Question

( a x^{3}+6 x^{2}+c x+a )
( f^{prime}(x)=39 x^{2}+26 x+c+f^{2}+6 x^{2}+1 )
[
2>0
]
( 46^{2}-4(3 9 1 longdiv { text { C) } } > 0 )
[
6^{2}-39 c>0
]
( left[b^{2}>3 a cright] )
He only forl can have nod nusts
4
牛 ( longrightarrow )

# 24. Prove that if ax3 + bx2 + cx + d = 0 has three real and distinct roots, then b2 > 3ac.

Solution