How do you solve #\log _ { 2} ( x ^ { 3} + x ^ { 2} + 1) = 2#?
1 Answer
Real solution:
#x = 1/3(-1+root(3)((79+9sqrt(77))/2)+root(3)((79-9sqrt(77))/2))#
and related complex solutions.
Explanation:
Given:
#log_2(x^3+x^2+1) = 2#
Raise
#x^3+x^2+1 = 4#
Subtract
#x^3+x^2-3 = 0#
Discriminant
The discriminant
#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#
In our example,
#Delta = 0+0+12-243+0 = -231#
Since
Tschirnhaus transformation
To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.
#0=27f(x)=27x^3+27x^2-81#
#=(3x+1)^3-3(3x+1)-79#
#=t^3-3t-79#
where
Cardano's method
We want to solve:
#t^3-3t-79=0#
Let
Then:
#u^3+v^3+3(uv-1)(u+v)-79=0#
Add the constraint
#u^3+1/u^3-79=0#
Multiply through by
#(u^3)^2-79(u^3)+1=0#
Use the quadratic formula to find:
#u^3=(79+-sqrt((-79)^2-4(1)(1)))/(2*1)#
#=(79+-sqrt(6241-4))/2#
#=(79+-sqrt(6237))/2#
#=(79+-9sqrt(77))/2#
Since this is Real and the derivation is symmetric in
#t_1=root(3)((79+9sqrt(77))/2)+root(3)((79-9sqrt(77))/2)#
and related Complex roots:
#t_2=omega root(3)((79+9sqrt(77))/2)+omega^2 root(3)((79-9sqrt(77))/2)#
#t_3=omega^2 root(3)((79+9sqrt(77))/2)+omega root(3)((79-9sqrt(77))/2)#
where
Now
#x_1 = 1/3(-1+root(3)((79+9sqrt(77))/2)+root(3)((79-9sqrt(77))/2))#
#x_2 = 1/3(-1+omega root(3)((79+9sqrt(77))/2)+omega^2 root(3)((79-9sqrt(77))/2))#
#x_3 = 1/3(-1+omega^2 root(3)((79+9sqrt(77))/2)+omega root(3)((79-9sqrt(77))/2))#