- #1
anemone
Gold Member
MHB
POTW Director
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Hi MHB,
This is the second headache problem that I wish to get some insight from MHB today...
Problem:
It is known that the equation \(\displaystyle ax^3+bx^2+cx+d=0\) has three distinct real roots. How many real roots does the following equation have?
\(\displaystyle 4(ax^3+bx^2+cx+d)(3ax+b)=(3ax^2+2bx+c)^2\)
Attempt:
I observed that if we let \(\displaystyle f(x)=ax^3+bx^2+cx+d\), then the given second equation can be rewritten as
\(\displaystyle 6f(x)f''(x)=(f'(x))^2\)
That is, the LHS of the equation has a total of 4 roots, and the RHS of the equation has a total of 2 repeated roots and the range of it is $[0, \infty)$. But I know I need to figure out how many times the curve of the LHS function cuts the curve of the RHS function, but not finding out their roots.
I don't see even a single way to proceed from there and I feel so supremely stupid now.
Can anyone please help me?
Thanks.
This is the second headache problem that I wish to get some insight from MHB today...
Problem:
It is known that the equation \(\displaystyle ax^3+bx^2+cx+d=0\) has three distinct real roots. How many real roots does the following equation have?
\(\displaystyle 4(ax^3+bx^2+cx+d)(3ax+b)=(3ax^2+2bx+c)^2\)
Attempt:
I observed that if we let \(\displaystyle f(x)=ax^3+bx^2+cx+d\), then the given second equation can be rewritten as
\(\displaystyle 6f(x)f''(x)=(f'(x))^2\)
That is, the LHS of the equation has a total of 4 roots, and the RHS of the equation has a total of 2 repeated roots and the range of it is $[0, \infty)$. But I know I need to figure out how many times the curve of the LHS function cuts the curve of the RHS function, but not finding out their roots.
I don't see even a single way to proceed from there and I feel so supremely stupid now.
Can anyone please help me?
Thanks.