What is the value of (b^2/2a)+(a^2/2b) when a^2=3a+5, b^2=3b+5, and a≠b?

[Albert1]

$a^2=3a+5$
$b^2=3b+5$
$a\neq b$
$\dfrac {b^2}{2a}+\dfrac {a^2}{2b}=?$

 

[anemone]

Re: find :(b^2/2a)+(a^2/2b)

Since we're given two quadratic equations $$a^2=3a+5$$ and $$b^2=3b+5$$ and that $$a\ne b$$, we can tell by quadratic formula that

$$a=\frac{3+ \sqrt{29}}{2}$$ and $$b=\frac{3- \sqrt{29}}{2}$$.

Thus, $$a+b=3$$ and $$ab=-5$$.

Therefore,

$$\frac{b^2}{2a}+\frac{a^2}{2b}$$

$$=\frac{b^3}{2ab}+\frac{a^3}{2ab}$$

$$=\frac{a^3+b^3}{2ab}$$

$$=\frac{(a+b)(a^2+b^2-ab)}{2ab}$$

$$=\frac{(3)(3(3)+10-(-5))}{2(-5)}$$

$$=-7.2$$

P.S. The values for a and b are interchangeable.