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f. \( 4 n^{4}+8 n^{2}-9=0 \)

Ask by Sullivan Maxwell. in El Salvador
Feb 09,2025

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Las soluciones son \(n_{1}=-\frac{\sqrt{-4+2\sqrt{13}}}{2}\) y \(n_{2}=\frac{\sqrt{-4+2\sqrt{13}}}{2}\).

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Beyond the Answer

To solve the equation \( 4n^{4} + 8n^{2} - 9 = 0 \), let's first substitute \( x = n^{2} \). This transforms the equation into \( 4x^{2} + 8x - 9 = 0 \), which is a quadratic equation. Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \), we can find the roots. Plugging in \( a = 4 \), \( b = 8 \), and \( c = -9 \), we find the discriminant \( b^2 - 4ac = 64 + 144 = 208 \). Now calculate \( x = \frac{-8 \pm \sqrt{208}}{8} = \frac{-8 \pm 4\sqrt{13}}{8} = \frac{-2 \pm \sqrt{13}}{2} \). Thus, our solutions for \( n^2 \) lead to both positive and negative \( n \) after we take square roots from the positive values of \( x \). But watch out! Common mistakes include forgetting that both \( n \) and \( -n \) count as solutions when dealing with squares. Finding the roots might seem tricky, but with a little confidence and the quadratic formula, you'll solve this like a pro! Plus, always double-check your discriminant to avoid surprises. Happy solving!

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