x^4-25^2+144=0
x^4-25^2+144=0 u=x^2 u^2-25u+144=0 (u-9)(u-16)=0 u=9 u=16 x^2=9 x^2=16 solve
\[x^4-25x^2+144=0\] is this ur equation??
oh the 25^2 is supposed to be 25x^2.....sorry.
Don't forget to substitute back for u
yes @TheViper
\[x^4-9x^2-16x^2+144\]
\[x^2(x^2-9)-16(x^2-9)\]
\[=>(x^2-16)(x^2-9)\]
That's ur ans.:)
those aren't my numbers?........or was that an example of the same?
What do u want to say??
the numbers you replied with aren't the same from the question..
No ans should be this do u know the ans if yes than reply it plz:)
@TheViper You can further factor that sinc eit says factor completely
There are 2 difference in squares there.
Actually, @.Sam. has done most for you.. x^4-25x^2+144=0 (x^2-9)(x^2-16)=0 (x^2-9)=0 or (x^2-16)=0 x^2 = 9 or x^2 =16 \(x=\pm 3\) or \(x=\pm4\)
Oh I factor that eq. u want to solve for x
Thanks @RolyPoly @Calcmathlete @TheViper@.Sam.
U r welcome @teun7840
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