Question

split in middle term x^2-x+1/4
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Answer 1

It's easier to split the middle term -x as,

$\displaystyle\longrightarrow\sf{-x=(u+v)x+(u-v)x}$

$\displaystyle\longrightarrow\sf{-1=u+v+u-v}$

$\displaystyle\longrightarrow\sf{u=-\dfrac{1}{2}}$

We know the middle term should be split into two in such a way that their product must be equal to the term in $\displaystyle\sf{x^2}$ whose coefficient is equal to the product of the coefficient of $\displaystyle\sf{x^2}$ in the given polynomial and the constant term. Thus,

$\displaystyle\longrightarrow\sf{[(u+v)x][(u-v)x]=\dfrac{1}{4}x^2}$

$\displaystyle\longrightarrow\sf{(u+v)(u-v)=\dfrac{1}{4}}$

$\displaystyle\longrightarrow\sf{u^2-v^2=\dfrac{1}{4}}$

$\displaystyle\longrightarrow\sf{\left(-\dfrac{1}{2}\right)^2-v^2=\dfrac{1}{4}}$

$\displaystyle\longrightarrow\sf{\dfrac{1}{4}-v^2=\dfrac{1}{4}}$

$\displaystyle\longrightarrow\sf{v=0}$

Therefore,

$\displaystyle\longrightarrow\sf{u+v=-\dfrac{1}{2}}$

$\displaystyle\longrightarrow\sf{u-v=-\dfrac{1}{2}}$

Hence the middle term should be split as,

$\displaystyle\longrightarrow\sf{\underline{\underline{-x=-\dfrac{1}{2}x-\dfrac{1}{2}x}}}$

Let me factorise the polynomial.

$\displaystyle\longrightarrow\sf{x^2-x+\dfrac{1}{4}=x^2-\dfrac{1}{2}x-\dfrac{1}{2}x+\dfrac{1}{4}}$

$\displaystyle\longrightarrow\sf{x^2-x+\dfrac{1}{4}=x\left(x-\dfrac{1}{2}\right)-\dfrac{1}{2}\left(x-\dfrac{1}{2}\right)}$

$\displaystyle\longrightarrow\sf{\underline{\underline{x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2}}}$

Answer 2

Answer:

solution is in the attachment

Step-by-step explanation: