solve a*x^4 + b*x = c for x
asked by guest
on Jan 23, 2025 at 2:37 am
You asked:
Solve the equation
ax4+bx=c for the unknown
x.
MathBot Answer:
xxxx=⎩⎨⎧−2−3−a2b2−23−a2b2+a−3−a2b22b−22327a3c3+256a4b4+16a2b2−3a327a3c3+256a4b4+16a2b22c−2−2327a3c3+256a4b4+16a2b2+a2327a3c3+256a4b4+16a2b2−3a327a3c3+256a4b4+16a2b22c2b+3a327a3c3+256a4b4+16a2b22cforac=0otherwise=⎩⎨⎧−2−3−a2b2+23−a2b2+a−3−a2b22b−22327a3c3+256a4b4+16a2b2−3a327a3c3+256a4b4+16a2b22c+2−2327a3c3+256a4b4+16a2b2+a2327a3c3+256a4b4+16a2b2−3a327a3c3+256a4b4+16a2b22c2b+3a327a3c3+256a4b4+16a2b22cforac=0otherwise=⎩⎨⎧2−3−a2b2−23−a2b2−a−3−a2b22b22327a3c3+256a4b4+16a2b2−3a327a3c3+256a4b4+16a2b22c−2−2327a3c3+256a4b4+16a2b2−a2327a3c3+256a4b4+16a2b2−3a327a3c3+256a4b4+16a2b22c2b+3a327a3c3+256a4b4+16a2b22cforac=0otherwise=⎩⎨⎧2−3−a2b2+23−a2b2−a−3−a2b22b22327a3c3+256a4b4+16a2b2−3a327a3c3+256a4b4+16a2b22c+2−2327a3c3+256a4b4+16a2b2−a2327a3c3+256a4b4+16a2b2−3a327a3c3+256a4b4+16a2b22c2b+3a327a3c3+256a4b4+16a2b22cforac=0otherwise