Question 1046294
Kindly help me solving it 
What are the roots obtained by solving equation 
{{{z^2-6z+9=4 (z^2-6z+6)^(1/2)}}}?
Well i tried hard and simplified it upto
{{{z^4-12z^3+38z^2-12z+63=0}}}
But i can't go further, so please guide me.
<pre>{{{z^2 - 6z + 9 = (z^2 - 6z + 6)^(1/2)}}}
Your equation is incorrect! Sorry! 

After changing {{{matrix(1,3, (z^2 - 6z + 6)^(1/2), to, sqrt(z^2 - 6z + 6))}}}, squaring both sides, and combining like-terms, the following equation is derived: {{{z^4 - 12z^3 + 38z^2 - 12z - 15 = 0}}}
Using the rational root theorem, we see that z = 1 and z = 5 are REAL ROOTS of the equation. Using the trinomial divisor of the roots, or {{{z^2 - 6z + 5}}} (z - 1)(z - 5),
we get the following quotient: {{{z^2 - 6z - 3}}}. Using whichever method to solve this quadratic will give you 2 IRRATIONAL REAL roots: 6.464101615, and - 0.4641