 at 9:41
@Wuestenfux I guess here, both the solutions might not be positive. Most likely one is positive and other one is negative.
– newstudent
Nov 28 at 9:41
Add the condition that the discriminant is positive. This should give you a condition on the coefficients.
– Wuestenfux
Nov 28 at 9:42
Add the condition that the discriminant is positive. This should give you a condition on the coefficients.
– Wuestenfux
Nov 28 at 9:42
What do you mean by "depict possible solutions"?
– Todor Markov
Nov 28 at 9:43
What do you mean by "depict possible solutions"?
– Todor Markov
Nov 28 at 9:43
@TodorMarkov I would like to plot the possible values of m as a function of $c_1$ and $c_2$ or $b$.
– newstudent
Nov 28 at 9:47
@TodorMarkov I would like to plot the possible values of m as a function of $c_1$ and $c_2$ or $b$.
– newstudent
Nov 28 at 9:47
|
show 4 more comments
1 Answer
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Let $f(x) = ax^2 + bx + (c_1^2 +c_2^2)$.
If $Delta gt 0$, then
(1) $f(x) = 0$ has two real roots ($alpha$ and $beta$, say) and this implies the graph of $y = f(x)$ will cut the x-axis at two distinct points [$(alpha, 0)$ and $(beta, 0)$].
(2) $b^2 gt 4a(c_1^2 + c_2^2)$. From $dfrac {b^2 }{4a }gt c_1^2 + c_2^2$, we can say $a gt 0$. This further means the graph of $y = f(x)$ concaves upward. Since $alpha times beta = dfrac {c_1^2 + c_2^2}{a}$, which is positive, the two points will lie on the same side of the y-axis (either left or right but not both).
Since $f(0) = (c_1^2 +c_2^2) gt 0$, $y = f(x)$ passes through $(0, k)$ where $k = c_1^2 +c_2^2$, which is positive.
Because we know nothing about the sign of b, there are two possible solutions for the plot as shown.
Even from the fact that the minimum occurs at $(dfrac {-b}{2a}, f[dfrac {-b}{2a}])$ with $f[dfrac {-b}{2a}] < 0$, we can at the most deduce $dfrac {(c_1^2 + c_2^2)}{a} < (dfrac {-b}{2a})^2$. But we still cannot make any further deduction.
answered Nov 28 at 15:31
Mick
11.7k21641
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $f(x) = ax^2 + bx + (c_1^2 +c_2^2)$.
If $Delta gt 0$, then
(1) $f(x) = 0$ has two real roots ($alpha$ and $beta$, say) and this implies the graph of $y = f(x)$ will cut the x-axis at two distinct points [$(alpha, 0)$ and $(beta, 0)$].
(2) $b^2 gt 4a(c_1^2 + c_2^2)$. From $dfrac {b^2 }{4a }gt c_1^2 + c_2^2$, we can say $a gt 0$. This further means the graph of $y = f(x)$ concaves upward. Since $alpha times beta = dfrac {c_1^2 + c_2^2}{a}$, which is positive, the two points will lie on the same side of the y-axis (either left or right but not both).
Since $f(0) = (c_1^2 +c_2^2) gt 0$, $y = f(x)$ passes through $(0, k)$ where $k = c_1^2 +c_2^2$, which is positive.
Because we know nothing about the sign of b, there are two possible solutions for the plot as shown.
Even from the fact that the minimum occurs at $(dfrac {-b}{2a}, f[dfrac {-b}{2a}])$ with $f[dfrac {-b}{2a}] < 0$, we can at the most deduce $dfrac {(c_1^2 + c_2^2)}{a} < (dfrac {-b}{2a})^2$. But we still cannot make any further deduction.
answered Nov 28 at 15:31
Mick
11.7k21641
add a comment |
up vote
1
down vote
accepted
Let $f(x) = ax^2 + bx + (c_1^2 +c_2^2)$.
If $Delta gt 0$, then
(1) $f(x) = 0$ has two real roots ($alpha$ and $beta$, say) and this implies the graph of $y = f(x)$ will cut the x-axis at two distinct points [$(alpha, 0)$ and $(beta, 0)$].
(2) $b^2 gt 4a(c_1^2 + c_2^2)$. From $dfrac {b^2 }{4a }gt c_1^2 + c_2^2$, we can say $a gt 0$. This further means the graph of $y = f(x)$ concaves upward. Since $alpha times beta = dfrac {c_1^2 + c_2^2}{a}$, which is positive, the two points will lie on the same side of the y-axis (either left or right but not both).
Since $f(0) = (c_1^2 +c_2^2) gt 0$, $y = f(x)$ passes through $(0, k)$ where $k = c_1^2 +c_2^2$, which is positive.
Because we know nothing about the sign of b, there are two possible solutions for the plot as shown.
Even from the fact that the minimum occurs at $(dfrac {-b}{2a}, f[dfrac {-b}{2a}])$ with $f[dfrac {-b}{2a}] < 0$, we can at the most deduce $dfrac {(c_1^2 + c_2^2)}{a} < (dfrac {-b}{2a})^2$. But we still cannot make any further deduction.
answered Nov 28 at 15:31
Mick
11.7k21641
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $f(x) = ax^2 + bx + (c_1^2 +c_2^2)$.
If $Delta gt 0$, then
(1) $f(x) = 0$ has two real roots ($alpha$ and $beta$, say) and this implies the graph of $y = f(x)$ will cut the x-axis at two distinct points [$(alpha, 0)$ and $(beta, 0)$].
(2) $b^2 gt 4a(c_1^2 + c_2^2)$. From $dfrac {b^2 }{4a }gt c_1^2 + c_2^2$, we can say $a gt 0$. This further means the graph of $y = f(x)$ concaves upward. Since $alpha times beta = dfrac {c_1^2 + c_2^2}{a}$, which is positive, the two points will lie on the same side of the y-axis (either left or right but not both).
Since $f(0) = (c_1^2 +c_2^2) gt 0$, $y = f(x)$ passes through $(0, k)$ where $k = c_1^2 +c_2^2$, which is positive.
Because we know nothing about the sign of b, there are two possible solutions for the plot as shown.
Even from the fact that the minimum occurs at $(dfrac {-b}{2a}, f[dfrac {-b}{2a}])$ with $f[dfrac {-b}{2a}] < 0$, we can at the most deduce $dfrac {(c_1^2 + c_2^2)}{a} < (dfrac {-b}{2a})^2$. But we still cannot make any further deduction.
answered Nov 28 at 15:31
Mick
11.7k21641
Let $f(x) = ax^2 + bx + (c_1^2 +c_2^2)$.
If $Delta gt 0$, then
(1) $f(x) = 0$ has two real roots ($alpha$ and $beta$, say) and this implies the graph of $y = f(x)$ will cut the x-axis at two distinct points [$(alpha, 0)$ and $(beta, 0)$].
(2) $b^2 gt 4a(c_1^2 + c_2^2)$. From $dfrac {b^2 }{4a }gt c_1^2 + c_2^2$, we can say $a gt 0$. This further means the graph of $y = f(x)$ concaves upward. Since $alpha times beta = dfrac {c_1^2 + c_2^2}{a}$, which is positive, the two points will lie on the same side of the y-axis (either left or right but not both).
Since $f(0) = (c_1^2 +c_2^2) gt 0$, $y = f(x)$ passes through $(0, k)$ where $k = c_1^2 +c_2^2$, which is positive.
Because we know nothing about the sign of b, there are two possible solutions for the plot as shown.
Even from the fact that the minimum occurs at $(dfrac {-b}{2a}, f[dfrac {-b}{2a}])$ with $f[dfrac {-b}{2a}] < 0$, we can at the most deduce $dfrac {(c_1^2 + c_2^2)}{a} < (dfrac {-b}{2a})^2$. But we still cannot make any further deduction.
answered Nov 28 at 15:31
Mick
11.7k21641
edited Nov 29 at 2:34
answered Nov 28 at 15:31
Mick
11.7k21641
answered Nov 28 at 15:31
Mick
11.7k21641
answered Nov 28 at 15:31
Mick
11.7k21641
11.7k21641
add a comment |
add a comment |
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There is a formula for the two (positive) solutions $m_{1,2}$.
– Wuestenfux
Nov 28 at 9:26
@Wuestenfux I guess here, both the solutions might not be positive. Most likely one is positive and other one is negative.
– newstudent
Nov 28 at 9:41
Add the condition that the discriminant is positive. This should give you a condition on the coefficients.
– Wuestenfux
Nov 28 at 9:42
What do you mean by "depict possible solutions"?
– Todor Markov
Nov 28 at 9:43
@TodorMarkov I would like to plot the possible values of m as a function of $c_1$ and $c_2$ or $b$.
– newstudent
Nov 28 at 9:47