Finding value of k for which fg(x)=k has equal roots?
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I've been going through this community and I Find this really helpful. About me(I know I should be precise but ya), I'm just a highschool student who can't afford any coachings/schools. Self schooling being my only option I'm trying to teach myself mathematics from some torrented books. I am on functions and their graphs and stuck with one question
The functions f and g are defined for x ∈ R by f(x) =4x − 2x^2; g(x)= 5x + 3.
(i) Find the range of f. (ii)  Find the value of the constant k for which the equation gf(x) = k has equal roots.
Now, I do understand the composition of functions but I just don't understand what they are asking in this case,
g(fx()=k would result in
20x-10x^2 +3=k
Now this is a quadratic equation of second degree which should have 2 roots/solutions, but that's the case if the right hand side was zero and not k. I have absolutely no idea how to tackle this question and what's being asked in part ii of the question. Anyhelp would be highly appreciated
functions
asked Aug 27 '16 at 6:40
mathemagician
12
add a comment |Â
up vote
0
down vote
favorite
I've been going through this community and I Find this really helpful. About me(I know I should be precise but ya), I'm just a highschool student who can't afford any coachings/schools. Self schooling being my only option I'm trying to teach myself mathematics from some torrented books. I am on functions and their graphs and stuck with one question
The functions f and g are defined for x ∈ R by f(x) =4x − 2x^2; g(x)= 5x + 3.
(i) Find the range of f. (ii)  Find the value of the constant k for which the equation gf(x) = k has equal roots.
Now, I do understand the composition of functions but I just don't understand what they are asking in this case,
g(fx()=k would result in
20x-10x^2 +3=k
Now this is a quadratic equation of second degree which should have 2 roots/solutions, but that's the case if the right hand side was zero and not k. I have absolutely no idea how to tackle this question and what's being asked in part ii of the question. Anyhelp would be highly appreciated
functions
asked Aug 27 '16 at 6:40
mathemagician
12
1
Notation $gf(x)=k$ is ambiguous. Is it $g(f(x))=k$ or $g(x)f(x)=k$?
– Jean Marie
Aug 27 '16 at 8:11
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I've been going through this community and I Find this really helpful. About me(I know I should be precise but ya), I'm just a highschool student who can't afford any coachings/schools. Self schooling being my only option I'm trying to teach myself mathematics from some torrented books. I am on functions and their graphs and stuck with one question
The functions f and g are defined for x ∈ R by f(x) =4x − 2x^2; g(x)= 5x + 3.
(i) Find the range of f. (ii)  Find the value of the constant k for which the equation gf(x) = k has equal roots.
Now, I do understand the composition of functions but I just don't understand what they are asking in this case,
g(fx()=k would result in
20x-10x^2 +3=k
Now this is a quadratic equation of second degree which should have 2 roots/solutions, but that's the case if the right hand side was zero and not k. I have absolutely no idea how to tackle this question and what's being asked in part ii of the question. Anyhelp would be highly appreciated
functions
asked Aug 27 '16 at 6:40
mathemagician
12
I've been going through this community and I Find this really helpful. About me(I know I should be precise but ya), I'm just a highschool student who can't afford any coachings/schools. Self schooling being my only option I'm trying to teach myself mathematics from some torrented books. I am on functions and their graphs and stuck with one question
The functions f and g are defined for x ∈ R by f(x) =4x − 2x^2; g(x)= 5x + 3.
(i) Find the range of f. (ii)  Find the value of the constant k for which the equation gf(x) = k has equal roots.
Now, I do understand the composition of functions but I just don't understand what they are asking in this case,
g(fx()=k would result in
20x-10x^2 +3=k
Now this is a quadratic equation of second degree which should have 2 roots/solutions, but that's the case if the right hand side was zero and not k. I have absolutely no idea how to tackle this question and what's being asked in part ii of the question. Anyhelp would be highly appreciated
functions
asked Aug 27 '16 at 6:40
mathemagician
12
asked Aug 27 '16 at 6:40
mathemagician
12
asked Aug 27 '16 at 6:40
mathemagician
12
asked Aug 27 '16 at 6:40
mathemagician
12
12
1
Notation $gf(x)=k$ is ambiguous. Is it $g(f(x))=k$ or $g(x)f(x)=k$?
– Jean Marie
Aug 27 '16 at 8:11
add a comment |Â
1
Notation $gf(x)=k$ is ambiguous. Is it $g(f(x))=k$ or $g(x)f(x)=k$?
– Jean Marie
Aug 27 '16 at 8:11
1
1
Notation $gf(x)=k$ is ambiguous. Is it $g(f(x))=k$ or $g(x)f(x)=k$?
– Jean Marie
Aug 27 '16 at 8:11
Notation $gf(x)=k$ is ambiguous. Is it $g(f(x))=k$ or $g(x)f(x)=k$?
– Jean Marie
Aug 27 '16 at 8:11
add a comment |Â
1 Answer
1
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up vote
0
down vote
Your equation could be rewritten as
$$10x^2-20x+(k-3)=0.$$
Recall that the roots of a quadratic equation $ax^2+bx+c=0$ are given by $$frac-bpmsqrtb^2-4ac2a.$$ So the two roots are equal when $b^2-4ac=0$. Can you apply this to your problem?
answered Aug 27 '16 at 6:50
pi66
3,6981037
Hey, Thank you for the answer! I really appreciate that. Now back to the question b^2 -4ac=0 would result in k=10.3 (I Just isolated the k) There are a number of questions I have here, I am equally embarrassed of my ignorance but here it goes How did you know that the two roots would be equal when b^2-4ac=0 ? The two roots are equal when the equation is a perfect square since (x+2)^2 can be written as (x+2)(x+2), that's just my intuition but how did you know that b^2-4ac =0 is a case for that too? Secondly, I thought(and I know I've been wrong) that they were asking a value (cont.)
– mathemagician
Aug 27 '16 at 6:59
(continued) a value of k for which there would be k roots, my bad
– mathemagician
Aug 27 '16 at 6:59
1
Your $k=10.3$ is not quite correct. You're solving $20^2-4cdot 10cdot (k-3)=0$.
– pi66
Aug 27 '16 at 7:01
1
If you look at the root formula that I wrote, you can see that the two roots are equal only if the term $sqrtb^2-4ac$ is zero. This corresponds to the case where you can write the polynomial as $(x+d)(x+d)$, where $d$ is the root that appears twice.
– pi66
Aug 27 '16 at 7:02
Oops, my bad! k=13. I see, since adding and subtracting zero would be the same. Thank you for this. That was really helpful :)
– mathemagician
Aug 27 '16 at 7:15
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Your equation could be rewritten as
$$10x^2-20x+(k-3)=0.$$
Recall that the roots of a quadratic equation $ax^2+bx+c=0$ are given by $$frac-bpmsqrtb^2-4ac2a.$$ So the two roots are equal when $b^2-4ac=0$. Can you apply this to your problem?
answered Aug 27 '16 at 6:50
pi66
3,6981037
Hey, Thank you for the answer! I really appreciate that. Now back to the question b^2 -4ac=0 would result in k=10.3 (I Just isolated the k) There are a number of questions I have here, I am equally embarrassed of my ignorance but here it goes How did you know that the two roots would be equal when b^2-4ac=0 ? The two roots are equal when the equation is a perfect square since (x+2)^2 can be written as (x+2)(x+2), that's just my intuition but how did you know that b^2-4ac =0 is a case for that too? Secondly, I thought(and I know I've been wrong) that they were asking a value (cont.)
– mathemagician
Aug 27 '16 at 6:59
(continued) a value of k for which there would be k roots, my bad
– mathemagician
Aug 27 '16 at 6:59
1
Your $k=10.3$ is not quite correct. You're solving $20^2-4cdot 10cdot (k-3)=0$.
– pi66
Aug 27 '16 at 7:01
1
If you look at the root formula that I wrote, you can see that the two roots are equal only if the term $sqrtb^2-4ac$ is zero. This corresponds to the case where you can write the polynomial as $(x+d)(x+d)$, where $d$ is the root that appears twice.
– pi66
Aug 27 '16 at 7:02
Oops, my bad! k=13. I see, since adding and subtracting zero would be the same. Thank you for this. That was really helpful :)
– mathemagician
Aug 27 '16 at 7:15
add a comment |Â
up vote
0
down vote
Your equation could be rewritten as
$$10x^2-20x+(k-3)=0.$$
Recall that the roots of a quadratic equation $ax^2+bx+c=0$ are given by $$frac-bpmsqrtb^2-4ac2a.$$ So the two roots are equal when $b^2-4ac=0$. Can you apply this to your problem?
answered Aug 27 '16 at 6:50
pi66
3,6981037
Hey, Thank you for the answer! I really appreciate that. Now back to the question b^2 -4ac=0 would result in k=10.3 (I Just isolated the k) There are a number of questions I have here, I am equally embarrassed of my ignorance but here it goes How did you know that the two roots would be equal when b^2-4ac=0 ? The two roots are equal when the equation is a perfect square since (x+2)^2 can be written as (x+2)(x+2), that's just my intuition but how did you know that b^2-4ac =0 is a case for that too? Secondly, I thought(and I know I've been wrong) that they were asking a value (cont.)
– mathemagician
Aug 27 '16 at 6:59
(continued) a value of k for which there would be k roots, my bad
– mathemagician
Aug 27 '16 at 6:59
1
Your $k=10.3$ is not quite correct. You're solving $20^2-4cdot 10cdot (k-3)=0$.
– pi66
Aug 27 '16 at 7:01
1
If you look at the root formula that I wrote, you can see that the two roots are equal only if the term $sqrtb^2-4ac$ is zero. This corresponds to the case where you can write the polynomial as $(x+d)(x+d)$, where $d$ is the root that appears twice.
– pi66
Aug 27 '16 at 7:02
Oops, my bad! k=13. I see, since adding and subtracting zero would be the same. Thank you for this. That was really helpful :)
– mathemagician
Aug 27 '16 at 7:15
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your equation could be rewritten as
$$10x^2-20x+(k-3)=0.$$
Recall that the roots of a quadratic equation $ax^2+bx+c=0$ are given by $$frac-bpmsqrtb^2-4ac2a.$$ So the two roots are equal when $b^2-4ac=0$. Can you apply this to your problem?
answered Aug 27 '16 at 6:50
pi66
3,6981037
Your equation could be rewritten as
$$10x^2-20x+(k-3)=0.$$
Recall that the roots of a quadratic equation $ax^2+bx+c=0$ are given by $$frac-bpmsqrtb^2-4ac2a.$$ So the two roots are equal when $b^2-4ac=0$. Can you apply this to your problem?
answered Aug 27 '16 at 6:50
pi66
3,6981037
answered Aug 27 '16 at 6:50
pi66
3,6981037
answered Aug 27 '16 at 6:50
pi66
3,6981037
answered Aug 27 '16 at 6:50
pi66
3,6981037
3,6981037
Hey, Thank you for the answer! I really appreciate that. Now back to the question b^2 -4ac=0 would result in k=10.3 (I Just isolated the k) There are a number of questions I have here, I am equally embarrassed of my ignorance but here it goes How did you know that the two roots would be equal when b^2-4ac=0 ? The two roots are equal when the equation is a perfect square since (x+2)^2 can be written as (x+2)(x+2), that's just my intuition but how did you know that b^2-4ac =0 is a case for that too? Secondly, I thought(and I know I've been wrong) that they were asking a value (cont.)
– mathemagician
Aug 27 '16 at 6:59
(continued) a value of k for which there would be k roots, my bad
– mathemagician
Aug 27 '16 at 6:59
1
Your $k=10.3$ is not quite correct. You're solving $20^2-4cdot 10cdot (k-3)=0$.
– pi66
Aug 27 '16 at 7:01
1
If you look at the root formula that I wrote, you can see that the two roots are equal only if the term $sqrtb^2-4ac$ is zero. This corresponds to the case where you can write the polynomial as $(x+d)(x+d)$, where $d$ is the root that appears twice.
– pi66
Aug 27 '16 at 7:02
Oops, my bad! k=13. I see, since adding and subtracting zero would be the same. Thank you for this. That was really helpful :)
– mathemagician
Aug 27 '16 at 7:15
add a comment |Â
Hey, Thank you for the answer! I really appreciate that. Now back to the question b^2 -4ac=0 would result in k=10.3 (I Just isolated the k) There are a number of questions I have here, I am equally embarrassed of my ignorance but here it goes How did you know that the two roots would be equal when b^2-4ac=0 ? The two roots are equal when the equation is a perfect square since (x+2)^2 can be written as (x+2)(x+2), that's just my intuition but how did you know that b^2-4ac =0 is a case for that too? Secondly, I thought(and I know I've been wrong) that they were asking a value (cont.)
– mathemagician
Aug 27 '16 at 6:59
(continued) a value of k for which there would be k roots, my bad
– mathemagician
Aug 27 '16 at 6:59
1
Your $k=10.3$ is not quite correct. You're solving $20^2-4cdot 10cdot (k-3)=0$.
– pi66
Aug 27 '16 at 7:01
1
If you look at the root formula that I wrote, you can see that the two roots are equal only if the term $sqrtb^2-4ac$ is zero. This corresponds to the case where you can write the polynomial as $(x+d)(x+d)$, where $d$ is the root that appears twice.
– pi66
Aug 27 '16 at 7:02
Oops, my bad! k=13. I see, since adding and subtracting zero would be the same. Thank you for this. That was really helpful :)
– mathemagician
Aug 27 '16 at 7:15
Hey, Thank you for the answer! I really appreciate that. Now back to the question b^2 -4ac=0 would result in k=10.3 (I Just isolated the k) There are a number of questions I have here, I am equally embarrassed of my ignorance but here it goes How did you know that the two roots would be equal when b^2-4ac=0 ? The two roots are equal when the equation is a perfect square since (x+2)^2 can be written as (x+2)(x+2), that's just my intuition but how did you know that b^2-4ac =0 is a case for that too? Secondly, I thought(and I know I've been wrong) that they were asking a value (cont.)
– mathemagician
Aug 27 '16 at 6:59
Hey, Thank you for the answer! I really appreciate that. Now back to the question b^2 -4ac=0 would result in k=10.3 (I Just isolated the k) There are a number of questions I have here, I am equally embarrassed of my ignorance but here it goes How did you know that the two roots would be equal when b^2-4ac=0 ? The two roots are equal when the equation is a perfect square since (x+2)^2 can be written as (x+2)(x+2), that's just my intuition but how did you know that b^2-4ac =0 is a case for that too? Secondly, I thought(and I know I've been wrong) that they were asking a value (cont.)
– mathemagician
Aug 27 '16 at 6:59
(continued) a value of k for which there would be k roots, my bad
– mathemagician
Aug 27 '16 at 6:59
(continued) a value of k for which there would be k roots, my bad
– mathemagician
Aug 27 '16 at 6:59
1
1
Your $k=10.3$ is not quite correct. You're solving $20^2-4cdot 10cdot (k-3)=0$.
– pi66
Aug 27 '16 at 7:01
Your $k=10.3$ is not quite correct. You're solving $20^2-4cdot 10cdot (k-3)=0$.
– pi66
Aug 27 '16 at 7:01
1
1
If you look at the root formula that I wrote, you can see that the two roots are equal only if the term $sqrtb^2-4ac$ is zero. This corresponds to the case where you can write the polynomial as $(x+d)(x+d)$, where $d$ is the root that appears twice.
– pi66
Aug 27 '16 at 7:02
If you look at the root formula that I wrote, you can see that the two roots are equal only if the term $sqrtb^2-4ac$ is zero. This corresponds to the case where you can write the polynomial as $(x+d)(x+d)$, where $d$ is the root that appears twice.
– pi66
Aug 27 '16 at 7:02
Oops, my bad! k=13. I see, since adding and subtracting zero would be the same. Thank you for this. That was really helpful :)
– mathemagician
Aug 27 '16 at 7:15
Oops, my bad! k=13. I see, since adding and subtracting zero would be the same. Thank you for this. That was really helpful :)
– mathemagician
Aug 27 '16 at 7:15
add a comment |Â
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1
Notation $gf(x)=k$ is ambiguous. Is it $g(f(x))=k$ or $g(x)f(x)=k$?
– Jean Marie
Aug 27 '16 at 8:11