How do i solve (1000-x)^2/(x^2)

Holy h**l, that was hard to solve haha!

i manage to multiply 0.47x^2-2000x+1000000 with 100

then divide 47x^2-200000x+100 000 000 with 47

then i just solved it with pq- formula which gave me x = 3675 & x= 579 

 

Thanks alot for your help, couldnt solve it without you <3 

i assume (1000-x)^2 = 1 000 000 - x^2 ?

the equation says =>   0.53 = (1000-x)^2 / x^2

Not quite anon  :)

 

$${\left({\mathtt{1\,000}}{\mathtt{\,-\,}}{\mathtt{x}}\right)}^{{\mathtt{2}}} = \left(\left({\mathtt{1\,000}}{\mathtt{\,-\,}}{\mathtt{x}}\right){\mathtt{\,\times\,}}\left({\mathtt{1\,000}}{\mathtt{\,-\,}}{\mathtt{x}}\right)\right)$$

 

$${\mathtt{1\,000}}{\mathtt{\,\times\,}}\left({\mathtt{1\,000}}{\mathtt{\,-\,}}{\mathtt{x}}\right){\mathtt{\,-\,}}{x}{\left({\mathtt{1\,000}}{\mathtt{\,-\,}}{\mathtt{x}}\right)}$$

 

$$1000000-1000x-1000x+x^2$$

 

$$x^2-2000x+1000000$$

 

Now you cannot solve anything because there is no equal sign.

really you cannot even simplify because the numerator and the denominator have no common factors.

You can expand the top if you want but that is all you can do :)

Thank you for your answer, this is actully an chemist question but invloves math with 2nd grade equation..

 

i want to find out x

 

0.53 = (1000-x)*(1000-x)    /  x^2

 

this is the real euqation. But i struggle to find get rid of the x^2 in the denominator

 

would be so gratefull if you understand this :-)

$$\\0.53=\frac{(1000-x)^2}{x^2}\\\\
0.53x^2=(1000-x)^2\\\\
0.53x^2=1x^2-2000x+1000000\\\\
0=1x^2-0.53x^2-2000x+1000000\\\\
0=0.47x^2-2000x+1000000\\\\$$

 

now you can use the quadratic formula - I will use the site calc.

 

$${\mathtt{0.47}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2\,000}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,000\,000}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{10\,000}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{53}}}}{\mathtt{\,-\,}}{\mathtt{100\,000}}\right)}{{\mathtt{47}}}}\\
{\mathtt{x}} = {\frac{\left({\mathtt{10\,000}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{53}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{100\,000}}\right)}{{\mathtt{47}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{578.700\: \!023\: \!557\: \!336\: \!538\: \!1}}\\
{\mathtt{x}} = {\mathtt{3\,676.619\: \!125\: \!378\: \!833\: \!674\: \!7}}\\
\end{array} \right\}$$

so

$$x\approx 579\qquad or \qquad x\approx 3677$$

 

Does that help?

Thank you so much !

it did help alot and i got the correct answer  with 579.

But i struggle myself when trying to solve the equation with pq-formula since it has to be x^2+px-q ( cant be 0.47 x^2) could i possibly multiply so x^2 becomes integer ?

you can change it by multoplying everything by 100 if you want to.  :))

 

$$\\0=0.47x^2-2000x+1000000\\\\
0=47x^2-200000x+100000000\\\\$$