Basic Algebra: square numbers

Question

I am new to calculus. Can you help me with this?

$a+\sqrt{a}=4$

$5a+a\sqrt{a}=? $

Answer 1

$$a+\sqrt{a}=4\iff \sqrt a=4-a\underbrace{\implies}_{\mathrm{squaring}} a=a^2-8a+16.$$

$$a+\sqrt{a}\implies a=4-\sqrt a\underbrace{\implies}_{\times a} \color{red}{ a^2=4a-a\sqrt a}.$$ Substitute $a^2$ in the first equality and get

$$a=a^2-8a+16=\color{red}{4a-a\sqrt{a}}-8a+16.$$ This is equivalent to say $$5a+a\sqrt{a}=16$$ and you are done.

Answer 2

Set $a=x^2$, $x=\sqrt a$ to obtain basic quadratic equations.