SOLUTION: 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+6

Question 1046294: Kindly help me solving it
What are the roots obtained by solving equation
?
Well i tried hard and simplified it upto

But i can't go further, so please guide me.
Found 3 solutions by josgarithmetic, ikleyn, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
SEE NOTE BELOW - YOU MADE SOME ALGEBRA STEPS MISTAKE.

Rational Roots Theorem will tell you that the possible roots to check for are the pluses and minuses of 1,3,7,9,21. This would be best done using synthetic division; and Factor Theorem tells you that if remainder is zero, that possible checked root IS a root.

Each time a root is found, the quotient of coefficients gives you the new arrangment of coefficients to check, meaning a factor has been taken care of. Be aware that some roots may be repeated, depending on what you find.

I took your resulting equation and used a graphing tool as a quick method to see any roots. No intersections with the x-axis found, so NO REAL ROOTS. You might expect all of the possible rational roots to check using synthetic division to give NON-ZERO remainders.

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~~I did not yet check if you made algebraic mistakes in getting your general form equation.~~

NOTE: Not showing my actual steps, I worked through from your original equation to simplify and put into general form, and find . This is very different from your initial result. You may find that finding the roots to be easier now. The possible real RATIONAL roots to test for are the pluses and minuses of 1, 3, 5, 15.
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Graphing tool indicates two real but irrational roots. You may need to use some numerical approximation method to find the roots. Something near 12.4 and something else near -0.8.
Answer by ikleyn(52777)   (Show Source): You can put this solution on YOUR website!
.
Kindly help me solving it
What are the roots obtained by solving equation
?
Well i tried hard and simplified it upto

But i can't go further, so please guide me.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

= Rewrite equivalently = . Square both sides: = . Introduce new variable x = . The last equation takes the form = , or = . Solve by any way (factoring or the quadratic formula). You will get the roots x = 12 and/or x = 4. Thus you have two options: 1. {z-3)^2 = 4 ---> z-3 = 2 or z-3 = -2 ---> z = 5 or z = 1. 2. (z-3)^2 = 12 ---> z-3 = +/- = +/- ---> z = and/or z = . Check that all these roots are the solutions of the original equation. Answer. The solutions are z = 1, z = 5, z = , z = .

Plots y = (red), y = (green)

Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!
Kindly help me solving it
What are the roots obtained by solving equation
?
Well i tried hard and simplified it upto

But i can't go further, so please guide me.

Your equation is incorrect! Sorry!
After changing , squaring both sides, and combining like-terms, the following equation is derived:
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 - 1)(z - 5),
we get the following quotient: . Using whichever method to solve this quadratic will give you 2 IRRATIONAL REAL roots: 6.464101615, and - 0.4641
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