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
112
add a comment |
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
112
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 |
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
112
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
functions
asked Aug 27 '16 at 6:40
mathemagician
112
asked Aug 27 '16 at 6:40
mathemagician
112
asked Aug 27 '16 at 6:40
mathemagician
112
asked Aug 27 '16 at 6:40
mathemagician
112
asked Aug 27 '16 at 6:40
mathemagician
112
112
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 |
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 |
1 Answer
1
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 $$frac{-bpmsqrt{b^2-4ac}}{2a}.$$ 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
3,9551238
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 $sqrt{b^2-4ac}$ 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 |
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 $$frac{-bpmsqrt{b^2-4ac}}{2a}.$$ 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
3,9551238
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 $sqrt{b^2-4ac}$ 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 |
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 $$frac{-bpmsqrt{b^2-4ac}}{2a}.$$ 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
3,9551238
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 $sqrt{b^2-4ac}$ 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 |
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 $$frac{-bpmsqrt{b^2-4ac}}{2a}.$$ 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
3,9551238
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 $$frac{-bpmsqrt{b^2-4ac}}{2a}.$$ 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
3,9551238
answered Aug 27 '16 at 6:50
pi66
3,9551238
answered Aug 27 '16 at 6:50
pi66
3,9551238
answered Aug 27 '16 at 6:50
pi66
3,9551238
3,9551238
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 $sqrt{b^2-4ac}$ 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 |
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 $sqrt{b^2-4ac}$ 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 $sqrt{b^2-4ac}$ 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 $sqrt{b^2-4ac}$ 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 |
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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