- Thread starter
- #1

#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

$a^2=3a+5$

$b^2=3b+5$

$a\neq b$

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

$b^2=3b+5$

$a\neq b$

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

- Thread starter Albert
- Start date

- Thread starter
- #1

- Jan 25, 2013

- 1,225

$a^2=3a+5$

$b^2=3b+5$

$a\neq b$

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

$b^2=3b+5$

$a\neq b$

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

- Admin
- #2

- Feb 14, 2012

- 3,909

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

\(\displaystyle a=\frac{3+ \sqrt{29}}{2}\) and \(\displaystyle b=\frac{3- \sqrt{29}}{2}\).

Thus, \(\displaystyle a+b=3\) and \(\displaystyle ab=-5\).

Therefore,

\(\displaystyle \frac{b^2}{2a}+\frac{a^2}{2b}\)

\(\displaystyle =\frac{b^3}{2ab}+\frac{a^3}{2ab}\)

\(\displaystyle =\frac{a^3+b^3}{2ab}\)

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

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

\(\displaystyle =-7.2\)

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