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?!
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?!
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 } $
In general: $ax^2+bx+c=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}$
$x^2-\frac{(5+m)}{2}x+\frac{3+m}{2}=0$
$x^2-2\frac{(5+m)}{4}x+\frac{3+m}{2}=0$
$x^2-2\left(\frac{5+m}{4}\right)x+\left(\frac{5+m}{4}\right)^2-\left(\frac{5+m}{4}\right)^2+\frac{3+m}{2}=0$
$\left(x-\frac{5+m}{4}\right)^2-\left(\frac{(5+m)^2}{16}-\frac{3+m}{2}\right)=0$