Question

Factorise:x^2-x+1/24

Answer 1

$\textbf{Given:}$

$x^2-x+\frac{1}{24}$

$\textbf{To find:}$

$\text{Factors of x^2-x+\frac{1}{24} }$

$\textbf{Solution:}$

$\text{Consider,}$

$x^2-x+\frac{1}{24}$

$\text{First we make part of the polynomial as a perfect square}$

$=(x^2-x+\frac{1}{4})-\frac{1}{4}+\frac{1}{24}$

$=(x-\frac{1}{2})^2-\frac{5}{24}$

$=(x-\frac{1}{2})^2-(\frac{\sqrt{5}}{\sqrt{24}})^2$

$\text{Using the identity,}\boxed{\bf\,a^2-b^2=(a+b)(a-b)}$

$\text{we get}$

$=(x-\frac{1}{2}+\frac{\sqrt{5}}{\sqrt{24}})(x-\frac{1}{2}-\frac{\sqrt{5}}{\sqrt{24}})$

$=(x-\frac{1}{2}+\frac{\sqrt{5}}{2\sqrt{6}})(x-\frac{1}{2}-\frac{\sqrt{5}}{2\sqrt{6}})$

$\therefore\textbf{The factors of \bf\,x^2-x+\frac{1}{24} are \bf\,(x-\frac{1}{2}+\frac{\sqrt{5}}{2\sqrt{6}})(x-\frac{1}{2}-\frac{\sqrt{5}}{2\sqrt{6}}) }$

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Answer 2

The roots of the equation are ($x=\frac{1-\sqrt{\frac{5}{6} } }{2}$ , $x=\frac{1+\sqrt{\frac{5}{6} } }{2}$ ).

Step-by-step explanation:

We are given with the following equation to factorize;

$x^{2} -x+(\frac{1}{24})$

To factorize the above equation, we will use the $D=b^{2} -4ac$ method;

In the above equation; a = 1 , b = (-1) and c = $\frac{1}{24}$.

$D=(-1)^{2} -4\times (1)\times (\frac{1}{24})$

     =  $1-\frac{1}{6}$

     =  $\frac{6-1}{6}=\frac{5}{6}$

Now, the roots of the equation are given by;

$x=\frac{-b-\sqrt{D} }{2a}$            and        $x=\frac{-b+\sqrt{D} }{2a}$  

$x=\frac{-(-1)-\sqrt{\frac{5}{6} } }{2\times 1}$       and        $x=\frac{-(-1)+\sqrt{\frac{5}{6} } }{2\times 1}$

$x=\frac{1-\sqrt{\frac{5}{6} } }{2}$             and        $x=\frac{1+\sqrt{\frac{5}{6} } }{2}$

Hence, the roots of the equation are ($x=\frac{1-\sqrt{\frac{5}{6} } }{2}$ , $x=\frac{1+\sqrt{\frac{5}{6} } }{2}$ ).