Solving $x^2-2x-3<0$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Solving inequalities with absolute valuesQuestions about solving inequality: $2 < frac3x+12x+4$Solving an inequality and multiplying by -1Why does the sign have to be flipped in this inequality?Why not always use the quadratic equationSolving trigonometric inequality for positive valuesStating that either root is zero in solving a quadratic equationSolving for $x$ in Quadratic EquationsSolving one quadratic equals anotherDoubt in law of reciprocal
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Solving $x^2-2x-3
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Solving inequalities with absolute valuesQuestions about solving inequality: $2 < frac3x+12x+4$Solving an inequality and multiplying by -1Why does the sign have to be flipped in this inequality?Why not always use the quadratic equationSolving trigonometric inequality for positive valuesStating that either root is zero in solving a quadratic equationSolving for $x$ in Quadratic EquationsSolving one quadratic equals anotherDoubt in law of reciprocal
$begingroup$
If i have to solve $x^2-2x-3<0$ I would do
$$x+1 < 0, quad x-3<0$$
and end up getting $x<-1$ and $x<3$. Why is it wrong to use the same inequality sign? Shouldn’t both of the linear equations formed from factorising the quadratic use the same inequality sign as the quadratic?
proof-verification inequality quadratics problem-solving
edited Apr 12 at 11:50
YuiTo Cheng
2,58641037
asked Apr 12 at 10:38
Ubaid HassanUbaid Hassan
1125
$endgroup$
add a comment |
$begingroup$
If i have to solve $x^2-2x-3<0$ I would do
$$x+1 < 0, quad x-3<0$$
and end up getting $x<-1$ and $x<3$. Why is it wrong to use the same inequality sign? Shouldn’t both of the linear equations formed from factorising the quadratic use the same inequality sign as the quadratic?
proof-verification inequality quadratics problem-solving
edited Apr 12 at 11:50
YuiTo Cheng
2,58641037
asked Apr 12 at 10:38
Ubaid HassanUbaid Hassan
1125
$endgroup$
$begingroup$
Did you try plugging some $x<-1$ into the original inequality? If you do that, I think you will agree that it doesn't work.
$endgroup$
– Hans Lundmark
Apr 12 at 10:51
$begingroup$
The answer you accepted doesn't address your question !
$endgroup$
– Yves Daoust
Apr 12 at 12:15
add a comment |
$begingroup$
If i have to solve $x^2-2x-3<0$ I would do
$$x+1 < 0, quad x-3<0$$
and end up getting $x<-1$ and $x<3$. Why is it wrong to use the same inequality sign? Shouldn’t both of the linear equations formed from factorising the quadratic use the same inequality sign as the quadratic?
proof-verification inequality quadratics problem-solving
edited Apr 12 at 11:50
YuiTo Cheng
2,58641037
asked Apr 12 at 10:38
Ubaid HassanUbaid Hassan
1125
$endgroup$
If i have to solve $x^2-2x-3<0$ I would do
$$x+1 < 0, quad x-3<0$$
and end up getting $x<-1$ and $x<3$. Why is it wrong to use the same inequality sign? Shouldn’t both of the linear equations formed from factorising the quadratic use the same inequality sign as the quadratic?
proof-verification inequality quadratics problem-solving
proof-verification inequality quadratics problem-solving
edited Apr 12 at 11:50
YuiTo Cheng
2,58641037
asked Apr 12 at 10:38
Ubaid HassanUbaid Hassan
1125
edited Apr 12 at 11:50
YuiTo Cheng
2,58641037
asked Apr 12 at 10:38
Ubaid HassanUbaid Hassan
1125
edited Apr 12 at 11:50
YuiTo Cheng
2,58641037
edited Apr 12 at 11:50
YuiTo Cheng
2,58641037
edited Apr 12 at 11:50
YuiTo Cheng
2,58641037
2,58641037
asked Apr 12 at 10:38
Ubaid HassanUbaid Hassan
1125
asked Apr 12 at 10:38
Ubaid HassanUbaid Hassan
1125
asked Apr 12 at 10:38
Ubaid HassanUbaid Hassan
1125
1125
$begingroup$
Did you try plugging some $x<-1$ into the original inequality? If you do that, I think you will agree that it doesn't work.
$endgroup$
– Hans Lundmark
Apr 12 at 10:51
$begingroup$
The answer you accepted doesn't address your question !
$endgroup$
– Yves Daoust
Apr 12 at 12:15
add a comment |
$begingroup$
Did you try plugging some $x<-1$ into the original inequality? If you do that, I think you will agree that it doesn't work.
$endgroup$
– Hans Lundmark
Apr 12 at 10:51
$begingroup$
The answer you accepted doesn't address your question !
$endgroup$
– Yves Daoust
Apr 12 at 12:15
$begingroup$
Did you try plugging some $x<-1$ into the original inequality? If you do that, I think you will agree that it doesn't work.
$endgroup$
– Hans Lundmark
Apr 12 at 10:51
$begingroup$
Did you try plugging some $x<-1$ into the original inequality? If you do that, I think you will agree that it doesn't work.
$endgroup$
– Hans Lundmark
Apr 12 at 10:51
$begingroup$
The answer you accepted doesn't address your question !
$endgroup$
– Yves Daoust
Apr 12 at 12:15
$begingroup$
The answer you accepted doesn't address your question !
$endgroup$
– Yves Daoust
Apr 12 at 12:15
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Ok I get your point.
Let's first assume
$$x^2-2x-3=0$$
$$(x+1)(x-3)=0$$
This is true when $x=-1$ or when $x=3$. That is, by equating each of the linear term with $0$ one by one.
Now let's come to
$$(x+1)(x-3)<0$$
So you are asking, to solve this why don't we do the same thing again, i.e. one by one making both the linear terms less than $0$.
But that's not the correct way to do it, because multiplying two negative numbers yields a positive number.
Suppose there are two variables $A$ and $B$ and it is given that,
$AB<0$
This is true when $A<0$ and $B>0$ because $-vetimes +ve=-ve$ or when $A>0$ and $B<0$ because $+vetimes -ve=-ve$.
Similarly
$$(x+1)(x-3)<0$$
Either when $x+1<0$ and $x-3>0$ or when $x+1>0$ and $x-3<0$. The former statement is absurd as there exist no real number which is less than $-1$ as well as greater than $3$. Hence the answer is $xin (-1,3)$
answered Apr 12 at 10:56
Loop BackLoop Back
44312
$endgroup$
$begingroup$
This finally starting to make sense, but I have one more question : in the statement (x+1(x-3)<0 is it correct to divide both sides by x to get -3<0?
$endgroup$
– Ubaid Hassan
Apr 12 at 13:22
$begingroup$
@UbaidHassan its not sensible to divide by $x$ until and unless you know whether $x$ is positive, negative or zero. And even if you somehow know what x is, i don't know how you will get -3<0?
$endgroup$
– Loop Back
Apr 13 at 7:16
add a comment |
$begingroup$
Hint: Sketch the graph of $x²-2x -3$. Notice that it is zero when $x=-1$ and $x=3$. Then consider the following areas: $(-infty,-1)$, $(-1,3)$ and $(3,infty)$.
answered Apr 12 at 10:42
Tim DiklandTim Dikland
44319
$endgroup$
add a comment |
$begingroup$
Option:
Completing the square:
$(x-1)^2-4<0;$
$(x-1)^2 <4;$
$|x-1| <2;$
$-1 < x <3.$
answered Apr 12 at 11:08
Peter SzilasPeter Szilas
12k2822
$endgroup$
add a comment |
$begingroup$
No, because of the rule of signs.
Anyway, there's a theorem on the sign of quadratic polynomials:
Let $ax^2+bx+censpace (ane0$) a quadrattic polynomial with real coefficients. Then $ax^2+bx+c$ has the sign of $a$, except between its (real) roots, if any.
answered Apr 12 at 10:44
BernardBernard
124k742117
$endgroup$
add a comment |
$begingroup$
$$ab<0iff a<0land b<0$$ is wrong. You can easily find counterexamples.
The truth is
$$ab<0iff (a<0land b>0)lor(a>0land b<0).$$
A product is negative when the factors have opposite signs.
answered Apr 12 at 12:18
Yves DaoustYves Daoust
133k676232
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Ok I get your point.
Let's first assume
$$x^2-2x-3=0$$
$$(x+1)(x-3)=0$$
This is true when $x=-1$ or when $x=3$. That is, by equating each of the linear term with $0$ one by one.
Now let's come to
$$(x+1)(x-3)<0$$
So you are asking, to solve this why don't we do the same thing again, i.e. one by one making both the linear terms less than $0$.
But that's not the correct way to do it, because multiplying two negative numbers yields a positive number.
Suppose there are two variables $A$ and $B$ and it is given that,
$AB<0$
This is true when $A<0$ and $B>0$ because $-vetimes +ve=-ve$ or when $A>0$ and $B<0$ because $+vetimes -ve=-ve$.
Similarly
$$(x+1)(x-3)<0$$
Either when $x+1<0$ and $x-3>0$ or when $x+1>0$ and $x-3<0$. The former statement is absurd as there exist no real number which is less than $-1$ as well as greater than $3$. Hence the answer is $xin (-1,3)$
answered Apr 12 at 10:56
Loop BackLoop Back
44312
$endgroup$
$begingroup$
This finally starting to make sense, but I have one more question : in the statement (x+1(x-3)<0 is it correct to divide both sides by x to get -3<0?
$endgroup$
– Ubaid Hassan
Apr 12 at 13:22
$begingroup$
@UbaidHassan its not sensible to divide by $x$ until and unless you know whether $x$ is positive, negative or zero. And even if you somehow know what x is, i don't know how you will get -3<0?
$endgroup$
– Loop Back
Apr 13 at 7:16
add a comment |
$begingroup$
Ok I get your point.
Let's first assume
$$x^2-2x-3=0$$
$$(x+1)(x-3)=0$$
This is true when $x=-1$ or when $x=3$. That is, by equating each of the linear term with $0$ one by one.
Now let's come to
$$(x+1)(x-3)<0$$
So you are asking, to solve this why don't we do the same thing again, i.e. one by one making both the linear terms less than $0$.
But that's not the correct way to do it, because multiplying two negative numbers yields a positive number.
Suppose there are two variables $A$ and $B$ and it is given that,
$AB<0$
This is true when $A<0$ and $B>0$ because $-vetimes +ve=-ve$ or when $A>0$ and $B<0$ because $+vetimes -ve=-ve$.
Similarly
$$(x+1)(x-3)<0$$
Either when $x+1<0$ and $x-3>0$ or when $x+1>0$ and $x-3<0$. The former statement is absurd as there exist no real number which is less than $-1$ as well as greater than $3$. Hence the answer is $xin (-1,3)$
answered Apr 12 at 10:56
Loop BackLoop Back
44312
$endgroup$
$begingroup$
This finally starting to make sense, but I have one more question : in the statement (x+1(x-3)<0 is it correct to divide both sides by x to get -3<0?
$endgroup$
– Ubaid Hassan
Apr 12 at 13:22
$begingroup$
@UbaidHassan its not sensible to divide by $x$ until and unless you know whether $x$ is positive, negative or zero. And even if you somehow know what x is, i don't know how you will get -3<0?
$endgroup$
– Loop Back
Apr 13 at 7:16
add a comment |
$begingroup$
Ok I get your point.
Let's first assume
$$x^2-2x-3=0$$
$$(x+1)(x-3)=0$$
This is true when $x=-1$ or when $x=3$. That is, by equating each of the linear term with $0$ one by one.
Now let's come to
$$(x+1)(x-3)<0$$
So you are asking, to solve this why don't we do the same thing again, i.e. one by one making both the linear terms less than $0$.
But that's not the correct way to do it, because multiplying two negative numbers yields a positive number.
Suppose there are two variables $A$ and $B$ and it is given that,
$AB<0$
This is true when $A<0$ and $B>0$ because $-vetimes +ve=-ve$ or when $A>0$ and $B<0$ because $+vetimes -ve=-ve$.
Similarly
$$(x+1)(x-3)<0$$
Either when $x+1<0$ and $x-3>0$ or when $x+1>0$ and $x-3<0$. The former statement is absurd as there exist no real number which is less than $-1$ as well as greater than $3$. Hence the answer is $xin (-1,3)$
answered Apr 12 at 10:56
Loop BackLoop Back
44312
$endgroup$
Ok I get your point.
Let's first assume
$$x^2-2x-3=0$$
$$(x+1)(x-3)=0$$
This is true when $x=-1$ or when $x=3$. That is, by equating each of the linear term with $0$ one by one.
Now let's come to
$$(x+1)(x-3)<0$$
So you are asking, to solve this why don't we do the same thing again, i.e. one by one making both the linear terms less than $0$.
But that's not the correct way to do it, because multiplying two negative numbers yields a positive number.
Suppose there are two variables $A$ and $B$ and it is given that,
$AB<0$
This is true when $A<0$ and $B>0$ because $-vetimes +ve=-ve$ or when $A>0$ and $B<0$ because $+vetimes -ve=-ve$.
Similarly
$$(x+1)(x-3)<0$$
Either when $x+1<0$ and $x-3>0$ or when $x+1>0$ and $x-3<0$. The former statement is absurd as there exist no real number which is less than $-1$ as well as greater than $3$. Hence the answer is $xin (-1,3)$
answered Apr 12 at 10:56
Loop BackLoop Back
44312
edited Apr 12 at 11:01
answered Apr 12 at 10:56
Loop BackLoop Back
44312
answered Apr 12 at 10:56
Loop BackLoop Back
44312
answered Apr 12 at 10:56
Loop BackLoop Back
44312
44312
$begingroup$
This finally starting to make sense, but I have one more question : in the statement (x+1(x-3)<0 is it correct to divide both sides by x to get -3<0?
$endgroup$
– Ubaid Hassan
Apr 12 at 13:22
$begingroup$
@UbaidHassan its not sensible to divide by $x$ until and unless you know whether $x$ is positive, negative or zero. And even if you somehow know what x is, i don't know how you will get -3<0?
$endgroup$
– Loop Back
Apr 13 at 7:16
add a comment |
$begingroup$
This finally starting to make sense, but I have one more question : in the statement (x+1(x-3)<0 is it correct to divide both sides by x to get -3<0?
$endgroup$
– Ubaid Hassan
Apr 12 at 13:22
$begingroup$
@UbaidHassan its not sensible to divide by $x$ until and unless you know whether $x$ is positive, negative or zero. And even if you somehow know what x is, i don't know how you will get -3<0?
$endgroup$
– Loop Back
Apr 13 at 7:16
$begingroup$
This finally starting to make sense, but I have one more question : in the statement (x+1(x-3)<0 is it correct to divide both sides by x to get -3<0?
$endgroup$
– Ubaid Hassan
Apr 12 at 13:22
$begingroup$
This finally starting to make sense, but I have one more question : in the statement (x+1(x-3)<0 is it correct to divide both sides by x to get -3<0?
$endgroup$
– Ubaid Hassan
Apr 12 at 13:22
$begingroup$
@UbaidHassan its not sensible to divide by $x$ until and unless you know whether $x$ is positive, negative or zero. And even if you somehow know what x is, i don't know how you will get -3<0?
$endgroup$
– Loop Back
Apr 13 at 7:16
$begingroup$
@UbaidHassan its not sensible to divide by $x$ until and unless you know whether $x$ is positive, negative or zero. And even if you somehow know what x is, i don't know how you will get -3<0?
$endgroup$
– Loop Back
Apr 13 at 7:16
add a comment |
$begingroup$
Hint: Sketch the graph of $x²-2x -3$. Notice that it is zero when $x=-1$ and $x=3$. Then consider the following areas: $(-infty,-1)$, $(-1,3)$ and $(3,infty)$.
answered Apr 12 at 10:42
Tim DiklandTim Dikland
44319
$endgroup$
add a comment |
$begingroup$
Hint: Sketch the graph of $x²-2x -3$. Notice that it is zero when $x=-1$ and $x=3$. Then consider the following areas: $(-infty,-1)$, $(-1,3)$ and $(3,infty)$.
answered Apr 12 at 10:42
Tim DiklandTim Dikland
44319
$endgroup$
add a comment |
$begingroup$
Hint: Sketch the graph of $x²-2x -3$. Notice that it is zero when $x=-1$ and $x=3$. Then consider the following areas: $(-infty,-1)$, $(-1,3)$ and $(3,infty)$.
answered Apr 12 at 10:42
Tim DiklandTim Dikland
44319
$endgroup$
Hint: Sketch the graph of $x²-2x -3$. Notice that it is zero when $x=-1$ and $x=3$. Then consider the following areas: $(-infty,-1)$, $(-1,3)$ and $(3,infty)$.
answered Apr 12 at 10:42
Tim DiklandTim Dikland
44319
answered Apr 12 at 10:42
Tim DiklandTim Dikland
44319
answered Apr 12 at 10:42
Tim DiklandTim Dikland
44319
answered Apr 12 at 10:42
Tim DiklandTim Dikland
44319
44319
add a comment |
add a comment |
$begingroup$
Option:
Completing the square:
$(x-1)^2-4<0;$
$(x-1)^2 <4;$
$|x-1| <2;$
$-1 < x <3.$
answered Apr 12 at 11:08
Peter SzilasPeter Szilas
12k2822
$endgroup$
add a comment |
$begingroup$
Option:
Completing the square:
$(x-1)^2-4<0;$
$(x-1)^2 <4;$
$|x-1| <2;$
$-1 < x <3.$
answered Apr 12 at 11:08
Peter SzilasPeter Szilas
12k2822
$endgroup$
add a comment |
$begingroup$
Option:
Completing the square:
$(x-1)^2-4<0;$
$(x-1)^2 <4;$
$|x-1| <2;$
$-1 < x <3.$
answered Apr 12 at 11:08
Peter SzilasPeter Szilas
12k2822
$endgroup$
Option:
Completing the square:
$(x-1)^2-4<0;$
$(x-1)^2 <4;$
$|x-1| <2;$
$-1 < x <3.$
answered Apr 12 at 11:08
Peter SzilasPeter Szilas
12k2822
answered Apr 12 at 11:08
Peter SzilasPeter Szilas
12k2822
answered Apr 12 at 11:08
Peter SzilasPeter Szilas
12k2822
answered Apr 12 at 11:08
Peter SzilasPeter Szilas
12k2822
12k2822
add a comment |
add a comment |
$begingroup$
No, because of the rule of signs.
Anyway, there's a theorem on the sign of quadratic polynomials:
Let $ax^2+bx+censpace (ane0$) a quadrattic polynomial with real coefficients. Then $ax^2+bx+c$ has the sign of $a$, except between its (real) roots, if any.
answered Apr 12 at 10:44
BernardBernard
124k742117
$endgroup$
add a comment |
$begingroup$
No, because of the rule of signs.
Anyway, there's a theorem on the sign of quadratic polynomials:
Let $ax^2+bx+censpace (ane0$) a quadrattic polynomial with real coefficients. Then $ax^2+bx+c$ has the sign of $a$, except between its (real) roots, if any.
answered Apr 12 at 10:44
BernardBernard
124k742117
$endgroup$
add a comment |
$begingroup$
No, because of the rule of signs.
Anyway, there's a theorem on the sign of quadratic polynomials:
Let $ax^2+bx+censpace (ane0$) a quadrattic polynomial with real coefficients. Then $ax^2+bx+c$ has the sign of $a$, except between its (real) roots, if any.
answered Apr 12 at 10:44
BernardBernard
124k742117
$endgroup$
No, because of the rule of signs.
Anyway, there's a theorem on the sign of quadratic polynomials:
Let $ax^2+bx+censpace (ane0$) a quadrattic polynomial with real coefficients. Then $ax^2+bx+c$ has the sign of $a$, except between its (real) roots, if any.
answered Apr 12 at 10:44
BernardBernard
124k742117
answered Apr 12 at 10:44
BernardBernard
124k742117
answered Apr 12 at 10:44
BernardBernard
124k742117
answered Apr 12 at 10:44
BernardBernard
124k742117
124k742117
add a comment |
add a comment |
$begingroup$
$$ab<0iff a<0land b<0$$ is wrong. You can easily find counterexamples.
The truth is
$$ab<0iff (a<0land b>0)lor(a>0land b<0).$$
A product is negative when the factors have opposite signs.
answered Apr 12 at 12:18
Yves DaoustYves Daoust
133k676232
$endgroup$
add a comment |
$begingroup$
$$ab<0iff a<0land b<0$$ is wrong. You can easily find counterexamples.
The truth is
$$ab<0iff (a<0land b>0)lor(a>0land b<0).$$
A product is negative when the factors have opposite signs.
answered Apr 12 at 12:18
Yves DaoustYves Daoust
133k676232
$endgroup$
add a comment |
$begingroup$
$$ab<0iff a<0land b<0$$ is wrong. You can easily find counterexamples.
The truth is
$$ab<0iff (a<0land b>0)lor(a>0land b<0).$$
A product is negative when the factors have opposite signs.
answered Apr 12 at 12:18
Yves DaoustYves Daoust
133k676232
$endgroup$
$$ab<0iff a<0land b<0$$ is wrong. You can easily find counterexamples.
The truth is
$$ab<0iff (a<0land b>0)lor(a>0land b<0).$$
A product is negative when the factors have opposite signs.
answered Apr 12 at 12:18
Yves DaoustYves Daoust
133k676232
answered Apr 12 at 12:18
Yves DaoustYves Daoust
133k676232
answered Apr 12 at 12:18
Yves DaoustYves Daoust
133k676232
answered Apr 12 at 12:18
Yves DaoustYves Daoust
133k676232
133k676232
add a comment |
add a comment |
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$begingroup$
Did you try plugging some $x<-1$ into the original inequality? If you do that, I think you will agree that it doesn't work.
$endgroup$
– Hans Lundmark
Apr 12 at 10:51
$begingroup$
The answer you accepted doesn't address your question !
$endgroup$
– Yves Daoust
Apr 12 at 12:15