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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 somone please describe this method for this particular equation?!

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Look at the second answer on this question math.stackexchange.com/questions/20597/…; –  Chris K. Caldwell Apr 12 '12 at 13:38
    
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. –  André Nicolas Apr 12 '12 at 14:41
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3 Answers 3

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To complete the square on the left hand side of your equation, first factor out the $2$: $$\tag{1} 2x^2-x(5+m)+(3+m) =2\bigl( {x^2-{\textstyle{5+m\over2}} x+{\textstyle{3+m\over2}} }\bigr). $$ Write down what you want: $$\tag{2} {x^2\color{maroon}{-\textstyle{5+m\over2} x}+\textstyle\color{darkgreen}{3+m\over2}} = (x-h)^2+k. $$ Write the right hand side of $(2)$ as $$\tag{3} x^2\color{maroon}{-2 h x}+\color{darkgreen}{h^2+k}. $$ 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 $h$: $$\textstyle \color{maroon}{-{5+m\over 2} x} =\color{maroon}{-2 hx}\ \Longrightarrow\ h={5+m\over 4} $$ Now use equations $(2)$ and $(3)$ to find the value of $k$: $$\textstyle \color{darkgreen}{{3+m\over2}}= \color{darkgreen}{h^2+k}\ \Longrightarrow\ k={3+m\over 2}-h^2\ \Longrightarrow k ={3+m\over2}-({5+m\over 4})^2 $$

So $$ x^2-{\textstyle{5+m\over2}} x+{\textstyle{3+m\over2}}=(x-h)^2+k= \bigl( x-{\textstyle {5+m\over4}}\bigr)^2 + {\textstyle{3+m\over2}-({5+m\over 4}})^2 $$ and $$\eqalign{ 2x^2-x(5+m)+(3+m) &=2\bigl( {x^2-{\textstyle{5+m\over2}} x+{\textstyle{3+m\over2}} }\bigr) \cr &=2\Bigl(\bigl( x-{\textstyle {5+m\over4}}\bigr)^2 + {\textstyle{3+m\over2}-({5+m\over 4}})^2 \Bigr)\cr &=2\bigl( x-{\textstyle {5+m\over4}}\bigr)^2 + {3+m }-2\bigl({\textstyle{5+m\over 4}}\bigr)^2. \cr } $$

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In general: $$ax^2+bx+c=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}$$

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Thanks, I am still confused about what to do with $m$ in my equation? –  Sean87 Apr 12 '12 at 13:52
    
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 $x_{1,2}=\frac{-b \pm \sqrt(D)}{2a}=\frac{5+m \pm (m+1)}{4}$ –  chemeng Apr 12 '12 at 14:03
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$$x^2-\frac{(5+m)}{2}x+\frac{3+m}{2}=0$$

$$x^2-2\frac{(5+m)}{4}x+\frac{3+m}{2}=0$$

$$x^2-2\left(\frac{5+m}{4}\right)x+\left(\frac{5+m}{4}\right)^2-\left(\frac{5+m}{4}\right)^2+\frac{3+m}{2}=0$$

$$\left(x-\frac{5+m}{4}\right)^2-\left(\frac{(5+m)^2}{16}-\frac{3+m}{2}\right)=0$$

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