Let u and v be the solutions to 3x^2 + 5x + 7 = 0. Find $\frac{1}{u^2} + \frac{1}{v^2}$.
+14913 Find \(\dfrac{1}{u^2} + \dfrac{1}{v^2}.\)
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\(3x^2 + 5x + 7 = 0\\ x^2+\frac{5}{3}x+\frac{7}{3}=0\\ x=-\frac{5}{6}\pm \sqrt{\frac{25}{36}-\frac{7}{3}}\\ x=-\frac{5}{6}\pm \sqrt{\frac{25}{36}-\frac{84}{36}}\\\)
\(u=\dfrac{-5+i\sqrt{59}}{6}\\ v=\dfrac{-5-i\sqrt{59}}{6}\\\)
\(\dfrac{1}{u^2}+\dfrac{1}{v^2}=\dfrac{36}{ ( -5+i\sqrt{59})^2}+\dfrac{36}{ ( -5-i\sqrt{59})^2}=-\dfrac{17}{49}\)
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