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How is discriminant related to real $x$?
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Question in my text book
Solve for range of the function, $$y=fracx^2+4x-13x^2+12x +20$$
Text book says, cross multiply and express the obtained equation as a quadratic equation in $x$
So I get $ (3y-1)x^2 + (12y -4)x + (20y-1) = 0$
Now it says find discriminant $D$, so we have,
$$ D = -4 ( 3y-1)( 8y+3)$$
Now it says, set $D≥0$ as $x$ is real.
Wait what?
Isn't $x in mathbbR$ the domain for a quadratic function? Meaning "$x$" is always real? What does a discriminant got anything to do with $x$ being real, when all discriminant tells us is whether or not the ROOTS are real? Help please.
To be more specific about my doubt, here's an edit.
EDIT : I'm confused, setting discriminant $≥0$ would tell whether or not roots are real, meaning whether the graph of the quadratic function cuts/touches X axis. Now tell me what does this got anything to do with range? As far as I know, quadratic fucntions that don't have real roots are also continous throughout the X axis, meaning there SHOULD always be a corresponding $y-value$
functions quadratics
asked Jul 17 at 16:41
William
801214
add a comment |Â
up vote
2
down vote
favorite
Question in my text book
Solve for range of the function, $$y=fracx^2+4x-13x^2+12x +20$$
Text book says, cross multiply and express the obtained equation as a quadratic equation in $x$
So I get $ (3y-1)x^2 + (12y -4)x + (20y-1) = 0$
Now it says find discriminant $D$, so we have,
$$ D = -4 ( 3y-1)( 8y+3)$$
Now it says, set $D≥0$ as $x$ is real.
Wait what?
Isn't $x in mathbbR$ the domain for a quadratic function? Meaning "$x$" is always real? What does a discriminant got anything to do with $x$ being real, when all discriminant tells us is whether or not the ROOTS are real? Help please.
To be more specific about my doubt, here's an edit.
EDIT : I'm confused, setting discriminant $≥0$ would tell whether or not roots are real, meaning whether the graph of the quadratic function cuts/touches X axis. Now tell me what does this got anything to do with range? As far as I know, quadratic fucntions that don't have real roots are also continous throughout the X axis, meaning there SHOULD always be a corresponding $y-value$
functions quadratics
asked Jul 17 at 16:41
William
801214
Can you allow any imaginary value of $x$ in the function?
– Entrepreneur
Jul 17 at 16:46
2
That's right, but you're solving for $x$. You're asking for what values of $y$ is the there a real value of $x$ such that the given rational function takes the value $y$ at $x$?
– saulspatz
Jul 17 at 16:48
2
The question has overloaded the meaning of variable $x$ and caused your confusion. What happens is given a $x$, you can compute a $y$ and the original $x$ will be a root for the quadratic polynomial $(3y-1)t^2 + (12y -4)t + (20y-1) = 0$ in $t$. Since this quadratic polynomial (assume $y ne frac13$) has a real root at $t = x$. Its discriminant $D ge 0$.
– achille hui
Jul 17 at 16:51
@saulspatz please see the question, I've edited
– William
Jul 17 at 16:56
@achillehui I read all these answers, even though they answered it in their own way, they still weren't being clear as to WHY should the discriminant be non-negative. I read your comment and Riley Jacob's answer and yall answered exactly what I was asking, thank you so much man, you're the man!! I was getting confused between roots of the equation and roots of the functions. I hope I don't make this mistake ever again, ty again.
– William
Jul 17 at 17:49
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Question in my text book
Solve for range of the function, $$y=fracx^2+4x-13x^2+12x +20$$
Text book says, cross multiply and express the obtained equation as a quadratic equation in $x$
So I get $ (3y-1)x^2 + (12y -4)x + (20y-1) = 0$
Now it says find discriminant $D$, so we have,
$$ D = -4 ( 3y-1)( 8y+3)$$
Now it says, set $D≥0$ as $x$ is real.
Wait what?
Isn't $x in mathbbR$ the domain for a quadratic function? Meaning "$x$" is always real? What does a discriminant got anything to do with $x$ being real, when all discriminant tells us is whether or not the ROOTS are real? Help please.
To be more specific about my doubt, here's an edit.
EDIT : I'm confused, setting discriminant $≥0$ would tell whether or not roots are real, meaning whether the graph of the quadratic function cuts/touches X axis. Now tell me what does this got anything to do with range? As far as I know, quadratic fucntions that don't have real roots are also continous throughout the X axis, meaning there SHOULD always be a corresponding $y-value$
functions quadratics
asked Jul 17 at 16:41
William
801214
Question in my text book
Solve for range of the function, $$y=fracx^2+4x-13x^2+12x +20$$
Text book says, cross multiply and express the obtained equation as a quadratic equation in $x$
So I get $ (3y-1)x^2 + (12y -4)x + (20y-1) = 0$
Now it says find discriminant $D$, so we have,
$$ D = -4 ( 3y-1)( 8y+3)$$
Now it says, set $D≥0$ as $x$ is real.
Wait what?
Isn't $x in mathbbR$ the domain for a quadratic function? Meaning "$x$" is always real? What does a discriminant got anything to do with $x$ being real, when all discriminant tells us is whether or not the ROOTS are real? Help please.
To be more specific about my doubt, here's an edit.
EDIT : I'm confused, setting discriminant $≥0$ would tell whether or not roots are real, meaning whether the graph of the quadratic function cuts/touches X axis. Now tell me what does this got anything to do with range? As far as I know, quadratic fucntions that don't have real roots are also continous throughout the X axis, meaning there SHOULD always be a corresponding $y-value$
functions quadratics
asked Jul 17 at 16:41
William
801214
edited Jul 17 at 16:56
asked Jul 17 at 16:41
William
801214
asked Jul 17 at 16:41
William
801214
asked Jul 17 at 16:41
William
801214
801214
Can you allow any imaginary value of $x$ in the function?
– Entrepreneur
Jul 17 at 16:46
2
That's right, but you're solving for $x$. You're asking for what values of $y$ is the there a real value of $x$ such that the given rational function takes the value $y$ at $x$?
– saulspatz
Jul 17 at 16:48
2
The question has overloaded the meaning of variable $x$ and caused your confusion. What happens is given a $x$, you can compute a $y$ and the original $x$ will be a root for the quadratic polynomial $(3y-1)t^2 + (12y -4)t + (20y-1) = 0$ in $t$. Since this quadratic polynomial (assume $y ne frac13$) has a real root at $t = x$. Its discriminant $D ge 0$.
– achille hui
Jul 17 at 16:51
@saulspatz please see the question, I've edited
– William
Jul 17 at 16:56
@achillehui I read all these answers, even though they answered it in their own way, they still weren't being clear as to WHY should the discriminant be non-negative. I read your comment and Riley Jacob's answer and yall answered exactly what I was asking, thank you so much man, you're the man!! I was getting confused between roots of the equation and roots of the functions. I hope I don't make this mistake ever again, ty again.
– William
Jul 17 at 17:49
add a comment |Â
Can you allow any imaginary value of $x$ in the function?
– Entrepreneur
Jul 17 at 16:46
2
That's right, but you're solving for $x$. You're asking for what values of $y$ is the there a real value of $x$ such that the given rational function takes the value $y$ at $x$?
– saulspatz
Jul 17 at 16:48
2
The question has overloaded the meaning of variable $x$ and caused your confusion. What happens is given a $x$, you can compute a $y$ and the original $x$ will be a root for the quadratic polynomial $(3y-1)t^2 + (12y -4)t + (20y-1) = 0$ in $t$. Since this quadratic polynomial (assume $y ne frac13$) has a real root at $t = x$. Its discriminant $D ge 0$.
– achille hui
Jul 17 at 16:51
@saulspatz please see the question, I've edited
– William
Jul 17 at 16:56
@achillehui I read all these answers, even though they answered it in their own way, they still weren't being clear as to WHY should the discriminant be non-negative. I read your comment and Riley Jacob's answer and yall answered exactly what I was asking, thank you so much man, you're the man!! I was getting confused between roots of the equation and roots of the functions. I hope I don't make this mistake ever again, ty again.
– William
Jul 17 at 17:49
Can you allow any imaginary value of $x$ in the function?
– Entrepreneur
Jul 17 at 16:46
Can you allow any imaginary value of $x$ in the function?
– Entrepreneur
Jul 17 at 16:46
2
2
That's right, but you're solving for $x$. You're asking for what values of $y$ is the there a real value of $x$ such that the given rational function takes the value $y$ at $x$?
– saulspatz
Jul 17 at 16:48
That's right, but you're solving for $x$. You're asking for what values of $y$ is the there a real value of $x$ such that the given rational function takes the value $y$ at $x$?
– saulspatz
Jul 17 at 16:48
2
2
The question has overloaded the meaning of variable $x$ and caused your confusion. What happens is given a $x$, you can compute a $y$ and the original $x$ will be a root for the quadratic polynomial $(3y-1)t^2 + (12y -4)t + (20y-1) = 0$ in $t$. Since this quadratic polynomial (assume $y ne frac13$) has a real root at $t = x$. Its discriminant $D ge 0$.
– achille hui
Jul 17 at 16:51
The question has overloaded the meaning of variable $x$ and caused your confusion. What happens is given a $x$, you can compute a $y$ and the original $x$ will be a root for the quadratic polynomial $(3y-1)t^2 + (12y -4)t + (20y-1) = 0$ in $t$. Since this quadratic polynomial (assume $y ne frac13$) has a real root at $t = x$. Its discriminant $D ge 0$.
– achille hui
Jul 17 at 16:51
@saulspatz please see the question, I've edited
– William
Jul 17 at 16:56
@saulspatz please see the question, I've edited
– William
Jul 17 at 16:56
@achillehui I read all these answers, even though they answered it in their own way, they still weren't being clear as to WHY should the discriminant be non-negative. I read your comment and Riley Jacob's answer and yall answered exactly what I was asking, thank you so much man, you're the man!! I was getting confused between roots of the equation and roots of the functions. I hope I don't make this mistake ever again, ty again.
– William
Jul 17 at 17:49
@achillehui I read all these answers, even though they answered it in their own way, they still weren't being clear as to WHY should the discriminant be non-negative. I read your comment and Riley Jacob's answer and yall answered exactly what I was asking, thank you so much man, you're the man!! I was getting confused between roots of the equation and roots of the functions. I hope I don't make this mistake ever again, ty again.
– William
Jul 17 at 17:49
add a comment |Â
5 Answers
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up vote
2
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Here's an example that may clear up your confusion. Suppose we ask whether $1$ is in the range. We set the fraction equal to $1,$ cross multiply, and arrive at the equation $$
2x^2+8x+19=0,$$ which has roots $$x=frac-8pm sqrt64-1524$$
So, in order for $1$ to be in the range, $x$ must be non-real, and since $x$ is restricted to be real in the problem statement, $1$ is not in the range.
The method suggested in the book, is exactly this, with a general $y$ in place of $1$.
answered Jul 17 at 17:05
saulspatz
10.7k21323
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up vote
2
down vote
The domain of a quadratic polynomial with real coefficients need not be restricted to the reals.
The discriminant of a polynomial is a function of that polynomial's coefficients, and it can be used to deduce some properties of the roots (such as whether they are real). In the case of a quadratic, the roots will be real and distinct for $D>0$, real (but the same) if $D = 0$, but complex conjugates if $D <0$. That is to say, a negative discriminant implies that the roots of the quadratic will both be complex (i.e. there are no real roots).
What does a discriminant got anything to do with x being real, when all discriminant tells us is whether or not the ROOTS are real?
The $x$ in your above equation is the root(s), as you have set your expression equal to 0.
EDIT: I see your edit about the range, and will address it. Essentially, the problem asks for you to find the range of the function given the knowledge that its domain must be real. So, we can use the knowledge that the discriminant must be non-negative to solve for allowable y: $$(12y-4)^2 -4(3y-1)(20y+1) geq 0$$
$$implies -frac58 leq y < frac13$$
which is the range of your function.
EDIT 2: OP, note that the quadratic and therefore discriminant you calculated above is incorrect. Your "c" term should be $(20y+1)$, not $(20y-1)$.
answered Jul 17 at 16:53
Riley Jacob
1212
Thank you so much, this is exactly what I was asking.
– William
Jul 17 at 17:51
@William If you say so, check this answer.
– Takahiro Waki
Jul 30 at 0:38
add a comment |Â
up vote
1
down vote
$$fracx^2+4x-13x^2+12x+20=y$$
After cross multiplying, we obtain
$$(3y-1)x^2+(12y-4)x+(20y+1)=0$$
If the discriminant is nonnegative, then we can find real $x$ as solution to the quadratic equation, and hence given a particular value of $y$ that makes the discriminant nonnegative, we can find $x$ that satisfy the original equation.
However, if the discriminant is negative, then $x$ that satisfies the quadratic equation is not real.
answered Jul 17 at 16:51
Siong Thye Goh
77.7k134796
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up vote
0
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The $y$-values making the discriminant non-negative are the $y$-values the function can produce (at least one real solution $x$ exists) . Therefore the set of $y$ making the discriminant non-negative is the range.
answered Jul 17 at 16:50
Peter
45.1k939119
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$$fracx^2+4x-13x^2+12x +20 = fracx^2+4x + frac2033x^2+12x +20 - frac frac2333x^2+12x +20 = frac13- frac frac2333x^2+12x +20$$
Meanwhile, $$ 3x^2 + 12 x + 20 = 3 (x+2)^2 + 8 $$ is always positive
with a minimum value of $8$ when $x=-2,$ so the minimum value of the whole thing is
$$ frac13- frac frac2338 = frac13-frac2324= -frac58 $$
It has an upper bound of $frac13$ but never achieves the bound, it just gets very close when $|x|$ is large.
answered Jul 17 at 17:16
Will Jagy
97.2k594196
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Here's an example that may clear up your confusion. Suppose we ask whether $1$ is in the range. We set the fraction equal to $1,$ cross multiply, and arrive at the equation $$
2x^2+8x+19=0,$$ which has roots $$x=frac-8pm sqrt64-1524$$
So, in order for $1$ to be in the range, $x$ must be non-real, and since $x$ is restricted to be real in the problem statement, $1$ is not in the range.
The method suggested in the book, is exactly this, with a general $y$ in place of $1$.
answered Jul 17 at 17:05
saulspatz
10.7k21323
add a comment |Â
up vote
2
down vote
Here's an example that may clear up your confusion. Suppose we ask whether $1$ is in the range. We set the fraction equal to $1,$ cross multiply, and arrive at the equation $$
2x^2+8x+19=0,$$ which has roots $$x=frac-8pm sqrt64-1524$$
So, in order for $1$ to be in the range, $x$ must be non-real, and since $x$ is restricted to be real in the problem statement, $1$ is not in the range.
The method suggested in the book, is exactly this, with a general $y$ in place of $1$.
answered Jul 17 at 17:05
saulspatz
10.7k21323
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Here's an example that may clear up your confusion. Suppose we ask whether $1$ is in the range. We set the fraction equal to $1,$ cross multiply, and arrive at the equation $$
2x^2+8x+19=0,$$ which has roots $$x=frac-8pm sqrt64-1524$$
So, in order for $1$ to be in the range, $x$ must be non-real, and since $x$ is restricted to be real in the problem statement, $1$ is not in the range.
The method suggested in the book, is exactly this, with a general $y$ in place of $1$.
answered Jul 17 at 17:05
saulspatz
10.7k21323
Here's an example that may clear up your confusion. Suppose we ask whether $1$ is in the range. We set the fraction equal to $1,$ cross multiply, and arrive at the equation $$
2x^2+8x+19=0,$$ which has roots $$x=frac-8pm sqrt64-1524$$
So, in order for $1$ to be in the range, $x$ must be non-real, and since $x$ is restricted to be real in the problem statement, $1$ is not in the range.
The method suggested in the book, is exactly this, with a general $y$ in place of $1$.
answered Jul 17 at 17:05
saulspatz
10.7k21323
edited Jul 17 at 17:11
answered Jul 17 at 17:05
saulspatz
10.7k21323
answered Jul 17 at 17:05
saulspatz
10.7k21323
answered Jul 17 at 17:05
saulspatz
10.7k21323
10.7k21323
add a comment |Â
add a comment |Â
up vote
2
down vote
The domain of a quadratic polynomial with real coefficients need not be restricted to the reals.
The discriminant of a polynomial is a function of that polynomial's coefficients, and it can be used to deduce some properties of the roots (such as whether they are real). In the case of a quadratic, the roots will be real and distinct for $D>0$, real (but the same) if $D = 0$, but complex conjugates if $D <0$. That is to say, a negative discriminant implies that the roots of the quadratic will both be complex (i.e. there are no real roots).
What does a discriminant got anything to do with x being real, when all discriminant tells us is whether or not the ROOTS are real?
The $x$ in your above equation is the root(s), as you have set your expression equal to 0.
EDIT: I see your edit about the range, and will address it. Essentially, the problem asks for you to find the range of the function given the knowledge that its domain must be real. So, we can use the knowledge that the discriminant must be non-negative to solve for allowable y: $$(12y-4)^2 -4(3y-1)(20y+1) geq 0$$
$$implies -frac58 leq y < frac13$$
which is the range of your function.
EDIT 2: OP, note that the quadratic and therefore discriminant you calculated above is incorrect. Your "c" term should be $(20y+1)$, not $(20y-1)$.
answered Jul 17 at 16:53
Riley Jacob
1212
Thank you so much, this is exactly what I was asking.
– William
Jul 17 at 17:51
@William If you say so, check this answer.
– Takahiro Waki
Jul 30 at 0:38
add a comment |Â
up vote
2
down vote
The domain of a quadratic polynomial with real coefficients need not be restricted to the reals.
The discriminant of a polynomial is a function of that polynomial's coefficients, and it can be used to deduce some properties of the roots (such as whether they are real). In the case of a quadratic, the roots will be real and distinct for $D>0$, real (but the same) if $D = 0$, but complex conjugates if $D <0$. That is to say, a negative discriminant implies that the roots of the quadratic will both be complex (i.e. there are no real roots).
What does a discriminant got anything to do with x being real, when all discriminant tells us is whether or not the ROOTS are real?
The $x$ in your above equation is the root(s), as you have set your expression equal to 0.
EDIT: I see your edit about the range, and will address it. Essentially, the problem asks for you to find the range of the function given the knowledge that its domain must be real. So, we can use the knowledge that the discriminant must be non-negative to solve for allowable y: $$(12y-4)^2 -4(3y-1)(20y+1) geq 0$$
$$implies -frac58 leq y < frac13$$
which is the range of your function.
EDIT 2: OP, note that the quadratic and therefore discriminant you calculated above is incorrect. Your "c" term should be $(20y+1)$, not $(20y-1)$.
answered Jul 17 at 16:53
Riley Jacob
1212
Thank you so much, this is exactly what I was asking.
– William
Jul 17 at 17:51
@William If you say so, check this answer.
– Takahiro Waki
Jul 30 at 0:38
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The domain of a quadratic polynomial with real coefficients need not be restricted to the reals.
The discriminant of a polynomial is a function of that polynomial's coefficients, and it can be used to deduce some properties of the roots (such as whether they are real). In the case of a quadratic, the roots will be real and distinct for $D>0$, real (but the same) if $D = 0$, but complex conjugates if $D <0$. That is to say, a negative discriminant implies that the roots of the quadratic will both be complex (i.e. there are no real roots).
What does a discriminant got anything to do with x being real, when all discriminant tells us is whether or not the ROOTS are real?
The $x$ in your above equation is the root(s), as you have set your expression equal to 0.
EDIT: I see your edit about the range, and will address it. Essentially, the problem asks for you to find the range of the function given the knowledge that its domain must be real. So, we can use the knowledge that the discriminant must be non-negative to solve for allowable y: $$(12y-4)^2 -4(3y-1)(20y+1) geq 0$$
$$implies -frac58 leq y < frac13$$
which is the range of your function.
EDIT 2: OP, note that the quadratic and therefore discriminant you calculated above is incorrect. Your "c" term should be $(20y+1)$, not $(20y-1)$.
answered Jul 17 at 16:53
Riley Jacob
1212
The domain of a quadratic polynomial with real coefficients need not be restricted to the reals.
The discriminant of a polynomial is a function of that polynomial's coefficients, and it can be used to deduce some properties of the roots (such as whether they are real). In the case of a quadratic, the roots will be real and distinct for $D>0$, real (but the same) if $D = 0$, but complex conjugates if $D <0$. That is to say, a negative discriminant implies that the roots of the quadratic will both be complex (i.e. there are no real roots).
What does a discriminant got anything to do with x being real, when all discriminant tells us is whether or not the ROOTS are real?
The $x$ in your above equation is the root(s), as you have set your expression equal to 0.
EDIT: I see your edit about the range, and will address it. Essentially, the problem asks for you to find the range of the function given the knowledge that its domain must be real. So, we can use the knowledge that the discriminant must be non-negative to solve for allowable y: $$(12y-4)^2 -4(3y-1)(20y+1) geq 0$$
$$implies -frac58 leq y < frac13$$
which is the range of your function.
EDIT 2: OP, note that the quadratic and therefore discriminant you calculated above is incorrect. Your "c" term should be $(20y+1)$, not $(20y-1)$.
answered Jul 17 at 16:53
Riley Jacob
1212
edited Jul 17 at 18:26
answered Jul 17 at 16:53
Riley Jacob
1212
answered Jul 17 at 16:53
Riley Jacob
1212
answered Jul 17 at 16:53
Riley Jacob
1212
1212
Thank you so much, this is exactly what I was asking.
– William
Jul 17 at 17:51
@William If you say so, check this answer.
– Takahiro Waki
Jul 30 at 0:38
add a comment |Â
Thank you so much, this is exactly what I was asking.
– William
Jul 17 at 17:51
@William If you say so, check this answer.
– Takahiro Waki
Jul 30 at 0:38
Thank you so much, this is exactly what I was asking.
– William
Jul 17 at 17:51
Thank you so much, this is exactly what I was asking.
– William
Jul 17 at 17:51
@William If you say so, check this answer.
– Takahiro Waki
Jul 30 at 0:38
@William If you say so, check this answer.
– Takahiro Waki
Jul 30 at 0:38
add a comment |Â
up vote
1
down vote
$$fracx^2+4x-13x^2+12x+20=y$$
After cross multiplying, we obtain
$$(3y-1)x^2+(12y-4)x+(20y+1)=0$$
If the discriminant is nonnegative, then we can find real $x$ as solution to the quadratic equation, and hence given a particular value of $y$ that makes the discriminant nonnegative, we can find $x$ that satisfy the original equation.
However, if the discriminant is negative, then $x$ that satisfies the quadratic equation is not real.
answered Jul 17 at 16:51
Siong Thye Goh
77.7k134796
add a comment |Â
up vote
1
down vote
$$fracx^2+4x-13x^2+12x+20=y$$
After cross multiplying, we obtain
$$(3y-1)x^2+(12y-4)x+(20y+1)=0$$
If the discriminant is nonnegative, then we can find real $x$ as solution to the quadratic equation, and hence given a particular value of $y$ that makes the discriminant nonnegative, we can find $x$ that satisfy the original equation.
However, if the discriminant is negative, then $x$ that satisfies the quadratic equation is not real.
answered Jul 17 at 16:51
Siong Thye Goh
77.7k134796
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$fracx^2+4x-13x^2+12x+20=y$$
After cross multiplying, we obtain
$$(3y-1)x^2+(12y-4)x+(20y+1)=0$$
If the discriminant is nonnegative, then we can find real $x$ as solution to the quadratic equation, and hence given a particular value of $y$ that makes the discriminant nonnegative, we can find $x$ that satisfy the original equation.
However, if the discriminant is negative, then $x$ that satisfies the quadratic equation is not real.
answered Jul 17 at 16:51
Siong Thye Goh
77.7k134796
$$fracx^2+4x-13x^2+12x+20=y$$
After cross multiplying, we obtain
$$(3y-1)x^2+(12y-4)x+(20y+1)=0$$
If the discriminant is nonnegative, then we can find real $x$ as solution to the quadratic equation, and hence given a particular value of $y$ that makes the discriminant nonnegative, we can find $x$ that satisfy the original equation.
However, if the discriminant is negative, then $x$ that satisfies the quadratic equation is not real.
answered Jul 17 at 16:51
Siong Thye Goh
77.7k134796
edited Jul 17 at 19:41
answered Jul 17 at 16:51
Siong Thye Goh
77.7k134796
answered Jul 17 at 16:51
Siong Thye Goh
77.7k134796
answered Jul 17 at 16:51
Siong Thye Goh
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The $y$-values making the discriminant non-negative are the $y$-values the function can produce (at least one real solution $x$ exists) . Therefore the set of $y$ making the discriminant non-negative is the range.
answered Jul 17 at 16:50
Peter
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The $y$-values making the discriminant non-negative are the $y$-values the function can produce (at least one real solution $x$ exists) . Therefore the set of $y$ making the discriminant non-negative is the range.
answered Jul 17 at 16:50
Peter
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up vote
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The $y$-values making the discriminant non-negative are the $y$-values the function can produce (at least one real solution $x$ exists) . Therefore the set of $y$ making the discriminant non-negative is the range.
answered Jul 17 at 16:50
Peter
45.1k939119
The $y$-values making the discriminant non-negative are the $y$-values the function can produce (at least one real solution $x$ exists) . Therefore the set of $y$ making the discriminant non-negative is the range.
answered Jul 17 at 16:50
Peter
45.1k939119
answered Jul 17 at 16:50
Peter
45.1k939119
answered Jul 17 at 16:50
Peter
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answered Jul 17 at 16:50
Peter
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$$fracx^2+4x-13x^2+12x +20 = fracx^2+4x + frac2033x^2+12x +20 - frac frac2333x^2+12x +20 = frac13- frac frac2333x^2+12x +20$$
Meanwhile, $$ 3x^2 + 12 x + 20 = 3 (x+2)^2 + 8 $$ is always positive
with a minimum value of $8$ when $x=-2,$ so the minimum value of the whole thing is
$$ frac13- frac frac2338 = frac13-frac2324= -frac58 $$
It has an upper bound of $frac13$ but never achieves the bound, it just gets very close when $|x|$ is large.
answered Jul 17 at 17:16
Will Jagy
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$$fracx^2+4x-13x^2+12x +20 = fracx^2+4x + frac2033x^2+12x +20 - frac frac2333x^2+12x +20 = frac13- frac frac2333x^2+12x +20$$
Meanwhile, $$ 3x^2 + 12 x + 20 = 3 (x+2)^2 + 8 $$ is always positive
with a minimum value of $8$ when $x=-2,$ so the minimum value of the whole thing is
$$ frac13- frac frac2338 = frac13-frac2324= -frac58 $$
It has an upper bound of $frac13$ but never achieves the bound, it just gets very close when $|x|$ is large.
answered Jul 17 at 17:16
Will Jagy
97.2k594196
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up vote
0
down vote
up vote
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down vote
$$fracx^2+4x-13x^2+12x +20 = fracx^2+4x + frac2033x^2+12x +20 - frac frac2333x^2+12x +20 = frac13- frac frac2333x^2+12x +20$$
Meanwhile, $$ 3x^2 + 12 x + 20 = 3 (x+2)^2 + 8 $$ is always positive
with a minimum value of $8$ when $x=-2,$ so the minimum value of the whole thing is
$$ frac13- frac frac2338 = frac13-frac2324= -frac58 $$
It has an upper bound of $frac13$ but never achieves the bound, it just gets very close when $|x|$ is large.
answered Jul 17 at 17:16
Will Jagy
97.2k594196
$$fracx^2+4x-13x^2+12x +20 = fracx^2+4x + frac2033x^2+12x +20 - frac frac2333x^2+12x +20 = frac13- frac frac2333x^2+12x +20$$
Meanwhile, $$ 3x^2 + 12 x + 20 = 3 (x+2)^2 + 8 $$ is always positive
with a minimum value of $8$ when $x=-2,$ so the minimum value of the whole thing is
$$ frac13- frac frac2338 = frac13-frac2324= -frac58 $$
It has an upper bound of $frac13$ but never achieves the bound, it just gets very close when $|x|$ is large.
answered Jul 17 at 17:16
Will Jagy
97.2k594196
answered Jul 17 at 17:16
Will Jagy
97.2k594196
answered Jul 17 at 17:16
Will Jagy
97.2k594196
answered Jul 17 at 17:16
Will Jagy
97.2k594196
97.2k594196
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Can you allow any imaginary value of $x$ in the function?
– Entrepreneur
Jul 17 at 16:46
2
That's right, but you're solving for $x$. You're asking for what values of $y$ is the there a real value of $x$ such that the given rational function takes the value $y$ at $x$?
– saulspatz
Jul 17 at 16:48
2
The question has overloaded the meaning of variable $x$ and caused your confusion. What happens is given a $x$, you can compute a $y$ and the original $x$ will be a root for the quadratic polynomial $(3y-1)t^2 + (12y -4)t + (20y-1) = 0$ in $t$. Since this quadratic polynomial (assume $y ne frac13$) has a real root at $t = x$. Its discriminant $D ge 0$.
– achille hui
Jul 17 at 16:51
@saulspatz please see the question, I've edited
– William
Jul 17 at 16:56
@achillehui I read all these answers, even though they answered it in their own way, they still weren't being clear as to WHY should the discriminant be non-negative. I read your comment and Riley Jacob's answer and yall answered exactly what I was asking, thank you so much man, you're the man!! I was getting confused between roots of the equation and roots of the functions. I hope I don't make this mistake ever again, ty again.
– William
Jul 17 at 17:49