Question 1178897
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Let *[tex \Large u\ =\ x^2], then *[tex \Large u^2\ -\ 13u\ -\ 48] is equivalent to *[tex \Large x^4\ -\ 13x^2\ -\ 48\ =\ 0]


Use the quadratic formula to solve *[tex \Large u^2\ -\ 13u\ -\ 48] for the two values of *[tex \Large u].  Then solve *[tex \Large x^2\ =\ u] for each value of *[tex \Large u] giving you four distinct roots as is expected from a quartic equation.  Hint: *[tex \Large \sqrt{361}\ =\ 19]

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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