solving $y=4/x+\sqrt{x+0.2−5x}$

Question

it's actually

$y=\frac{4}{x}+\sqrt{x+0.2-5x}$ (see algebra problem)

$$y=\frac{4}{x}+\sqrt{x+0.2-5x}$$ if $x=\frac45$ what is y?

Answer 1

Since $\frac{4}{5} = 0.8$, $$ y = \frac{4}{x} + \sqrt{x + 0.2} - 5x = \frac{4}{4/5} + \sqrt{0.8+0.2} - 5 \cdot \frac{4}{5} = 5 + \sqrt{1} - 4 = 2 $$

Edit: Since you've modified the question, $$ y = \frac{4}{x} + \sqrt{x + 0.2 - 5x} = \frac{4}{4/5} + \sqrt{0.8 + 0.2 - 4} = 5 \pm \sqrt{3}i $$

Answer 2

$f(x) = \frac{4}{x} + \sqrt{x+.2-5x}$

$f(4/5) = \frac{4}{4/5}+ \sqrt{\frac{4}{5}+0.2-5(\frac{4}{5})} = 5 + \sqrt{3}i$