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

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The value of (b^2/2a) + (a^2/2b) is calculated using the quadratic equations a^2 = 3a + 5 and b^2 = 3b + 5, with the condition a ≠ b. The roots of these equations are a = (3 + √29)/2 and b = (3 - √29)/2, leading to a sum of a + b = 3 and a product ab = -5. By substituting these values into the expression, it simplifies to (a^3 + b^3) / (2ab), which further reduces to -7.2. The calculations confirm that the final result is -7.2, with a and b being interchangeable.
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$a^2=3a+5$
$b^2=3b+5$
$a\neq b$
$\dfrac {b^2}{2a}+\dfrac {a^2}{2b}=?$
 
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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.
 

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