For #x^2 - 6x +9 =0#, how many number of distinct solutions are there, and are they rational, irrational or non real complex numbers?

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Answer 1 · 2018-02-23T06:50:51

There is one distinct solution for this equation which is #3#. However, this is regarded as double root. The roots are rational

# x^2 -6x +9 = 0#

by factoring,

#(x-3)(x-3) = 0#

#( x-3)= 0# and #( x-3)= 0#

therefore, # x=3#

Answer 2 · 2018-02-23T06:52:03

There are #2# real rational equal roots.

The quadratic equation is

#x^2-6x+9=0#

Comparing this equation to the quadratic equation

#ax^2+bx+c=0#

Remember that the amount of solutions to a polynomial is equal to its degree. Here, since the degree is #2#, we have #2# solutions.

#a=1#

#b=-6#

#c=9#

The discriminant is

#Delta=b^2-4ac=(-6)^2-4*(1)*(9)=36-36=0#

The discriminant tells us about the nature of the roots:
When:

#Delta>0#, the roots are distinct and real.

#Delta=0#, the roots are real and equal.

#Delta>0#, the roots are imaginary.

As #Delta=0#, there are #2# real equal roots.

To prove the above:

#x=(-b+-sqrt(Delta))/(2*a)=(6+-0)/(2*1)=3#

We can also simply obtain those roots by factorizing.

#x^2-6x+9=(x-3)^2=0#