Question

The equation:
[
4 x^{3}+20 x^{2}-23 x+6=0
]
has a double root. Find all roots.
If the equation has a double root we must have a factor
[
(x-r)^{2}
]
for a root ( r )
since our equation has degree 3 and has real coefficients, the remaining root ( s )
must be real as well and we must have:
[
4 x^{3}+20 x^{2}-23 x+6=4(x-s)(x-r)^{2}
]
Multiply the right hand side:
[
4 x^{3}+x^{2}(-8 r-4 s)+x(r(8 s+4 r))-4 r^{2} s
]
We should have:
[
begin{array}{ll}
-8 r-4 s & =20
r(8 s+4 r) & =-23
-4 r^{2} s & =6
end{array}
]
One could express ( r ) in ( s ) using the first equation and substitute in the second to
obtain a quadratic equation in ( s ) :
[
(-5-s)(3 s-5)=-23
]
Giving:
[
begin{array}{ll}
s=-6 & , r=1 / 2
s=8 / 3 & , r=-23 / 6
end{array}
]
It is a simple exercise to show that the first combination generates the solution.
On inspection we see that an easier method would have been to try the divisors
of 6 to find ( s ).

# Find the roots of 4x3 + 20x2 - 23x + 6 = 0 if two of its roots are equal.

Solution