Question

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

Answer 1

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

Explanation:

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

by factoring,

`(x-3)(x-3) = 0`

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

therefore, `x=3`

Answer 2

There are `2` real rational equal roots.

Explanation:

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`