Mixing
Completing squares, what are the steps: $2x^2-x(5+m)+(3+m)=0$
$begingroup$
I have the following equation:
$$2x^2-x(5+m)+(3+m)=0.$$
I want to find values of $m$ but I forgot the completing squares procedures. Can someone please describe this method for this particular equation?!
algebra-precalculus quadratics completing-the-square
edited Feb 3 at 9:07
Martin Sleziak
45k10123277
asked Apr 12 '12 at 13:33
Sean87Sean87
3152823
$endgroup$
add a comment |
$begingroup$
I have the following equation:
$$2x^2-x(5+m)+(3+m)=0.$$
I want to find values of $m$ but I forgot the completing squares procedures. Can someone please describe this method for this particular equation?!
algebra-precalculus quadratics completing-the-square
edited Feb 3 at 9:07
Martin Sleziak
45k10123277
asked Apr 12 '12 at 13:33
Sean87Sean87
3152823
$endgroup$
$begingroup$
Look at the second answer on this question math.stackexchange.com/questions/20597/…"
$endgroup$
– Chris K. Caldwell
Apr 12 '12 at 13:38
$begingroup$
If you are trying to solve the equation, note that by a miracle (?) the left side factors as $(x-1)(2x-(3+m))$. Or else first note that the original equation has $x=1$ as an obvious root. But you say you want to find the values of $m$. That cannot be done just given the equation. Perhaps (in a separate post) you might ask the real question.
$endgroup$
– André Nicolas
Apr 12 '12 at 14:41
add a comment |
$begingroup$
I have the following equation:
$$2x^2-x(5+m)+(3+m)=0.$$
I want to find values of $m$ but I forgot the completing squares procedures. Can someone please describe this method for this particular equation?!
algebra-precalculus quadratics completing-the-square
edited Feb 3 at 9:07
Martin Sleziak
45k10123277
asked Apr 12 '12 at 13:33
Sean87Sean87
3152823
$endgroup$
I have the following equation:
$$2x^2-x(5+m)+(3+m)=0.$$
I want to find values of $m$ but I forgot the completing squares procedures. Can someone please describe this method for this particular equation?!
algebra-precalculus quadratics completing-the-square
algebra-precalculus quadratics completing-the-square
edited Feb 3 at 9:07
Martin Sleziak
45k10123277
asked Apr 12 '12 at 13:33
Sean87Sean87
3152823
edited Feb 3 at 9:07
Martin Sleziak
45k10123277
asked Apr 12 '12 at 13:33
Sean87Sean87
3152823
edited Feb 3 at 9:07
Martin Sleziak
45k10123277
edited Feb 3 at 9:07
Martin Sleziak
45k10123277
edited Feb 3 at 9:07
Martin Sleziak
45k10123277
45k10123277
asked Apr 12 '12 at 13:33
Sean87Sean87
3152823
asked Apr 12 '12 at 13:33
Sean87Sean87
3152823
asked Apr 12 '12 at 13:33
Sean87Sean87
3152823
3152823
$begingroup$
Look at the second answer on this question math.stackexchange.com/questions/20597/…"
$endgroup$
– Chris K. Caldwell
Apr 12 '12 at 13:38
$begingroup$
If you are trying to solve the equation, note that by a miracle (?) the left side factors as $(x-1)(2x-(3+m))$. Or else first note that the original equation has $x=1$ as an obvious root. But you say you want to find the values of $m$. That cannot be done just given the equation. Perhaps (in a separate post) you might ask the real question.
$endgroup$
– André Nicolas
Apr 12 '12 at 14:41
add a comment |
$begingroup$
Look at the second answer on this question math.stackexchange.com/questions/20597/…"
$endgroup$
– Chris K. Caldwell
Apr 12 '12 at 13:38
$begingroup$
If you are trying to solve the equation, note that by a miracle (?) the left side factors as $(x-1)(2x-(3+m))$. Or else first note that the original equation has $x=1$ as an obvious root. But you say you want to find the values of $m$. That cannot be done just given the equation. Perhaps (in a separate post) you might ask the real question.
$endgroup$
– André Nicolas
Apr 12 '12 at 14:41
$begingroup$
Look at the second answer on this question math.stackexchange.com/questions/20597/…"
$endgroup$
– Chris K. Caldwell
Apr 12 '12 at 13:38
$begingroup$
Look at the second answer on this question math.stackexchange.com/questions/20597/…"
$endgroup$
– Chris K. Caldwell
Apr 12 '12 at 13:38
$begingroup$
If you are trying to solve the equation, note that by a miracle (?) the left side factors as $(x-1)(2x-(3+m))$. Or else first note that the original equation has $x=1$ as an obvious root. But you say you want to find the values of $m$. That cannot be done just given the equation. Perhaps (in a separate post) you might ask the real question.
$endgroup$
– André Nicolas
Apr 12 '12 at 14:41
$begingroup$
If you are trying to solve the equation, note that by a miracle (?) the left side factors as $(x-1)(2x-(3+m))$. Or else first note that the original equation has $x=1$ as an obvious root. But you say you want to find the values of $m$. That cannot be done just given the equation. Perhaps (in a separate post) you might ask the real question.
$endgroup$
– André Nicolas
Apr 12 '12 at 14:41
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
To complete the square on the left hand side of your equation, first factor out the $2$:
$$tag{1}
2x^2-x(5+m)+(3+m) =2bigl(
{x^2-{textstyle{5+mover2}} x+{textstyle{3+mover2}} }bigr).
$$
Write down what you want:
$$tag{2}
{x^2color{maroon}{-textstyle{5+mover2} x}+textstylecolor{darkgreen}{3+mover2}} = (x-h)^2+k.
$$
Write the right hand side of $(2)$ as
$$tag{3}
x^2color{maroon}{-2 h x}+color{darkgreen}{h^2+k}.
$$
Looking at the $color{maroon}x$ term of the left hand side of $(2)$ with the $color{maroon}x$ term of $(3)$, write down an equation and solve for $h$:
$$textstyle
color{maroon}{-{5+mover 2} x} =color{maroon}{-2 hx} Longrightarrow h={5+mover 4}
$$
Now use equations $(2)$ and $(3)$ to find the value of $k$:
$$textstyle
color{darkgreen}{{3+mover2}}= color{darkgreen}{h^2+k} Longrightarrow
k={3+mover 2}-h^2 Longrightarrow k ={3+mover2}-({5+mover 4})^2
$$
So
$$
x^2-{textstyle{5+mover2}} x+{textstyle{3+mover2}}=(x-h)^2+k=
bigl( x-{textstyle {5+mover4}}bigr)^2 + {textstyle{3+mover2}-({5+mover 4}})^2
$$
and
$$eqalign{
2x^2-x(5+m)+(3+m)
&=2bigl(
{x^2-{textstyle{5+mover2}} x+{textstyle{3+mover2}} }bigr) cr
&=2Bigl(bigl( x-{textstyle {5+mover4}}bigr)^2 + {textstyle{3+mover2}-({5+mover 4}})^2 Bigr)cr
&=2bigl( x-{textstyle {5+mover4}}bigr)^2 + {3+m }-2bigl({textstyle{5+mover 4}}bigr)^2. cr
}
$$
answered Apr 12 '12 at 14:12
David MitraDavid Mitra
63.8k6102165
$endgroup$
add a comment |
$begingroup$
In general: $$ax^2+bx+c=a(x+frac{b}{2a})^2+c-frac{b^2}{4a}$$
answered Apr 12 '12 at 13:38
chemengchemeng
1,6831112
$endgroup$
$begingroup$
Thanks, I am still confused about what to do with $m$ in my equation?
$endgroup$
– Sean87
Apr 12 '12 at 13:52
$begingroup$
actually you don't need completing squares to solve this equation. You compute the discriminant: $D=b^2-4ac=(-5-m)^2-4*2*(3+m)=m^2+2m+1=(m+1)^2 geq 0$. Therefore, the solutions of your equation are $x_{1,2}=frac{-b pm sqrt(D)}{2a}=frac{5+m pm (m+1)}{4}$
$endgroup$
– chemeng
Apr 12 '12 at 14:03
add a comment |
$begingroup$
$$x^2-frac{(5+m)}{2}x+frac{3+m}{2}=0$$
$$x^2-2frac{(5+m)}{4}x+frac{3+m}{2}=0$$
$$x^2-2left(frac{5+m}{4}right)x+left(frac{5+m}{4}right)^2-left(frac{5+m}{4}right)^2+frac{3+m}{2}=0$$
$$left(x-frac{5+m}{4}right)^2-left(frac{(5+m)^2}{16}-frac{3+m}{2}right)=0$$
answered Apr 12 '12 at 14:04
Peđa TerzićPeđa Terzić
7,67922674
$endgroup$
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
$begingroup$
To complete the square on the left hand side of your equation, first factor out the $2$:
$$tag{1}
2x^2-x(5+m)+(3+m) =2bigl(
{x^2-{textstyle{5+mover2}} x+{textstyle{3+mover2}} }bigr).
$$
Write down what you want:
$$tag{2}
{x^2color{maroon}{-textstyle{5+mover2} x}+textstylecolor{darkgreen}{3+mover2}} = (x-h)^2+k.
$$
Write the right hand side of $(2)$ as
$$tag{3}
x^2color{maroon}{-2 h x}+color{darkgreen}{h^2+k}.
$$
Looking at the $color{maroon}x$ term of the left hand side of $(2)$ with the $color{maroon}x$ term of $(3)$, write down an equation and solve for $h$:
$$textstyle
color{maroon}{-{5+mover 2} x} =color{maroon}{-2 hx} Longrightarrow h={5+mover 4}
$$
Now use equations $(2)$ and $(3)$ to find the value of $k$:
$$textstyle
color{darkgreen}{{3+mover2}}= color{darkgreen}{h^2+k} Longrightarrow
k={3+mover 2}-h^2 Longrightarrow k ={3+mover2}-({5+mover 4})^2
$$
So
$$
x^2-{textstyle{5+mover2}} x+{textstyle{3+mover2}}=(x-h)^2+k=
bigl( x-{textstyle {5+mover4}}bigr)^2 + {textstyle{3+mover2}-({5+mover 4}})^2
$$
and
$$eqalign{
2x^2-x(5+m)+(3+m)
&=2bigl(
{x^2-{textstyle{5+mover2}} x+{textstyle{3+mover2}} }bigr) cr
&=2Bigl(bigl( x-{textstyle {5+mover4}}bigr)^2 + {textstyle{3+mover2}-({5+mover 4}})^2 Bigr)cr
&=2bigl( x-{textstyle {5+mover4}}bigr)^2 + {3+m }-2bigl({textstyle{5+mover 4}}bigr)^2. cr
}
$$
answered Apr 12 '12 at 14:12
David MitraDavid Mitra
63.8k6102165
$endgroup$
add a comment |
$begingroup$
To complete the square on the left hand side of your equation, first factor out the $2$:
$$tag{1}
2x^2-x(5+m)+(3+m) =2bigl(
{x^2-{textstyle{5+mover2}} x+{textstyle{3+mover2}} }bigr).
$$
Write down what you want:
$$tag{2}
{x^2color{maroon}{-textstyle{5+mover2} x}+textstylecolor{darkgreen}{3+mover2}} = (x-h)^2+k.
$$
Write the right hand side of $(2)$ as
$$tag{3}
x^2color{maroon}{-2 h x}+color{darkgreen}{h^2+k}.
$$
Looking at the $color{maroon}x$ term of the left hand side of $(2)$ with the $color{maroon}x$ term of $(3)$, write down an equation and solve for $h$:
$$textstyle
color{maroon}{-{5+mover 2} x} =color{maroon}{-2 hx} Longrightarrow h={5+mover 4}
$$
Now use equations $(2)$ and $(3)$ to find the value of $k$:
$$textstyle
color{darkgreen}{{3+mover2}}= color{darkgreen}{h^2+k} Longrightarrow
k={3+mover 2}-h^2 Longrightarrow k ={3+mover2}-({5+mover 4})^2
$$
So
$$
x^2-{textstyle{5+mover2}} x+{textstyle{3+mover2}}=(x-h)^2+k=
bigl( x-{textstyle {5+mover4}}bigr)^2 + {textstyle{3+mover2}-({5+mover 4}})^2
$$
and
$$eqalign{
2x^2-x(5+m)+(3+m)
&=2bigl(
{x^2-{textstyle{5+mover2}} x+{textstyle{3+mover2}} }bigr) cr
&=2Bigl(bigl( x-{textstyle {5+mover4}}bigr)^2 + {textstyle{3+mover2}-({5+mover 4}})^2 Bigr)cr
&=2bigl( x-{textstyle {5+mover4}}bigr)^2 + {3+m }-2bigl({textstyle{5+mover 4}}bigr)^2. cr
}
$$
answered Apr 12 '12 at 14:12
David MitraDavid Mitra
63.8k6102165
$endgroup$
add a comment |
$begingroup$
To complete the square on the left hand side of your equation, first factor out the $2$:
$$tag{1}
2x^2-x(5+m)+(3+m) =2bigl(
{x^2-{textstyle{5+mover2}} x+{textstyle{3+mover2}} }bigr).
$$
Write down what you want:
$$tag{2}
{x^2color{maroon}{-textstyle{5+mover2} x}+textstylecolor{darkgreen}{3+mover2}} = (x-h)^2+k.
$$
Write the right hand side of $(2)$ as
$$tag{3}
x^2color{maroon}{-2 h x}+color{darkgreen}{h^2+k}.
$$
Looking at the $color{maroon}x$ term of the left hand side of $(2)$ with the $color{maroon}x$ term of $(3)$, write down an equation and solve for $h$:
$$textstyle
color{maroon}{-{5+mover 2} x} =color{maroon}{-2 hx} Longrightarrow h={5+mover 4}
$$
Now use equations $(2)$ and $(3)$ to find the value of $k$:
$$textstyle
color{darkgreen}{{3+mover2}}= color{darkgreen}{h^2+k} Longrightarrow
k={3+mover 2}-h^2 Longrightarrow k ={3+mover2}-({5+mover 4})^2
$$
So
$$
x^2-{textstyle{5+mover2}} x+{textstyle{3+mover2}}=(x-h)^2+k=
bigl( x-{textstyle {5+mover4}}bigr)^2 + {textstyle{3+mover2}-({5+mover 4}})^2
$$
and
$$eqalign{
2x^2-x(5+m)+(3+m)
&=2bigl(
{x^2-{textstyle{5+mover2}} x+{textstyle{3+mover2}} }bigr) cr
&=2Bigl(bigl( x-{textstyle {5+mover4}}bigr)^2 + {textstyle{3+mover2}-({5+mover 4}})^2 Bigr)cr
&=2bigl( x-{textstyle {5+mover4}}bigr)^2 + {3+m }-2bigl({textstyle{5+mover 4}}bigr)^2. cr
}
$$
answered Apr 12 '12 at 14:12
David MitraDavid Mitra
63.8k6102165
$endgroup$
To complete the square on the left hand side of your equation, first factor out the $2$:
$$tag{1}
2x^2-x(5+m)+(3+m) =2bigl(
{x^2-{textstyle{5+mover2}} x+{textstyle{3+mover2}} }bigr).
$$
Write down what you want:
$$tag{2}
{x^2color{maroon}{-textstyle{5+mover2} x}+textstylecolor{darkgreen}{3+mover2}} = (x-h)^2+k.
$$
Write the right hand side of $(2)$ as
$$tag{3}
x^2color{maroon}{-2 h x}+color{darkgreen}{h^2+k}.
$$
Looking at the $color{maroon}x$ term of the left hand side of $(2)$ with the $color{maroon}x$ term of $(3)$, write down an equation and solve for $h$:
$$textstyle
color{maroon}{-{5+mover 2} x} =color{maroon}{-2 hx} Longrightarrow h={5+mover 4}
$$
Now use equations $(2)$ and $(3)$ to find the value of $k$:
$$textstyle
color{darkgreen}{{3+mover2}}= color{darkgreen}{h^2+k} Longrightarrow
k={3+mover 2}-h^2 Longrightarrow k ={3+mover2}-({5+mover 4})^2
$$
So
$$
x^2-{textstyle{5+mover2}} x+{textstyle{3+mover2}}=(x-h)^2+k=
bigl( x-{textstyle {5+mover4}}bigr)^2 + {textstyle{3+mover2}-({5+mover 4}})^2
$$
and
$$eqalign{
2x^2-x(5+m)+(3+m)
&=2bigl(
{x^2-{textstyle{5+mover2}} x+{textstyle{3+mover2}} }bigr) cr
&=2Bigl(bigl( x-{textstyle {5+mover4}}bigr)^2 + {textstyle{3+mover2}-({5+mover 4}})^2 Bigr)cr
&=2bigl( x-{textstyle {5+mover4}}bigr)^2 + {3+m }-2bigl({textstyle{5+mover 4}}bigr)^2. cr
}
$$
answered Apr 12 '12 at 14:12
David MitraDavid Mitra
63.8k6102165
answered Apr 12 '12 at 14:12
David MitraDavid Mitra
63.8k6102165
answered Apr 12 '12 at 14:12
David MitraDavid Mitra
63.8k6102165
answered Apr 12 '12 at 14:12
David MitraDavid Mitra
63.8k6102165
63.8k6102165
add a comment |
add a comment |
$begingroup$
In general: $$ax^2+bx+c=a(x+frac{b}{2a})^2+c-frac{b^2}{4a}$$
answered Apr 12 '12 at 13:38
chemengchemeng
1,6831112
$endgroup$
$begingroup$
Thanks, I am still confused about what to do with $m$ in my equation?
$endgroup$
– Sean87
Apr 12 '12 at 13:52
$begingroup$
actually you don't need completing squares to solve this equation. You compute the discriminant: $D=b^2-4ac=(-5-m)^2-4*2*(3+m)=m^2+2m+1=(m+1)^2 geq 0$. Therefore, the solutions of your equation are $x_{1,2}=frac{-b pm sqrt(D)}{2a}=frac{5+m pm (m+1)}{4}$
$endgroup$
– chemeng
Apr 12 '12 at 14:03
add a comment |
$begingroup$
In general: $$ax^2+bx+c=a(x+frac{b}{2a})^2+c-frac{b^2}{4a}$$
answered Apr 12 '12 at 13:38
chemengchemeng
1,6831112
$endgroup$
$begingroup$
Thanks, I am still confused about what to do with $m$ in my equation?
$endgroup$
– Sean87
Apr 12 '12 at 13:52
$begingroup$
actually you don't need completing squares to solve this equation. You compute the discriminant: $D=b^2-4ac=(-5-m)^2-4*2*(3+m)=m^2+2m+1=(m+1)^2 geq 0$. Therefore, the solutions of your equation are $x_{1,2}=frac{-b pm sqrt(D)}{2a}=frac{5+m pm (m+1)}{4}$
$endgroup$
– chemeng
Apr 12 '12 at 14:03
add a comment |
$begingroup$
In general: $$ax^2+bx+c=a(x+frac{b}{2a})^2+c-frac{b^2}{4a}$$
answered Apr 12 '12 at 13:38
chemengchemeng
1,6831112
$endgroup$
In general: $$ax^2+bx+c=a(x+frac{b}{2a})^2+c-frac{b^2}{4a}$$
answered Apr 12 '12 at 13:38
chemengchemeng
1,6831112
answered Apr 12 '12 at 13:38
chemengchemeng
1,6831112
answered Apr 12 '12 at 13:38
chemengchemeng
1,6831112
answered Apr 12 '12 at 13:38
chemengchemeng
1,6831112
1,6831112
$begingroup$
Thanks, I am still confused about what to do with $m$ in my equation?
$endgroup$
– Sean87
Apr 12 '12 at 13:52
$begingroup$
actually you don't need completing squares to solve this equation. You compute the discriminant: $D=b^2-4ac=(-5-m)^2-4*2*(3+m)=m^2+2m+1=(m+1)^2 geq 0$. Therefore, the solutions of your equation are $x_{1,2}=frac{-b pm sqrt(D)}{2a}=frac{5+m pm (m+1)}{4}$
$endgroup$
– chemeng
Apr 12 '12 at 14:03
add a comment |
$begingroup$
Thanks, I am still confused about what to do with $m$ in my equation?
$endgroup$
– Sean87
Apr 12 '12 at 13:52
$begingroup$
actually you don't need completing squares to solve this equation. You compute the discriminant: $D=b^2-4ac=(-5-m)^2-4*2*(3+m)=m^2+2m+1=(m+1)^2 geq 0$. Therefore, the solutions of your equation are $x_{1,2}=frac{-b pm sqrt(D)}{2a}=frac{5+m pm (m+1)}{4}$
$endgroup$
– chemeng
Apr 12 '12 at 14:03
$begingroup$
Thanks, I am still confused about what to do with $m$ in my equation?
$endgroup$
– Sean87
Apr 12 '12 at 13:52
$begingroup$
Thanks, I am still confused about what to do with $m$ in my equation?
$endgroup$
– Sean87
Apr 12 '12 at 13:52
$begingroup$
actually you don't need completing squares to solve this equation. You compute the discriminant: $D=b^2-4ac=(-5-m)^2-4*2*(3+m)=m^2+2m+1=(m+1)^2 geq 0$. Therefore, the solutions of your equation are $x_{1,2}=frac{-b pm sqrt(D)}{2a}=frac{5+m pm (m+1)}{4}$
$endgroup$
– chemeng
Apr 12 '12 at 14:03
$begingroup$
actually you don't need completing squares to solve this equation. You compute the discriminant: $D=b^2-4ac=(-5-m)^2-4*2*(3+m)=m^2+2m+1=(m+1)^2 geq 0$. Therefore, the solutions of your equation are $x_{1,2}=frac{-b pm sqrt(D)}{2a}=frac{5+m pm (m+1)}{4}$
$endgroup$
– chemeng
Apr 12 '12 at 14:03
add a comment |
$begingroup$
$$x^2-frac{(5+m)}{2}x+frac{3+m}{2}=0$$
$$x^2-2frac{(5+m)}{4}x+frac{3+m}{2}=0$$
$$x^2-2left(frac{5+m}{4}right)x+left(frac{5+m}{4}right)^2-left(frac{5+m}{4}right)^2+frac{3+m}{2}=0$$
$$left(x-frac{5+m}{4}right)^2-left(frac{(5+m)^2}{16}-frac{3+m}{2}right)=0$$
answered Apr 12 '12 at 14:04
Peđa TerzićPeđa Terzić
7,67922674
$endgroup$
add a comment |
$begingroup$
$$x^2-frac{(5+m)}{2}x+frac{3+m}{2}=0$$
$$x^2-2frac{(5+m)}{4}x+frac{3+m}{2}=0$$
$$x^2-2left(frac{5+m}{4}right)x+left(frac{5+m}{4}right)^2-left(frac{5+m}{4}right)^2+frac{3+m}{2}=0$$
$$left(x-frac{5+m}{4}right)^2-left(frac{(5+m)^2}{16}-frac{3+m}{2}right)=0$$
answered Apr 12 '12 at 14:04
Peđa TerzićPeđa Terzić
7,67922674
$endgroup$
add a comment |
$begingroup$
$$x^2-frac{(5+m)}{2}x+frac{3+m}{2}=0$$
$$x^2-2frac{(5+m)}{4}x+frac{3+m}{2}=0$$
$$x^2-2left(frac{5+m}{4}right)x+left(frac{5+m}{4}right)^2-left(frac{5+m}{4}right)^2+frac{3+m}{2}=0$$
$$left(x-frac{5+m}{4}right)^2-left(frac{(5+m)^2}{16}-frac{3+m}{2}right)=0$$
answered Apr 12 '12 at 14:04
Peđa TerzićPeđa Terzić
7,67922674
$endgroup$
$$x^2-frac{(5+m)}{2}x+frac{3+m}{2}=0$$
$$x^2-2frac{(5+m)}{4}x+frac{3+m}{2}=0$$
$$x^2-2left(frac{5+m}{4}right)x+left(frac{5+m}{4}right)^2-left(frac{5+m}{4}right)^2+frac{3+m}{2}=0$$
$$left(x-frac{5+m}{4}right)^2-left(frac{(5+m)^2}{16}-frac{3+m}{2}right)=0$$
answered Apr 12 '12 at 14:04
Peđa TerzićPeđa Terzić
7,67922674
answered Apr 12 '12 at 14:04
Peđa TerzićPeđa Terzić
7,67922674
answered Apr 12 '12 at 14:04
Peđa TerzićPeđa Terzić
7,67922674
answered Apr 12 '12 at 14:04
Peđa TerzićPeđa Terzić
7,67922674
7,67922674
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$begingroup$
Look at the second answer on this question math.stackexchange.com/questions/20597/…"
$endgroup$
– Chris K. Caldwell
Apr 12 '12 at 13:38
$begingroup$
If you are trying to solve the equation, note that by a miracle (?) the left side factors as $(x-1)(2x-(3+m))$. Or else first note that the original equation has $x=1$ as an obvious root. But you say you want to find the values of $m$. That cannot be done just given the equation. Perhaps (in a separate post) you might ask the real question.
$endgroup$
– André Nicolas
Apr 12 '12 at 14:41