How to find the coefficient of $a$ in a quadratic equation?
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If a root of the equation $$3x^2 + 4x + 12a + 9ax = 0$$ is greater than 6, then the correct statement of the coefficient $a$ is:
a) $a = 2$
b) $a> -2$
c) $a = -2$
d) $a <-2$
e) $-2le ale 2$
What I did was solve it as if a root was exactly $6$, and I find that $a$ is equal to $2$ ... But the premise says that a root has to be greater than $6$, not equal to $6$ ... and I do not know how find that, for now I would know that $a$ does not have to be $2$ ...
algebra-precalculus polynomials roots
edited Aug 10 at 18:31
TheSimpliFire
9,75761952
asked Aug 10 at 18:20
Strong
406
add a comment |Â
up vote
2
down vote
favorite
If a root of the equation $$3x^2 + 4x + 12a + 9ax = 0$$ is greater than 6, then the correct statement of the coefficient $a$ is:
a) $a = 2$
b) $a> -2$
c) $a = -2$
d) $a <-2$
e) $-2le ale 2$
What I did was solve it as if a root was exactly $6$, and I find that $a$ is equal to $2$ ... But the premise says that a root has to be greater than $6$, not equal to $6$ ... and I do not know how find that, for now I would know that $a$ does not have to be $2$ ...
algebra-precalculus polynomials roots
edited Aug 10 at 18:31
TheSimpliFire
9,75761952
asked Aug 10 at 18:20
Strong
406
What is the reason for the downvote?
– mfl
Aug 10 at 18:23
1
@mfl I suspect it has something to do with the question being written in Spanish.
– Mastrem
Aug 10 at 18:25
Why did you choose to write this post in Spanish? Reading your other posts it seems that you are perfectly capable of writing in English.
– TheSimpliFire
Aug 10 at 18:34
If $a = 2$, then the roots must be negative. Did you mean $a = -2$?
– N. F. Taussig
Aug 10 at 18:45
Solve $3x^2+4x+12a+9ax = 0$ for $a$ to get $a = -fracx3$ (note the solve step assumes $xnot= -frac43$ when we divide by $9x+12$, but you are next going to assume $x>6$ so this case is not a problem)
– Winther
Aug 10 at 20:44
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If a root of the equation $$3x^2 + 4x + 12a + 9ax = 0$$ is greater than 6, then the correct statement of the coefficient $a$ is:
a) $a = 2$
b) $a> -2$
c) $a = -2$
d) $a <-2$
e) $-2le ale 2$
What I did was solve it as if a root was exactly $6$, and I find that $a$ is equal to $2$ ... But the premise says that a root has to be greater than $6$, not equal to $6$ ... and I do not know how find that, for now I would know that $a$ does not have to be $2$ ...
algebra-precalculus polynomials roots
edited Aug 10 at 18:31
TheSimpliFire
9,75761952
asked Aug 10 at 18:20
Strong
406
If a root of the equation $$3x^2 + 4x + 12a + 9ax = 0$$ is greater than 6, then the correct statement of the coefficient $a$ is:
a) $a = 2$
b) $a> -2$
c) $a = -2$
d) $a <-2$
e) $-2le ale 2$
What I did was solve it as if a root was exactly $6$, and I find that $a$ is equal to $2$ ... But the premise says that a root has to be greater than $6$, not equal to $6$ ... and I do not know how find that, for now I would know that $a$ does not have to be $2$ ...
algebra-precalculus polynomials roots
edited Aug 10 at 18:31
TheSimpliFire
9,75761952
asked Aug 10 at 18:20
Strong
406
edited Aug 10 at 18:31
TheSimpliFire
9,75761952
edited Aug 10 at 18:31
TheSimpliFire
9,75761952
edited Aug 10 at 18:31
TheSimpliFire
9,75761952
9,75761952
asked Aug 10 at 18:20
Strong
406
asked Aug 10 at 18:20
Strong
406
asked Aug 10 at 18:20
Strong
406
406
What is the reason for the downvote?
– mfl
Aug 10 at 18:23
1
@mfl I suspect it has something to do with the question being written in Spanish.
– Mastrem
Aug 10 at 18:25
Why did you choose to write this post in Spanish? Reading your other posts it seems that you are perfectly capable of writing in English.
– TheSimpliFire
Aug 10 at 18:34
If $a = 2$, then the roots must be negative. Did you mean $a = -2$?
– N. F. Taussig
Aug 10 at 18:45
Solve $3x^2+4x+12a+9ax = 0$ for $a$ to get $a = -fracx3$ (note the solve step assumes $xnot= -frac43$ when we divide by $9x+12$, but you are next going to assume $x>6$ so this case is not a problem)
– Winther
Aug 10 at 20:44
add a comment |Â
What is the reason for the downvote?
– mfl
Aug 10 at 18:23
1
@mfl I suspect it has something to do with the question being written in Spanish.
– Mastrem
Aug 10 at 18:25
Why did you choose to write this post in Spanish? Reading your other posts it seems that you are perfectly capable of writing in English.
– TheSimpliFire
Aug 10 at 18:34
If $a = 2$, then the roots must be negative. Did you mean $a = -2$?
– N. F. Taussig
Aug 10 at 18:45
Solve $3x^2+4x+12a+9ax = 0$ for $a$ to get $a = -fracx3$ (note the solve step assumes $xnot= -frac43$ when we divide by $9x+12$, but you are next going to assume $x>6$ so this case is not a problem)
– Winther
Aug 10 at 20:44
What is the reason for the downvote?
– mfl
Aug 10 at 18:23
What is the reason for the downvote?
– mfl
Aug 10 at 18:23
1
1
@mfl I suspect it has something to do with the question being written in Spanish.
– Mastrem
Aug 10 at 18:25
@mfl I suspect it has something to do with the question being written in Spanish.
– Mastrem
Aug 10 at 18:25
Why did you choose to write this post in Spanish? Reading your other posts it seems that you are perfectly capable of writing in English.
– TheSimpliFire
Aug 10 at 18:34
Why did you choose to write this post in Spanish? Reading your other posts it seems that you are perfectly capable of writing in English.
– TheSimpliFire
Aug 10 at 18:34
If $a = 2$, then the roots must be negative. Did you mean $a = -2$?
– N. F. Taussig
Aug 10 at 18:45
If $a = 2$, then the roots must be negative. Did you mean $a = -2$?
– N. F. Taussig
Aug 10 at 18:45
Solve $3x^2+4x+12a+9ax = 0$ for $a$ to get $a = -fracx3$ (note the solve step assumes $xnot= -frac43$ when we divide by $9x+12$, but you are next going to assume $x>6$ so this case is not a problem)
– Winther
Aug 10 at 20:44
Solve $3x^2+4x+12a+9ax = 0$ for $a$ to get $a = -fracx3$ (note the solve step assumes $xnot= -frac43$ when we divide by $9x+12$, but you are next going to assume $x>6$ so this case is not a problem)
– Winther
Aug 10 at 20:44
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
Hint: $;3x^2 + 4x + 12a + 9ax = x(3x+4)+3a(3x+4)= (x+3a)(3x+4),$.
answered Aug 10 at 18:45
dxiv
54.7k64798
add a comment |Â
up vote
1
down vote
Hint
beginalign 3x^2+4x+12a+9ax = 0&iff 3x^2+(4+9a)x+12a= 0\ &iff x=dfrac-4-9apm sqrt(4+9a)^2-144a6.endalign So,
$$dfrac-4-9apm sqrt(4+9a)^2-144a6ge 6 iff -4-9apm sqrt(4+9a)^2-144age 36. $$
That is
$$-4-9apm sqrt16+72a -63a^2ge 36$$
So, you can check what is the correct answer.
answered Aug 10 at 18:28
mfl
24.6k12040
Llego a que $;2a^2 + 9a + 22<0,$
– Strong
Aug 10 at 22:57
add a comment |Â
up vote
0
down vote
Using @dxiv hint $(x+3a)(3x+4) = 0$ we have $x = -frac43 < 6$ and $x = -3a > 6$ Therefore, $a < -2$
answered Aug 14 at 20:34
GinoCHJ
794
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Hint: $;3x^2 + 4x + 12a + 9ax = x(3x+4)+3a(3x+4)= (x+3a)(3x+4),$.
answered Aug 10 at 18:45
dxiv
54.7k64798
add a comment |Â
up vote
3
down vote
Hint: $;3x^2 + 4x + 12a + 9ax = x(3x+4)+3a(3x+4)= (x+3a)(3x+4),$.
answered Aug 10 at 18:45
dxiv
54.7k64798
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Hint: $;3x^2 + 4x + 12a + 9ax = x(3x+4)+3a(3x+4)= (x+3a)(3x+4),$.
answered Aug 10 at 18:45
dxiv
54.7k64798
Hint: $;3x^2 + 4x + 12a + 9ax = x(3x+4)+3a(3x+4)= (x+3a)(3x+4),$.
answered Aug 10 at 18:45
dxiv
54.7k64798
answered Aug 10 at 18:45
dxiv
54.7k64798
answered Aug 10 at 18:45
dxiv
54.7k64798
answered Aug 10 at 18:45
dxiv
54.7k64798
54.7k64798
add a comment |Â
add a comment |Â
up vote
1
down vote
Hint
beginalign 3x^2+4x+12a+9ax = 0&iff 3x^2+(4+9a)x+12a= 0\ &iff x=dfrac-4-9apm sqrt(4+9a)^2-144a6.endalign So,
$$dfrac-4-9apm sqrt(4+9a)^2-144a6ge 6 iff -4-9apm sqrt(4+9a)^2-144age 36. $$
That is
$$-4-9apm sqrt16+72a -63a^2ge 36$$
So, you can check what is the correct answer.
answered Aug 10 at 18:28
mfl
24.6k12040
Llego a que $;2a^2 + 9a + 22<0,$
– Strong
Aug 10 at 22:57
add a comment |Â
up vote
1
down vote
Hint
beginalign 3x^2+4x+12a+9ax = 0&iff 3x^2+(4+9a)x+12a= 0\ &iff x=dfrac-4-9apm sqrt(4+9a)^2-144a6.endalign So,
$$dfrac-4-9apm sqrt(4+9a)^2-144a6ge 6 iff -4-9apm sqrt(4+9a)^2-144age 36. $$
That is
$$-4-9apm sqrt16+72a -63a^2ge 36$$
So, you can check what is the correct answer.
answered Aug 10 at 18:28
mfl
24.6k12040
Llego a que $;2a^2 + 9a + 22<0,$
– Strong
Aug 10 at 22:57
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint
beginalign 3x^2+4x+12a+9ax = 0&iff 3x^2+(4+9a)x+12a= 0\ &iff x=dfrac-4-9apm sqrt(4+9a)^2-144a6.endalign So,
$$dfrac-4-9apm sqrt(4+9a)^2-144a6ge 6 iff -4-9apm sqrt(4+9a)^2-144age 36. $$
That is
$$-4-9apm sqrt16+72a -63a^2ge 36$$
So, you can check what is the correct answer.
answered Aug 10 at 18:28
mfl
24.6k12040
Hint
beginalign 3x^2+4x+12a+9ax = 0&iff 3x^2+(4+9a)x+12a= 0\ &iff x=dfrac-4-9apm sqrt(4+9a)^2-144a6.endalign So,
$$dfrac-4-9apm sqrt(4+9a)^2-144a6ge 6 iff -4-9apm sqrt(4+9a)^2-144age 36. $$
That is
$$-4-9apm sqrt16+72a -63a^2ge 36$$
So, you can check what is the correct answer.
answered Aug 10 at 18:28
mfl
24.6k12040
edited Aug 10 at 18:44
answered Aug 10 at 18:28
mfl
24.6k12040
answered Aug 10 at 18:28
mfl
24.6k12040
answered Aug 10 at 18:28
mfl
24.6k12040
24.6k12040
Llego a que $;2a^2 + 9a + 22<0,$
– Strong
Aug 10 at 22:57
add a comment |Â
Llego a que $;2a^2 + 9a + 22<0,$
– Strong
Aug 10 at 22:57
Llego a que $;2a^2 + 9a + 22<0,$
– Strong
Aug 10 at 22:57
Llego a que $;2a^2 + 9a + 22<0,$
– Strong
Aug 10 at 22:57
add a comment |Â
up vote
0
down vote
Using @dxiv hint $(x+3a)(3x+4) = 0$ we have $x = -frac43 < 6$ and $x = -3a > 6$ Therefore, $a < -2$
answered Aug 14 at 20:34
GinoCHJ
794
add a comment |Â
up vote
0
down vote
Using @dxiv hint $(x+3a)(3x+4) = 0$ we have $x = -frac43 < 6$ and $x = -3a > 6$ Therefore, $a < -2$
answered Aug 14 at 20:34
GinoCHJ
794
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Using @dxiv hint $(x+3a)(3x+4) = 0$ we have $x = -frac43 < 6$ and $x = -3a > 6$ Therefore, $a < -2$
answered Aug 14 at 20:34
GinoCHJ
794
Using @dxiv hint $(x+3a)(3x+4) = 0$ we have $x = -frac43 < 6$ and $x = -3a > 6$ Therefore, $a < -2$
answered Aug 14 at 20:34
GinoCHJ
794
answered Aug 14 at 20:34
GinoCHJ
794
answered Aug 14 at 20:34
GinoCHJ
794
answered Aug 14 at 20:34
GinoCHJ
794
794
add a comment |Â
add a comment |Â
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What is the reason for the downvote?
– mfl
Aug 10 at 18:23
1
@mfl I suspect it has something to do with the question being written in Spanish.
– Mastrem
Aug 10 at 18:25
Why did you choose to write this post in Spanish? Reading your other posts it seems that you are perfectly capable of writing in English.
– TheSimpliFire
Aug 10 at 18:34
If $a = 2$, then the roots must be negative. Did you mean $a = -2$?
– N. F. Taussig
Aug 10 at 18:45
Solve $3x^2+4x+12a+9ax = 0$ for $a$ to get $a = -fracx3$ (note the solve step assumes $xnot= -frac43$ when we divide by $9x+12$, but you are next going to assume $x>6$ so this case is not a problem)
– Winther
Aug 10 at 20:44