1 as a common root of a quadratic equation
$begingroup$
$$ax^2 + bx + c = 0quadtext{and}quad
bx^2 + cx + a = 0$$
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 2 hours ago
Robert Z
101k1071144
asked 2 hours ago
user638507user638507
292
$endgroup$
add a comment |
$begingroup$
$$ax^2 + bx + c = 0quadtext{and}quad
bx^2 + cx + a = 0$$
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 2 hours ago
Robert Z
101k1071144
asked 2 hours ago
user638507user638507
292
$endgroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
Could you clarify if the common root is required to be a real root?
$endgroup$
– Servaes
50 mins ago
add a comment |
$begingroup$
$$ax^2 + bx + c = 0quadtext{and}quad
bx^2 + cx + a = 0$$
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 2 hours ago
Robert Z
101k1071144
asked 2 hours ago
user638507user638507
292
$endgroup$
$$ax^2 + bx + c = 0quadtext{and}quad
bx^2 + cx + a = 0$$
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 2 hours ago
Robert Z
101k1071144
asked 2 hours ago
user638507user638507
292
edited 2 hours ago
Robert Z
101k1071144
asked 2 hours ago
user638507user638507
292
edited 2 hours ago
Robert Z
101k1071144
edited 2 hours ago
Robert Z
101k1071144
edited 2 hours ago
Robert Z
101k1071144
101k1071144
asked 2 hours ago
user638507user638507
292
asked 2 hours ago
user638507user638507
292
asked 2 hours ago
user638507user638507
292
292
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
Could you clarify if the common root is required to be a real root?
$endgroup$
– Servaes
50 mins ago
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
Could you clarify if the common root is required to be a real root?
$endgroup$
– Servaes
50 mins ago
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
Could you clarify if the common root is required to be a real root?
$endgroup$
– Servaes
50 mins ago
$begingroup$
Could you clarify if the common root is required to be a real root?
$endgroup$
– Servaes
50 mins ago
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 2 hours ago
Maria MazurMaria Mazur
49.8k1361124
$endgroup$
$begingroup$
rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
$endgroup$
– zwim
2 hours ago
$begingroup$
@zwim but then it is not a quadratic
$endgroup$
– Maria Mazur
2 hours ago
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 1 hour ago
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$begingroup$
See also : math.stackexchange.com/questions/475354/…
$endgroup$
– lab bhattacharjee
1 hour ago
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 42 mins ago
farruhotafarruhota
21.8k2842
$endgroup$
add a comment |
Your Answer
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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 2 hours ago
Maria MazurMaria Mazur
49.8k1361124
$endgroup$
$begingroup$
rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
$endgroup$
– zwim
2 hours ago
$begingroup$
@zwim but then it is not a quadratic
$endgroup$
– Maria Mazur
2 hours ago
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 2 hours ago
Maria MazurMaria Mazur
49.8k1361124
$endgroup$
$begingroup$
rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
$endgroup$
– zwim
2 hours ago
$begingroup$
@zwim but then it is not a quadratic
$endgroup$
– Maria Mazur
2 hours ago
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 2 hours ago
Maria MazurMaria Mazur
49.8k1361124
$endgroup$
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 2 hours ago
Maria MazurMaria Mazur
49.8k1361124
answered 2 hours ago
Maria MazurMaria Mazur
49.8k1361124
answered 2 hours ago
Maria MazurMaria Mazur
49.8k1361124
answered 2 hours ago
Maria MazurMaria Mazur
49.8k1361124
49.8k1361124
$begingroup$
rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
$endgroup$
– zwim
2 hours ago
$begingroup$
@zwim but then it is not a quadratic
$endgroup$
– Maria Mazur
2 hours ago
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
2 hours ago
$begingroup$
@zwim but then it is not a quadratic
$endgroup$
– Maria Mazur
2 hours ago
$begingroup$
rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
$endgroup$
– zwim
2 hours ago
$begingroup$
rem if $a=0$ then $bx+c=0$ vs $bx^2+cx=0$, still has common root, not $1$ though.
$endgroup$
– zwim
2 hours ago
$begingroup$
@zwim but then it is not a quadratic
$endgroup$
– Maria Mazur
2 hours ago
$begingroup$
@zwim but then it is not a quadratic
$endgroup$
– Maria Mazur
2 hours ago
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 1 hour ago
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$begingroup$
See also : math.stackexchange.com/questions/475354/…
$endgroup$
– lab bhattacharjee
1 hour ago
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 1 hour ago
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$begingroup$
See also : math.stackexchange.com/questions/475354/…
$endgroup$
– lab bhattacharjee
1 hour ago
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 1 hour ago
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 1 hour ago
lab bhattacharjeelab bhattacharjee
228k15158279
answered 1 hour ago
lab bhattacharjeelab bhattacharjee
228k15158279
answered 1 hour ago
lab bhattacharjeelab bhattacharjee
228k15158279
answered 1 hour ago
lab bhattacharjeelab bhattacharjee
228k15158279
228k15158279
$begingroup$
See also : math.stackexchange.com/questions/475354/…
$endgroup$
– lab bhattacharjee
1 hour ago
add a comment |
$begingroup$
See also : math.stackexchange.com/questions/475354/…
$endgroup$
– lab bhattacharjee
1 hour ago
$begingroup$
See also : math.stackexchange.com/questions/475354/…
$endgroup$
– lab bhattacharjee
1 hour ago
$begingroup$
See also : math.stackexchange.com/questions/475354/…
$endgroup$
– lab bhattacharjee
1 hour ago
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 42 mins ago
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 42 mins ago
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 42 mins ago
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 42 mins ago
farruhotafarruhota
21.8k2842
answered 42 mins ago
farruhotafarruhota
21.8k2842
answered 42 mins ago
farruhotafarruhota
21.8k2842
answered 42 mins ago
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
Could you clarify if the common root is required to be a real root?
$endgroup$
– Servaes
50 mins ago