Icannot get the answer for the life of me! . the question said solve x=√13x-42

X=√(13x-42).
I believe (13x-42) is under the radicand. If so, square both sides:
x² =[√(13x-42)]²

x² = 13x - 42 or
x² - 13x + 42 = 0. Solve this quadratic equation:
x' =[-b + √(b²-4.a.c)]/2a   and x" =[-b - √(b²-4.a.c)]/2a  

x' = [13+√13² - 4(1)(42)]/2  and x" = [13-√13² - 4(1)(42)]/2
x' = (13+√1)/2     and x" = (13-√1)/2

x' = 7 and x" = 6

Answer to your question: a. the fundamental theorem of algebra tells you that the equation will have two complex roots since the degree of the polynomial is 2. the roots are ​ x=1±i7√2 .number of roots=degree of polynomialhere, degree of polynomial is 2 so number of roots will be 2.hope it *_*