+592 3x2 + 4x + 12 = x2 - 5x + 7
2x2 + 9x + 5 = 0
x2 + \(\frac{9}{2}\)x + \(\frac{5}{2}\) = 0
x2 + \(\frac{9}{2}\)x + \(\frac{81}{16}\) - \(\frac{81}{16}\) + \(\frac{5}{2}\) = 0
(x + \(\frac{9}{4}\) )2 - \(\frac{81}{16}\)+ \(\frac{40}{16}\) = 0
(x + \(\frac{9}{4}\) )2 = \(\frac{41}{16}\)
x + \(\frac{9}{4}\) = ± \(\frac{\sqrt{41}}{4}\)
x = ± \(\frac{\sqrt{41} - 9}{4}\)
r = \(\frac{\sqrt{41} - 9}{4}\)
s = \(\frac{9 - \sqrt{41}}{4}\)
r2 + s2 = \((\frac{\sqrt{41} - 9}{4})^2 + (\frac{9 - \sqrt{41}}{4})^2\)
= \(\frac{(\sqrt{41} - 9)^2 + (9 - \sqrt{41})^2}{16}\)
= \(\frac{122 - 18\sqrt{41} + 122 - 18\sqrt{41}}{16}\)
= \(\frac{2(122 - 18\sqrt{41})}{16}\)
= \(\frac{122 - 18\sqrt{41}}{8}\)
= \(\frac{61 - 9\sqrt{41}}{4}\)
tell me if I did anything wrong :)
+505 When you get to 2x^2 + 9x + 5 = 0, you could just use Vieta's formulas. The sum \(r+s=-\frac{b}{a}=-\frac{9}{2}\), which means that \((r+s)^2=r^2+2rs+s^2=(-\frac{9}{2})^2=\frac{81}{4}\). The product \(rs=\frac{c}{a}=\frac{5}{2}\), which means \(2rs=5\). Notice that \(r^2+2rs+s^2-2rs=r^2+s^2=\frac{81}{4}-\frac{20}{4}=\boxed{\frac{61}{4}}\)
