Answer:
Given the equation: [tex]-11x^2 = x+11[/tex]
or we can write above equation as:
[tex]-11x^2-x-11=0[/tex] or
[tex]11x^2+x+11 =0[/tex]
The general equation of quadratic formula for [tex]ax^2+bx+c=0[/tex]; where a, b and c are constant;
Use Discriminant formula: [tex]D =b^2-4ac[/tex]
If [tex]D> 0[/tex], then there are 2 roots.
If D = 0, then there is only 1 root.
If D <0, then there are no real roots.
Now, from the given equation [tex]11x^2+x+11 =0[/tex] we have
a =11, b= 1 and c =11
Then, using discriminant formula we have;
[tex]D =b^2-4ac =(1)^2 -4(11)(11) = 1-4\cdot 121 = 1-484= -483[/tex]
⇒D<0 [ No real roots]
Therefore, for the given equation [tex]-11x^2 = x+11[/tex] , there are no real solutions.
