$1$ as a common root of a quadratic equationHow to show that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+Bomega+Comega^2)(A+Bomega^2+Comega)$ indirectly?Unfoiling quadratic equationTriple Simultaneous Equations not resolving the Equation for a Quadratic FunctionQuadratic Equation Based Problem:Prove either $a = 2l$ & $b = m$ or $b + m = al$Question about quadratic equationQuadratic equations greater than zero, then we have at most one real rootThe equations $x^3+5x^2+px+q=0$and $x^3+7x^2+px+r=0$ have 2 common roots, then find the third root of both equationsExplanation on the square root property, and how it's derived.Making this equation into Quadratic form produces two different answersEquilibrium from linear to quadraticIf the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root,then $a+b+c$ can be equal to
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$1$ as a common root of a quadratic equation
How to show that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+Bomega+Comega^2)(A+Bomega^2+Comega)$ indirectly?Unfoiling quadratic equationTriple Simultaneous Equations not resolving the Equation for a Quadratic FunctionQuadratic Equation Based Problem:Prove either $a = 2l$ & $b = m$ or $b + m = al$Question about quadratic equationQuadratic equations greater than zero, then we have at most one real rootThe equations $x^3+5x^2+px+q=0$and $x^3+7x^2+px+r=0$ have 2 common roots, then find the third root of both equationsExplanation on the square root property, and how it's derived.Making this equation into Quadratic form produces two different answersEquilibrium from linear to quadraticIf the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root,then $a+b+c$ can be equal to
$begingroup$
$$ax^2 + bx + c = 0quadtextandquad
bx^2 + cx + a = 0$$
have a common root.
In my book, it says that 1 is a common root for those equation?
Is this correct.
When we plug 1 in both the equations, we get $a+b+c = 0$, it says nothing about 1 being a root. Since we don't know if LHS is zero or not.
Where I'm going wrong here?
quadratics
edited yesterday
TheSimpliFire
13.1k62464
asked yesterday
user638507user638507
342
$endgroup$
add a comment |
$begingroup$
$$ax^2 + bx + c = 0quadtextandquad
bx^2 + cx + a = 0$$
have a common root.
In my book, it says that 1 is a common root for those equation?
Is this correct.
When we plug 1 in both the equations, we get $a+b+c = 0$, it says nothing about 1 being a root. Since we don't know if LHS is zero or not.
Where I'm going wrong here?
quadratics
edited yesterday
TheSimpliFire
13.1k62464
asked yesterday
user638507user638507
342
$endgroup$
3
$begingroup$
Without further information, we cannot say that $1$ is a root. For example, if $a=b=c=1$, the two quadratic equations are $x^2+x+1=0$ and $x^2+x+1=0$. These clearly have a common root (as they are the same equation), but $1$ is not a root. From answers below though, we know that a common root $x_0$ must satisfy $$aleft(x_0^3-1right)=0,$$ or equivalently (if $ane 0$) $$x_0^3=1.$$ Therefore, if you are given that there is a common real root, then you can conclude that $x_0=1$ is a common root.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
It's possible of course that this question came up in a context where the reader is not expected to know about complex numbers (just real numbers). In that case, when they said there is a common root, it automatically meant real root and you can more or less ignore my comment above.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
Could you clarify if the common root is required to be a real root?
$endgroup$
– Servaes
yesterday
add a comment |
$begingroup$
$$ax^2 + bx + c = 0quadtextandquad
bx^2 + cx + a = 0$$
have a common root.
In my book, it says that 1 is a common root for those equation?
Is this correct.
When we plug 1 in both the equations, we get $a+b+c = 0$, it says nothing about 1 being a root. Since we don't know if LHS is zero or not.
Where I'm going wrong here?
quadratics
edited yesterday
TheSimpliFire
13.1k62464
asked yesterday
user638507user638507
342
$endgroup$
$$ax^2 + bx + c = 0quadtextandquad
bx^2 + cx + a = 0$$
have a common root.
In my book, it says that 1 is a common root for those equation?
Is this correct.
When we plug 1 in both the equations, we get $a+b+c = 0$, it says nothing about 1 being a root. Since we don't know if LHS is zero or not.
Where I'm going wrong here?
quadratics
quadratics
edited yesterday
TheSimpliFire
13.1k62464
asked yesterday
user638507user638507
342
edited yesterday
TheSimpliFire
13.1k62464
asked yesterday
user638507user638507
342
edited yesterday
TheSimpliFire
13.1k62464
edited yesterday
TheSimpliFire
13.1k62464
edited yesterday
TheSimpliFire
13.1k62464
13.1k62464
asked yesterday
user638507user638507
342
asked yesterday
user638507user638507
342
asked yesterday
user638507user638507
342
342
3
$begingroup$
Without further information, we cannot say that $1$ is a root. For example, if $a=b=c=1$, the two quadratic equations are $x^2+x+1=0$ and $x^2+x+1=0$. These clearly have a common root (as they are the same equation), but $1$ is not a root. From answers below though, we know that a common root $x_0$ must satisfy $$aleft(x_0^3-1right)=0,$$ or equivalently (if $ane 0$) $$x_0^3=1.$$ Therefore, if you are given that there is a common real root, then you can conclude that $x_0=1$ is a common root.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
It's possible of course that this question came up in a context where the reader is not expected to know about complex numbers (just real numbers). In that case, when they said there is a common root, it automatically meant real root and you can more or less ignore my comment above.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
Could you clarify if the common root is required to be a real root?
$endgroup$
– Servaes
yesterday
add a comment |
3
$begingroup$
Without further information, we cannot say that $1$ is a root. For example, if $a=b=c=1$, the two quadratic equations are $x^2+x+1=0$ and $x^2+x+1=0$. These clearly have a common root (as they are the same equation), but $1$ is not a root. From answers below though, we know that a common root $x_0$ must satisfy $$aleft(x_0^3-1right)=0,$$ or equivalently (if $ane 0$) $$x_0^3=1.$$ Therefore, if you are given that there is a common real root, then you can conclude that $x_0=1$ is a common root.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
It's possible of course that this question came up in a context where the reader is not expected to know about complex numbers (just real numbers). In that case, when they said there is a common root, it automatically meant real root and you can more or less ignore my comment above.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
Could you clarify if the common root is required to be a real root?
$endgroup$
– Servaes
yesterday
3
3
$begingroup$
Without further information, we cannot say that $1$ is a root. For example, if $a=b=c=1$, the two quadratic equations are $x^2+x+1=0$ and $x^2+x+1=0$. These clearly have a common root (as they are the same equation), but $1$ is not a root. From answers below though, we know that a common root $x_0$ must satisfy $$aleft(x_0^3-1right)=0,$$ or equivalently (if $ane 0$) $$x_0^3=1.$$ Therefore, if you are given that there is a common real root, then you can conclude that $x_0=1$ is a common root.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
Without further information, we cannot say that $1$ is a root. For example, if $a=b=c=1$, the two quadratic equations are $x^2+x+1=0$ and $x^2+x+1=0$. These clearly have a common root (as they are the same equation), but $1$ is not a root. From answers below though, we know that a common root $x_0$ must satisfy $$aleft(x_0^3-1right)=0,$$ or equivalently (if $ane 0$) $$x_0^3=1.$$ Therefore, if you are given that there is a common real root, then you can conclude that $x_0=1$ is a common root.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
It's possible of course that this question came up in a context where the reader is not expected to know about complex numbers (just real numbers). In that case, when they said there is a common root, it automatically meant real root and you can more or less ignore my comment above.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
It's possible of course that this question came up in a context where the reader is not expected to know about complex numbers (just real numbers). In that case, when they said there is a common root, it automatically meant real root and you can more or less ignore my comment above.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
Could you clarify if the common root is required to be a real root?
$endgroup$
– Servaes
yesterday
$begingroup$
Could you clarify if the common root is required to be a real root?
$endgroup$
– Servaes
yesterday
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
From first one you have $$c=-ax^2-bx$$ so put this in second eqaution and you get:$$ ax^3-a=0$$ Since $aneq 0$ we have $x=1$ or solution of $x^2+x+1=0$.
answered yesterday
Maria MazurMaria Mazur
50k1361124
$endgroup$
$begingroup$
rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
$endgroup$
– zwim
yesterday
1
$begingroup$
@zwim but then it is not a quadratic
$endgroup$
– Maria Mazur
yesterday
add a comment |
$begingroup$
Let $t$ is the common root,
$at^2+bt+c=0$
$bt^2+ct+a=0$
Solve for $t^2,t$
and use $t^2=(t)^2$ to eliminate $t$
and find $$0=a^3+b^3+c^3-3abc=(a+b+c)(?)$$
answered yesterday
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$begingroup$
See also : math.stackexchange.com/questions/475354/…
$endgroup$
– lab bhattacharjee
yesterday
add a comment |
$begingroup$
Let $t$ be the common root:
$$at^2+bt+c=0\
bt^2+ct+a=0$$
Subtract and factorize:
$$a(t^2-1)+b(t-t^2)+c(1-t)=0 iff \
a(t-1)(t+1)-bt(t-1)-c(t-1)=0 iff \
(t-1)(at+a-bt-c)=0 Rightarrow t=1$$
answered yesterday
farruhotafarruhota
21.8k2842
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From first one you have $$c=-ax^2-bx$$ so put this in second eqaution and you get:$$ ax^3-a=0$$ Since $aneq 0$ we have $x=1$ or solution of $x^2+x+1=0$.
answered yesterday
Maria MazurMaria Mazur
50k1361124
$endgroup$
$begingroup$
rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
$endgroup$
– zwim
yesterday
1
$begingroup$
@zwim but then it is not a quadratic
$endgroup$
– Maria Mazur
yesterday
add a comment |
$begingroup$
From first one you have $$c=-ax^2-bx$$ so put this in second eqaution and you get:$$ ax^3-a=0$$ Since $aneq 0$ we have $x=1$ or solution of $x^2+x+1=0$.
answered yesterday
Maria MazurMaria Mazur
50k1361124
$endgroup$
$begingroup$
rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
$endgroup$
– zwim
yesterday
1
$begingroup$
@zwim but then it is not a quadratic
$endgroup$
– Maria Mazur
yesterday
add a comment |
$begingroup$
From first one you have $$c=-ax^2-bx$$ so put this in second eqaution and you get:$$ ax^3-a=0$$ Since $aneq 0$ we have $x=1$ or solution of $x^2+x+1=0$.
answered yesterday
Maria MazurMaria Mazur
50k1361124
$endgroup$
From first one you have $$c=-ax^2-bx$$ so put this in second eqaution and you get:$$ ax^3-a=0$$ Since $aneq 0$ we have $x=1$ or solution of $x^2+x+1=0$.
answered yesterday
Maria MazurMaria Mazur
50k1361124
answered yesterday
Maria MazurMaria Mazur
50k1361124
answered yesterday
Maria MazurMaria Mazur
50k1361124
answered yesterday
Maria MazurMaria Mazur
50k1361124
50k1361124
$begingroup$
rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
$endgroup$
– zwim
yesterday
1
$begingroup$
@zwim but then it is not a quadratic
$endgroup$
– Maria Mazur
yesterday
add a comment |
$begingroup$
rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
$endgroup$
– zwim
yesterday
1
$begingroup$
@zwim but then it is not a quadratic
$endgroup$
– Maria Mazur
yesterday
$begingroup$
rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
$endgroup$
– zwim
yesterday
$begingroup$
rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
$endgroup$
– zwim
yesterday
1
1
$begingroup$
@zwim but then it is not a quadratic
$endgroup$
– Maria Mazur
yesterday
$begingroup$
@zwim but then it is not a quadratic
$endgroup$
– Maria Mazur
yesterday
add a comment |
$begingroup$
Let $t$ is the common root,
$at^2+bt+c=0$
$bt^2+ct+a=0$
Solve for $t^2,t$
and use $t^2=(t)^2$ to eliminate $t$
and find $$0=a^3+b^3+c^3-3abc=(a+b+c)(?)$$
answered yesterday
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$begingroup$
See also : math.stackexchange.com/questions/475354/…
$endgroup$
– lab bhattacharjee
yesterday
add a comment |
$begingroup$
Let $t$ is the common root,
$at^2+bt+c=0$
$bt^2+ct+a=0$
Solve for $t^2,t$
and use $t^2=(t)^2$ to eliminate $t$
and find $$0=a^3+b^3+c^3-3abc=(a+b+c)(?)$$
answered yesterday
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$begingroup$
See also : math.stackexchange.com/questions/475354/…
$endgroup$
– lab bhattacharjee
yesterday
add a comment |
$begingroup$
Let $t$ is the common root,
$at^2+bt+c=0$
$bt^2+ct+a=0$
Solve for $t^2,t$
and use $t^2=(t)^2$ to eliminate $t$
and find $$0=a^3+b^3+c^3-3abc=(a+b+c)(?)$$
answered yesterday
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
Let $t$ is the common root,
$at^2+bt+c=0$
$bt^2+ct+a=0$
Solve for $t^2,t$
and use $t^2=(t)^2$ to eliminate $t$
and find $$0=a^3+b^3+c^3-3abc=(a+b+c)(?)$$
answered yesterday
lab bhattacharjeelab bhattacharjee
228k15158279
answered yesterday
lab bhattacharjeelab bhattacharjee
228k15158279
answered yesterday
lab bhattacharjeelab bhattacharjee
228k15158279
answered yesterday
lab bhattacharjeelab bhattacharjee
228k15158279
228k15158279
$begingroup$
See also : math.stackexchange.com/questions/475354/…
$endgroup$
– lab bhattacharjee
yesterday
add a comment |
$begingroup$
See also : math.stackexchange.com/questions/475354/…
$endgroup$
– lab bhattacharjee
yesterday
$begingroup$
See also : math.stackexchange.com/questions/475354/…
$endgroup$
– lab bhattacharjee
yesterday
$begingroup$
See also : math.stackexchange.com/questions/475354/…
$endgroup$
– lab bhattacharjee
yesterday
add a comment |
$begingroup$
Let $t$ be the common root:
$$at^2+bt+c=0\
bt^2+ct+a=0$$
Subtract and factorize:
$$a(t^2-1)+b(t-t^2)+c(1-t)=0 iff \
a(t-1)(t+1)-bt(t-1)-c(t-1)=0 iff \
(t-1)(at+a-bt-c)=0 Rightarrow t=1$$
answered yesterday
farruhotafarruhota
21.8k2842
$endgroup$
add a comment |
$begingroup$
Let $t$ be the common root:
$$at^2+bt+c=0\
bt^2+ct+a=0$$
Subtract and factorize:
$$a(t^2-1)+b(t-t^2)+c(1-t)=0 iff \
a(t-1)(t+1)-bt(t-1)-c(t-1)=0 iff \
(t-1)(at+a-bt-c)=0 Rightarrow t=1$$
answered yesterday
farruhotafarruhota
21.8k2842
$endgroup$
add a comment |
$begingroup$
Let $t$ be the common root:
$$at^2+bt+c=0\
bt^2+ct+a=0$$
Subtract and factorize:
$$a(t^2-1)+b(t-t^2)+c(1-t)=0 iff \
a(t-1)(t+1)-bt(t-1)-c(t-1)=0 iff \
(t-1)(at+a-bt-c)=0 Rightarrow t=1$$
answered yesterday
farruhotafarruhota
21.8k2842
$endgroup$
Let $t$ be the common root:
$$at^2+bt+c=0\
bt^2+ct+a=0$$
Subtract and factorize:
$$a(t^2-1)+b(t-t^2)+c(1-t)=0 iff \
a(t-1)(t+1)-bt(t-1)-c(t-1)=0 iff \
(t-1)(at+a-bt-c)=0 Rightarrow t=1$$
answered yesterday
farruhotafarruhota
21.8k2842
answered yesterday
farruhotafarruhota
21.8k2842
answered yesterday
farruhotafarruhota
21.8k2842
answered yesterday
farruhotafarruhota
21.8k2842
21.8k2842
add a comment |
add a comment |
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$begingroup$
Without further information, we cannot say that $1$ is a root. For example, if $a=b=c=1$, the two quadratic equations are $x^2+x+1=0$ and $x^2+x+1=0$. These clearly have a common root (as they are the same equation), but $1$ is not a root. From answers below though, we know that a common root $x_0$ must satisfy $$aleft(x_0^3-1right)=0,$$ or equivalently (if $ane 0$) $$x_0^3=1.$$ Therefore, if you are given that there is a common real root, then you can conclude that $x_0=1$ is a common root.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
It's possible of course that this question came up in a context where the reader is not expected to know about complex numbers (just real numbers). In that case, when they said there is a common root, it automatically meant real root and you can more or less ignore my comment above.
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– Minus One-Twelfth
yesterday
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Could you clarify if the common root is required to be a real root?
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– Servaes
yesterday