How do I solve the inequality $x<x^2-12<4x$?
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0
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So first I considered
$x < x^2 -12$
so I get $0 < x^2 - x -12$
which is $0<(x+3)(x-4)$
after this I don't know where to go
Again, I considered $x^2 - 12< 4x$
which is $x^2 - 4x - 12<0$
so $(x+2)(x-6)<0$
Again same issue, I don't know where to after that.
inequality
edited Nov 17 at 14:38
mrtaurho
2,7391827
asked Nov 17 at 14:36
RiRi
143
add a comment |
up vote
0
down vote
favorite
So first I considered
$x < x^2 -12$
so I get $0 < x^2 - x -12$
which is $0<(x+3)(x-4)$
after this I don't know where to go
Again, I considered $x^2 - 12< 4x$
which is $x^2 - 4x - 12<0$
so $(x+2)(x-6)<0$
Again same issue, I don't know where to after that.
inequality
edited Nov 17 at 14:38
mrtaurho
2,7391827
asked Nov 17 at 14:36
RiRi
143
You may look here
– A.Γ.
Nov 17 at 14:58
1
After you got the inequality $0<(x+3)(x-4)$, the next step is to remember that a product of two real numbers is positive if and only if the two factors are either both positive or both negative. Also, in this particular problem, it might be useful to notice, before working on the quadratic inequalities, that you'll only be dealing with positive values of $x$ because any solution will have $x<4x$.
– Andreas Blass
Nov 17 at 15:00
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So first I considered
$x < x^2 -12$
so I get $0 < x^2 - x -12$
which is $0<(x+3)(x-4)$
after this I don't know where to go
Again, I considered $x^2 - 12< 4x$
which is $x^2 - 4x - 12<0$
so $(x+2)(x-6)<0$
Again same issue, I don't know where to after that.
inequality
edited Nov 17 at 14:38
mrtaurho
2,7391827
asked Nov 17 at 14:36
RiRi
143
So first I considered
$x < x^2 -12$
so I get $0 < x^2 - x -12$
which is $0<(x+3)(x-4)$
after this I don't know where to go
Again, I considered $x^2 - 12< 4x$
which is $x^2 - 4x - 12<0$
so $(x+2)(x-6)<0$
Again same issue, I don't know where to after that.
inequality
inequality
edited Nov 17 at 14:38
mrtaurho
2,7391827
asked Nov 17 at 14:36
RiRi
143
edited Nov 17 at 14:38
mrtaurho
2,7391827
asked Nov 17 at 14:36
RiRi
143
edited Nov 17 at 14:38
mrtaurho
2,7391827
edited Nov 17 at 14:38
mrtaurho
2,7391827
edited Nov 17 at 14:38
mrtaurho
2,7391827
2,7391827
asked Nov 17 at 14:36
RiRi
143
asked Nov 17 at 14:36
RiRi
143
asked Nov 17 at 14:36
RiRi
143
143
You may look here
– A.Γ.
Nov 17 at 14:58
1
After you got the inequality $0<(x+3)(x-4)$, the next step is to remember that a product of two real numbers is positive if and only if the two factors are either both positive or both negative. Also, in this particular problem, it might be useful to notice, before working on the quadratic inequalities, that you'll only be dealing with positive values of $x$ because any solution will have $x<4x$.
– Andreas Blass
Nov 17 at 15:00
add a comment |
You may look here
– A.Γ.
Nov 17 at 14:58
1
After you got the inequality $0<(x+3)(x-4)$, the next step is to remember that a product of two real numbers is positive if and only if the two factors are either both positive or both negative. Also, in this particular problem, it might be useful to notice, before working on the quadratic inequalities, that you'll only be dealing with positive values of $x$ because any solution will have $x<4x$.
– Andreas Blass
Nov 17 at 15:00
You may look here
– A.Γ.
Nov 17 at 14:58
You may look here
– A.Γ.
Nov 17 at 14:58
1
1
After you got the inequality $0<(x+3)(x-4)$, the next step is to remember that a product of two real numbers is positive if and only if the two factors are either both positive or both negative. Also, in this particular problem, it might be useful to notice, before working on the quadratic inequalities, that you'll only be dealing with positive values of $x$ because any solution will have $x<4x$.
– Andreas Blass
Nov 17 at 15:00
After you got the inequality $0<(x+3)(x-4)$, the next step is to remember that a product of two real numbers is positive if and only if the two factors are either both positive or both negative. Also, in this particular problem, it might be useful to notice, before working on the quadratic inequalities, that you'll only be dealing with positive values of $x$ because any solution will have $x<4x$.
– Andreas Blass
Nov 17 at 15:00
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
$$0 < (x+3)(x-4)$$
is equivalent to $x < -3$ or $x >4$. To see this sketch the convex quadratic curve and see where is it positive.
Similarly,
$$(x+2)(x-6) < 0$$ is equivalent to $$-2 < x<6.$$
Intersecting them, the region of interest is
$$4 < x< 6.$$
edited Nov 17 at 14:49
irchans
94239
answered Nov 17 at 14:40
Siong Thye Goh
96.2k1462116
add a comment |
up vote
1
down vote
You have to solve the system of quadratic inequalities
begin{cases}
x^2-x-12>0 ,\
x^2-4x-12<0 .
end{cases}
The first quadratic polynomial has roots $4$ and $-3$, hence the solutions of the first inequation is
$$S_1=(-infty,-3)cup(4,infty).$$
As to the second quadratic polynomial, it has roots $6$ and $-2$, so the solution of the second inequation is
$$S_2=(-2, 6),$$
and the solutions of the system is
$$S_1cap S_2=(4,6).$$
answered Nov 17 at 14:56
Bernard
116k637108
Thank you so much! the last part cleared my doubt completely!!
– RiRi
Nov 18 at 10:21
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$$0 < (x+3)(x-4)$$
is equivalent to $x < -3$ or $x >4$. To see this sketch the convex quadratic curve and see where is it positive.
Similarly,
$$(x+2)(x-6) < 0$$ is equivalent to $$-2 < x<6.$$
Intersecting them, the region of interest is
$$4 < x< 6.$$
edited Nov 17 at 14:49
irchans
94239
answered Nov 17 at 14:40
Siong Thye Goh
96.2k1462116
add a comment |
up vote
1
down vote
$$0 < (x+3)(x-4)$$
is equivalent to $x < -3$ or $x >4$. To see this sketch the convex quadratic curve and see where is it positive.
Similarly,
$$(x+2)(x-6) < 0$$ is equivalent to $$-2 < x<6.$$
Intersecting them, the region of interest is
$$4 < x< 6.$$
edited Nov 17 at 14:49
irchans
94239
answered Nov 17 at 14:40
Siong Thye Goh
96.2k1462116
add a comment |
up vote
1
down vote
up vote
1
down vote
$$0 < (x+3)(x-4)$$
is equivalent to $x < -3$ or $x >4$. To see this sketch the convex quadratic curve and see where is it positive.
Similarly,
$$(x+2)(x-6) < 0$$ is equivalent to $$-2 < x<6.$$
Intersecting them, the region of interest is
$$4 < x< 6.$$
edited Nov 17 at 14:49
irchans
94239
answered Nov 17 at 14:40
Siong Thye Goh
96.2k1462116
$$0 < (x+3)(x-4)$$
is equivalent to $x < -3$ or $x >4$. To see this sketch the convex quadratic curve and see where is it positive.
Similarly,
$$(x+2)(x-6) < 0$$ is equivalent to $$-2 < x<6.$$
Intersecting them, the region of interest is
$$4 < x< 6.$$
edited Nov 17 at 14:49
irchans
94239
answered Nov 17 at 14:40
Siong Thye Goh
96.2k1462116
edited Nov 17 at 14:49
irchans
94239
edited Nov 17 at 14:49
irchans
94239
edited Nov 17 at 14:49
irchans
94239
94239
answered Nov 17 at 14:40
Siong Thye Goh
96.2k1462116
answered Nov 17 at 14:40
Siong Thye Goh
96.2k1462116
answered Nov 17 at 14:40
Siong Thye Goh
96.2k1462116
96.2k1462116
add a comment |
add a comment |
up vote
1
down vote
You have to solve the system of quadratic inequalities
begin{cases}
x^2-x-12>0 ,\
x^2-4x-12<0 .
end{cases}
The first quadratic polynomial has roots $4$ and $-3$, hence the solutions of the first inequation is
$$S_1=(-infty,-3)cup(4,infty).$$
As to the second quadratic polynomial, it has roots $6$ and $-2$, so the solution of the second inequation is
$$S_2=(-2, 6),$$
and the solutions of the system is
$$S_1cap S_2=(4,6).$$
answered Nov 17 at 14:56
Bernard
116k637108
Thank you so much! the last part cleared my doubt completely!!
– RiRi
Nov 18 at 10:21
add a comment |
up vote
1
down vote
You have to solve the system of quadratic inequalities
begin{cases}
x^2-x-12>0 ,\
x^2-4x-12<0 .
end{cases}
The first quadratic polynomial has roots $4$ and $-3$, hence the solutions of the first inequation is
$$S_1=(-infty,-3)cup(4,infty).$$
As to the second quadratic polynomial, it has roots $6$ and $-2$, so the solution of the second inequation is
$$S_2=(-2, 6),$$
and the solutions of the system is
$$S_1cap S_2=(4,6).$$
answered Nov 17 at 14:56
Bernard
116k637108
Thank you so much! the last part cleared my doubt completely!!
– RiRi
Nov 18 at 10:21
add a comment |
up vote
1
down vote
up vote
1
down vote
You have to solve the system of quadratic inequalities
begin{cases}
x^2-x-12>0 ,\
x^2-4x-12<0 .
end{cases}
The first quadratic polynomial has roots $4$ and $-3$, hence the solutions of the first inequation is
$$S_1=(-infty,-3)cup(4,infty).$$
As to the second quadratic polynomial, it has roots $6$ and $-2$, so the solution of the second inequation is
$$S_2=(-2, 6),$$
and the solutions of the system is
$$S_1cap S_2=(4,6).$$
answered Nov 17 at 14:56
Bernard
116k637108
You have to solve the system of quadratic inequalities
begin{cases}
x^2-x-12>0 ,\
x^2-4x-12<0 .
end{cases}
The first quadratic polynomial has roots $4$ and $-3$, hence the solutions of the first inequation is
$$S_1=(-infty,-3)cup(4,infty).$$
As to the second quadratic polynomial, it has roots $6$ and $-2$, so the solution of the second inequation is
$$S_2=(-2, 6),$$
and the solutions of the system is
$$S_1cap S_2=(4,6).$$
answered Nov 17 at 14:56
Bernard
116k637108
answered Nov 17 at 14:56
Bernard
116k637108
answered Nov 17 at 14:56
Bernard
116k637108
answered Nov 17 at 14:56
Bernard
116k637108
116k637108
Thank you so much! the last part cleared my doubt completely!!
– RiRi
Nov 18 at 10:21
add a comment |
Thank you so much! the last part cleared my doubt completely!!
– RiRi
Nov 18 at 10:21
Thank you so much! the last part cleared my doubt completely!!
– RiRi
Nov 18 at 10:21
Thank you so much! the last part cleared my doubt completely!!
– RiRi
Nov 18 at 10:21
add a comment |
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– A.Γ.
Nov 17 at 14:58
1
After you got the inequality $0<(x+3)(x-4)$, the next step is to remember that a product of two real numbers is positive if and only if the two factors are either both positive or both negative. Also, in this particular problem, it might be useful to notice, before working on the quadratic inequalities, that you'll only be dealing with positive values of $x$ because any solution will have $x<4x$.
– Andreas Blass
Nov 17 at 15:00