Answer
$x=12cm$
Work Step by Step
We are asked to find $x$ which is the distance of the object from the lens.
We know that the focal length ($F$) is $4.8 cm$. And they told us that the distance of the image is 4cm closer to the lens than the object itself, that means $y=x-4$ because a shorter distance means a smaller number.
This is the formula:
$\frac{1}{F}=\frac{1}{x}+\frac{1}{y}$
Replace with the information we were given:
$\frac{1}{4.8}=\frac{1}{x}+\frac{1}{x-4}$
We find the LCD: $4.8x(x-4)$
$\frac{1}{4.8}\times4.8x(x-4)=(\frac{1}{x}+\frac{1}{x-4})\times4.8x(x-4)$
$x(x-4)=\frac{x-4+x}{x(x-4)}\times4.8x(x-4)$
$x^{2}-4x=(2x-4)\times4.8$
$x^{2}-4x=9.6x-19.2$
Let's move all the terms to the left side:
$x^{2}-4x-9.6x+19.2=0$
$x^{2}-13.6x+19.2=0$
Now we can use the quadratic formula to solve the equation:
$a=1$
$b=-13.6$
$c=19.2$
I'll solve it by parts:
$D=(-13.6)^{2}-4\times1\times19.2=184,96-76.8=108.16$
Since our Discriminant ($D$) is positive we now know we'll get two answers:
$x=\frac{13.6+-\sqrt 108.16}{2}$
$x=\frac{13.6+10.4}{2}$
$x=12$
and
$x=\frac{13.6-10.4}{2}$
$x=1.6$
As mentioned before the value of $y$ is 4 less than $x$, if we try it with the values we just calculated:
$y=x-4$
$y=1.6-4=-2.4$
$y=12-4=8$
We got $-2.4$ and $8$. Since we're calculating a distance, distance is never negative. Because of that we know for sure that $8$ is the real value of $y$. So $12$ is the value of $x$ that we're looking for.
