Question

8x^4= 23x^3 + 3x^2

Answer 1

\[8x ^{4}=23x ^{3}+3x ^{2} \]
\[8x ^{4}-23x ^{3}-3x ^{2}=0\]
\[x ^{2}(8x ^{2}-23x -3)=0\]
Now :
\[\Delta=23^{2}-4\times8\times(-3)= 625=25^{2}\]

\[x _{1} = (23-25)\div (2\times8)=-1\div8\]

\[x _{2} = (23+25)\div (2\times8)=3\]
\[Equation : 8x ^{2}\times(x+(1\div8))\times(x-3)=0\]