Question

The transformed equation of $x^3-\frac{1}{4}{x}^2+\frac{1}{3}x-\frac{1}{144}=0$ by removing fractional coefficients, is:

• A. $x^3-3x^2+48x-12=0$
• B. $x^3+3x^2+48x+12=0$
• C. $x^3-3x^2+48x+12=0$
• D. $x^3+3x^2+48x-12=0$

Answer 1

The correct option is A $x^3-3x^2+48x-12=0$

The given equation is $x^3-\frac{1}{4}{x}^2+\frac{1}{3}x-\frac{1}{144}=0$
Multiplying the roots of the given equation by $m$, the transformed equation is
$x^3-\frac{1}{4}m{x}^2+\frac{1}{3}{m}^{2}x-\frac{1}{144}{m}^3=0$

$\Rightarrow x^3-\frac{1}{2^2}m{x}^2+\frac{1}{3}{m}^{2}x-\frac{1}{2^4.3^2}{m}^3=0$
Clearly the least value of m to remove the fractional coefficient is $m=2^2.3=12$
Therefore putting this, we get

$x^3-\frac{1}{2^2}\left(2^2.3\right)x^2+\frac{1}{3}{\left(2^2.3\right)}^{2}x-\frac{1}{2^4.3^2}{\left(2^2.3\right)}^3=0\Rightarrow x^3-3x^2+48x-12=0$