Solving for $y$, given $x = \frac{-1 + \sqrt{1+ (4y/50)} }{2}$

Question

I have the following quadratic equation:

$$x = \frac{-1 + \sqrt{1+ (4y/50)} }{2}$$

in this case $y$ is a known variable so I can solve the equation like this for $y = 600$

$$x = \frac{-1 + \sqrt{1+ (4\times 600/50)} }{2}$$

and I was looking for help on how to refactor it so I can solve it while knowing $x$ instead of $y$, but I have no idea how to do it.

I would be looking for something like $y = x$???

Thanks!

Answer 1

It's simple algebra. $$y=\frac{50(2x+1)^2-50}{4}=50x(x+1)$$

Answer 2

You probably know the quadratic formula: $x_{1/2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

This is the formula for the quadratic function $f=ax^2+bx+c$. In this case we have $a=1$ and $b=1$. Comparing $4ac=4c$ with $-\frac{4y}{50}$ gives $c=-\frac{y}{50}$. Thus $f=x^2+x-\frac{y}{50}=0$.

Finally you can solve for y.

Answer 3

First see how the right hand expression is built, starting from $y$, using simple arithmetic operations

$$y\xrightarrow[\times \tfrac{4}{50}]{}\frac{4y}{50}\xrightarrow[+1]{}1+\frac{4y}{50}\xrightarrow[\sqrt{{}}]{}\sqrt{1+\frac{4y}{50}}\xrightarrow[-1]{}-1+\sqrt{1+\frac{4y}{50}}\xrightarrow[\div 2]{}\tfrac{-1+\sqrt{1+\frac{4y}{50}}}{2}=x$$

Now, starting from $x$, undo each of these 5 operations to get $y$

$$x\xrightarrow[\times 2]{{}}2x\xrightarrow[+1]{{}}2x+1\xrightarrow[{{()}^{2}}]{{}}{{(2x+1)}^{2}}\xrightarrow[-1]{{}}{{(2x+1)}^{2}}-1\xrightarrow[\times \tfrac{50}{4}]{{}}\frac{50}{4}\left[ {{(2x+1)}^{2}}-1 \right]=y $$