Question

Find the real roots of : 63/x+72/x+6-3=0

Answer 1

$\textbf{Given equation is}$

$\displaystyle\frac{63}{x}+\frac{72}{x+6}-3=0$

$\text{This can be written as}$

$\displaystyle\frac{63}{x}+\frac{72}{x+6}=3$

$\displaystyle\frac{21}{x}+\frac{24}{x+6}=1$

$\displaystyle\frac{21(x+6)+24x}{x(x+6)}=1$

$\displaystyle\frac{21x+126+24x}{x^2+6x}=1$

$\displaystyle\,126+45x=x^2+6x$

$\implies\displaystyle\,x^2-39x-126=0$

$\text{Spilitting the middle term, we get}$

$\displaystyle\,x^2-42x+3x-126=0$

$\displaystyle\,x(x-42)+3(x-42)=0$

$\displaystyle\,(x-42)(x+3)=0$

$\implies\displaystyle\,x=42,-3$

$\therefore\textbf{The real roots of the given equations are 42 and -3}$

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2.Find the roots of the following quadratic (if they exist) by the method of completing the square.

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Answer 2

The real roots are "x = 42" and "x = -3".

Step-by-step explanation:

$The \ given \ expression \ is:\\\\\frac{63}{x}+\frac{72}{x+6}-3=0\\\\ \frac{63}{x} +\frac{72}{x+6} =3\\\\\frac{21}{x}+\frac{36}{x+6}=1 \\\\On \ taking \ LCM,we \ get\\\\\frac{21(x+6)+36x}{x(x+6)} =1\\\\\frac{21x+126+36x}{x^2+6x} =1\\\\On \ applying \ cross \ multiplication,we \ get\\\\21x+36x+126=x^2+6x\\\\x^2+6x-21x-36x-126=0\\\\x^2-51x-126=0\\\\x^2-42x+3x-126=0\\\\x(x-42)+3(x-42)=0\\\\(x-42)(x+3)=0\\\\x-42=0\\\\x=42\\\\Or,\\\\x+3=0\\\\x=-3$

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