Question : Simplify the following equation. What is the difference between the two values of $x$?
$7 x+4\left\{x^2 \div(5 x \div 10)\right\}-3\left\{5 \frac{1}{3}-x^3 \div\left(3 x^2 \div x\right)\right\}=0$
Option 1: 8
Option 2: 16
Option 3: 5
Option 4: 17
Correct Answer: 17
Solution :
Given expression, $7 x+4\left\{x^2 \div(5 x \div 10)\right\}-3\left\{5 \frac{1}{3}-x^3 \div\left(3 x^2 \div x\right)\right\}=0$
⇒ $7x + 4\{\frac{x^2}{\frac{5x}{10}}\}-3\{\frac{16}{3}-\frac{x^3}{\frac{3x^2}{x}}\}=0$
⇒ $7x+4(\frac{x^2}{\frac{x}{2}})-3(\frac{16}{3}-\frac{x^3}{3x})=0$
⇒ $7x+4(2x)-3(\frac{16}{3}-\frac{x^2}{3})=0$
⇒ $7x+8x-(16-x^2)=0$
⇒ $x^2+15x-16=0$
⇒ $x^2 +x-16x-16=0$
⇒ $x(x+1)-16(x+1)=0$
⇒ $(x-16)(x+1)=0$
⇒ $x=16$ and $x=-1$
$\therefore$ Difference $=16-(-1) = 17$
Hence, the correct answer is 17.
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