Question: Find the real roots of the equation $\displaystyle \frac{18}{x^4} + \frac{1}{x^2} = 4$
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Can you solve $\displaystyle 18+x^2=4x^4$? Hint: $\displaystyle 4x^4 -x^2 -18 = (4x^2-9)(x^2+2)$.
Originally Posted by looi76 Question: Find the real roots of the equation $\displaystyle \frac{18}{x^4} + \frac{1}{x^2} = 4$ Multiply through by x^4 and re-arrange to get $\displaystyle 4x^4 - x^2 - 18 = 0$. Let $\displaystyle w = x^2$ and solve $\displaystyle 4w^2 - w - 18 = 0$ for w. Then solve $\displaystyle x^2 = w$ for x (you'll only use one of the solutions of w to get real roots - why?)
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